Arithmetic

These notes are notes on basic arithmetic. All other notes on this site assume familiarity with topics covered here.

  1. Speaking Mathematics
    1. Expressions
    2. The Domain of an Expression
    3. Truth Sets
    4. Open Sentences
    5. Families of Equations
      1. Linear Equations
      2. Polynomial Equations
        1. Quadratic Equations
      3. Trigonometric Equations
      4. Exponential Equations
      5. Radical Equations
      6. Rational Equations
      7. Conditional Equations
      8. Equivalent Equations
      9. Inequations
  2. Number Sets
    1. The Natural Numbers
    2. Integers
    3. Rational Numbers
    4. Irrational Numbers
    5. Real Numbers
    6. Complex Numbers
    7. Laws of Arithmetic
    8. Order of Operations
  3. Addition
    1. Identity Property of Real Addition
    2. Real Subtraction
    3. Commutativity of Real Addition
    4. Associative Law of Addition
    5. The Cancellation Law of Addition
  4. Closure
  5. Inequalities
  6. Multiplication
    1. Laws of Multiplication
      1. Commutative Law of Multiplication.
      2. Reciprocal Law of Multiplication
      3. Associative Law of Multiplication
      4. Multiplication by Zero
      5. Identity Property of Multiplication
  7. Distributivity
  8. Multivariable Relationships
    1. Expansions
    2. The Multiplication Algorithm
  9. Exponentiation
    1. Perfect Powers
  10. Radicals
    1. The Square Root
    2. The Quotient Rule of Square Roots
    3. Adding and Subtracting Square Roots
    4. nth Root
    5. Rational Exponents
  11. Ratios
    1. Percentages
    2. Proportions
    3. Rational Expressions
      1. Simplifying Rational Expressions
      2. Multiplying Rational Expressions
      3. Adding and Subtracting Rational Expressions
      4. Simplifying Nested Rational Expressions
      5. Rationalizing Radical Denominators
  12. Division
    1. Divisibility
    2. Composites & Primes
    3. Division & Rational Numbers
    4. The Division Algorithm
  13. Absolute Value
    1. Removing Absolute Value Symbols
    2. Absolute Values and Inequalities
  14. Logarithms
  15. The Binomial Theorem
  16. Solving Quadratic Equations
    1. Completing the Square
    2. Why Does Completing the Square Work?
    3. The Quadratic Formula
    4. The Discriminant
    5. The Product and Sum of Polynomial Roots
  17. The Factorial
  18. Binomial Coefficients
    1. Pascal's Triangle
  19. Invervals
    1. Neighborhoods
  20. The Cartesian Plane
  21. The Distance Formula
  22. Functions
    1. Properties of Functions
    2. Arrow Notation
  23. Function Graphs
    1. The Vertical Line Test
    2. Graphing Transformations
    3. Vertical Translation
    4. Horizontal Translation
  24. Slope
    1. Comparing Slopes
    2. The Point-Slope Formula
    3. Slope-Intercept Formula
    4. Parallel and Perpendicular Lines

Speaking Mathematics

Expressions

In algebra, we use letters like x,{x,} y,{y,} n,{n,} and m{m} to represent numbers. We call such letters variables. When we combine these variables with mathematical operations like addition, subtraction, multiplication, etc., we have an expression. For example, the following are all expressions:

  • a+2{a + 2}
  • x2{x^2}
  • xy{\sqrt{x} - y}
  • x1/3+x2a/b{\dfrac{x^{1/3} + x^2}{a/b}}

When we place an equal sign between two expressions, we have an equation:

a2+b2=c2. a^2 + b^2 = c^2.

In mathematics, the equal sign is a symbol that expresses the relation between mathematical objects. In the example above above, these objects are expressions. When we place an equal sign between two expressions, we are saying: The expression a2+b2{a^2 + b^2} is the same, in every way, as the expression c2.{c^2.} This is a proposition. Notice its gravity. There are very few things in the world that have that property — the same in every way.

We will see in later sections that equality is just one kind of relation. In fact, a significant portion of mathematics (if not most) concerns inequations — propositions like x2<y2.{x^2 < y^2.} We will revisit inequations (also called inequalities) once we've established familiarity with equations.

The Domain of an Expression

All mathematical expressions have a domain — the set of all values for which the expression is defined. Less formally, the domain of an expression is essentially the boundaries of what we're discussing. For example, if we were debating the costs and benefits of capitalism, the debate might be limited to economies that exist today, rather than the economies of ancient Egypt. The basic idea is that, outside of the domain, the expression is undefined — all bets are off.

For example, the expression 3x+43x + 4 has a domain of R\mathbb{R}, since it is defined for all real numbers. We denote the domain of 3x+43x + 4 with the notation: dom(3x+4)=Rdom(3x + 4) = \mathbb{R}. Example: What is the domain of 1(t1)(t+3)\dfrac{1}{(t-1)(t+3)}?

dom(1(t1)(t+3))={tRt1,t3}\text{dom}\left(\dfrac{1}{(t-1)(t+3)}\right) = \{t \in \mathbb{R} \mid t \neq 1, t \neq -3\}

We impose restrictions on the domain, since t=1t = 1 and t=3t = -3 lead to the denominator evaluating to 0, which is undefined. For example, what is the domain of 4x+3x14x + 3x^{-1}?

dom(4x+3x)={Rx0}\text{dom}\left(4x + \dfrac{3}{x}\right) = \{ \mathbb{R} \mid x \neq 0 \}

Truth Sets

Because expressions contain variables, an equation is true or false if, and only if, we substitute values into those variables. Until then, an equation is neither true nor false. If a value is inputted and the equation is true, we call that value the solution, or root, of the equation. The set of all roots to an equation is called the equation's truth set.

If an equation's truth set is the equation's domain, we say that equation is an identity. For example, the equations:

  • x29=(x3)(x+3){x^2 - 9 = (x - 3)(x + 3)}
  • 4x2x=x{\dfrac{4x^2}{x} = x}

are identities. Both equations have a domain of R\mathbb{R}, and substituting xx with any real number will render the equations true.

Open Sentences

Consider the following:

1+x=4 1 + x = 4

The above is clearly an equation, but more broadly, it is called an open sentence — a statement that contains a variable, and becomes a proposition when an element of the universal set is substituted for the variable. In this case, the universal set is the set of all real numbers, R.{\reals.} If we substitute 1,{1,} an element of the universal set, into the variable, we obtain 1+1=2,{1 + 1 = 2,} a false proposition. But, if we instead substituted 3, another element of the universal set, we obtain 1+3=4,{1 + 3 = 4,} a true proposition.

Like propostions, we denote open sentences with lower-case letters, with the one key difference — there is a subscript appended, corresponding to the variables used in the open sentence. For example, the open sentence above might be denoted px.{p_x.}

The rule of substituting remains applicable for variables that occur more than once in the sentence:

1+x+x=2 1 + x + x = 2

To return a true proposition for the sentence above, we must substitute a value into each x{x} in the open sentence.

We call these statements open sentences because they are in fact sentences. Equations, at the end of the day, are simply sentences. The equation 2x=3{2x = 3} is just another way of saying, “Two multiplied by some value is three.” The use of the symbols 2, x,{x,} =,{=,} and 3,{3,} and arranging those symbols in a particular way, is nothing more than a means of making things more concise and easier to read. Mathematicians invent and use symbols because certain propositions cannot be expressed concisely and clearly in plain English:

03.8π4xy(a+9c+8bcos(πzl)+18ccos(2πzl))sin2(πzl)l4dz \int_0^{3.8}\frac{\pi^4 x y (a+9 c+8 b \cos(\frac{\pi z}{l})+18 c \cos(\frac{2 \pi z}{l})) \sin^2(\frac{\pi z}{l})}{l^4} dz

“The ingegral with respect to z{z} from 0 to 3.8 of the quotient of the product of the product of the fourth power of Pi, x,{x,} and y{y} multiplied by the sum of a,{a,} the product of 8,{8,} b,{b,} and the cosine of the quotient of the product of Pi and z{z} divided by l,{l,} the product of 18,{18,} c,{c,} and the cosine of the quotient of 2, Pi, and z,{z,} and the square of the sine of the quotient of Pi and z{z} divided by l,{l,} all divided by the the fourth power of l.{l.}

Families of Equations

Equations fall into several types. Mathematicians have constructed these groupings because equations often share many properties. We will explore what these properties are carefully, but before doing so, we have some familiarity of the type names. The overview provided below is strictly informal. We will provide more formal definitions when we address each type of equation in turn.

Linear Equations

The linear equation is the simplest equation. Here are some examples of linear equations:

  • 12x2=0{12x - 2 = 0}
  • 19x=24{19x = 24}
  • 17x+4y=z{17x + 4y = z}
  • ax+by+nz=g{ax + by + nz = g}

Notice that the number of variables doesn't prevent an equation from being classified as a linear equation. Indeed, the only requirements for something to be a linear equation is that each term is either a constant or the product of a constant and a single variable. If the linear equation consists of just one variable, we say that the equation is a linear equation of one-variable. If there are two variables, it's a linear equation of two-variables, and so on.

Despite their simplicity, linear equations are extremely important in both pure and applied mathematics. In physics, the distance an object travels, called displacement, is given by a linear equation: s=xfxi,{s = x_f - x_i,} where s{s} is the displacement, xf{x_f} is the final position, and xi{x_i} is the initial position. The final velocity of an object can also be determined by the equation vf=v0+at,{v_f = v_0 + at,} where vf{v_f} is the final velocity, v0{v_0} is the initial velocity, a{a} is the acceleration, and t{t} is the time interval. These are both linear equations.

Polynomial Equations

Polynomial equations are equations consisting of polynomials — expressions that consist of variables and coefficients where the onperations involved are addition, subtraction, multiplication, and non-negative integer exponentiation. This is a bit of a mouthful, so here are some examples of polynomial expressions:

  • x24x+13{x^2 - 4x + 13}
  • x2+y2{x^2 + y^2}
  • bx3{bx^3}
  • 2x{2x}

and here are some examples of polynomial equations:

  • a2+b2=c2{a^2 + b^2 = c^2}
  • Δx=v0t+12at2{\Delta x = v_0t + \dfrac{1}{2}at^2}
  • 3x29=0{3x^2 - 9 = 0}
  • 2x1=0{2x - 1 = 0}

Wait a minute. Isn't that 2x1=0{2x - 1 = 0} a linear equation? It is in fact a linear equation. All linear equations are polynomial equations of degree-1. But, not every polynomial equation is a linear equation. For example, a polynomial equation of degree-2:

ax2+x+c=0 ax^2 + x + c = 0

Put simply, the degree of a polynomial equation is the leading exponent in the equation. A polynomial equation of degree-3:

ax3+bx2+cx+d=0 ax^3 + bx^2 + cx + d = 0

Because it's fairly cumbersome to write “of degree-n{n}” every time, we have some special names for certain degrees. A quadratic equation is a polynomial equation of degree-2. For example:

3x2+x9=0 3x^2 + x - 9 = 0

A cubic equation is a polynomial equation of degree-3:

4x3x=12 4x^3 - x = 12

A quartic equation is a polynomial equation of degree-4:

x4+3x39x2+12=0 x^4 + 3x^3 - 9x^2 + 12 = 0

Quartic equations are sometimes called biquadratic equations. We use the term quartic because it's shorter to write.

Polynomial equations are also classified by the number of terms the equation contains. A polynomial equation with only one term is called a monomial. For example, x2=0{x^2 = 0} is a monomial equation. A polynomial equation with two terms is called a binomial. For example, a2+b2=2{a^2 + b^2 = 2} is a binomial equation. A polynomial equation with three terms is called a trinomial — a2+b+c=0.{a^2 + b + c = 0.}

Quadratic Equations

Quadratic equations are equations in one variable that can be written in the form:

ax2+bx+c=0 ax^2 + bx + c = 0

If we seek to determine the the truth set of a quadratic equation (i.e., the set of all roots, or values of x{x} where the equation is true), our first course of action is to factor. And whenever we factor, we should be cognizant of the basic law of multiplication:

pq=0{p=0, or q=0, or p=0 and q=0pq = 0 \iff \begin{cases} p = 0, \text{ or } \\ q = 0, \text{ or } \\ p = 0 \text{ and } q = 0 \end{cases}

Example: What is the value of x{x} in 8x23=10x?{ 8x^2 - 3 = 10x?}

Subtracting 10x{10x} from both sides, we get:

8xx10x3=0 8x^x - 10x - 3 = 0

Factoring:

(2x+3)(4x+1)=0 (2x + 3)(4x + 1) = 0

Solving for x{x} for one of the expressions:

2x+3=02x=3x=32\begin{aligned} 2x + 3 &= 0 \\ 2x &= -3 \\ x &= \frac{-3}{2} \end{aligned}

And for the other expression:

4x+1=04x=1x=14\begin{aligned} 4x + 1 &= 0 \\ 4x &= -1 \\ x &= \frac{-1}{4} \\ \end{aligned}

Thus, we have:

x=32,14 x = \frac{-3}{2}, \frac{-1}{4}

Trigonometric Equations

Many equations represent cycles, and wherever there are cycles, there's likely a trigonometric operator involved. Such equations are called trigonometric equations. These equations might look something like:

  • sinx+cosy=z{\sin x + \cos y = z}
  • (tan2x+5)sinx=0{(\tan^2 x + 5) - \sin x = 0}
  • arctanzy2=x{\arctan z - y^2 = x}

Exponential Equations

Exponential equations look very much like polynomials, with the distinctive feature that their exponents have variables instead. For example:

  • y=ex{y = e^x}
  • z=yx24{z = y^x - 24}
  • 2a+c=b2{2^a + c = b^2}

Exponential equations are often seen in mathematical models for population growth and areas of finance.

Radical Equations

Radical equations have the distinctive feature of containing one or more radicands or fractional exponents:

  • y=x{y = \sqrt{x}}
  • a=b1/2{a = b^{1/2}}

Rational Equations

Finally, we have the rational equations. These are equations with rational expressions or negative exponents:

  • x4=(y3)217{\dfrac{x}{4} = \dfrac{(y - 3)^2}{17}}
  • y=1x{y = \dfrac{1}{x}}
  • f=n1{f = n^{-1}}

Conditional Equations

The opposite of an identity is a conditional equation. These are equations that are true only if some condition(s) is/are true. For example, the following are conditional equations:

  • 2x=10{2x = 10}
  • x=x+1{x = x + 1}

The equation 2x=102x = 10 is true only if it is true that x=5x = 5. The equation x=x+1x = x + 1 is true only if there is some number that equals one more than itself (a condition that can never be satisfied, so the equation can never be true).

Equivalent Equations

Equivalent expressions are represented by expressions. Equations themselves can be equivalent. Two equations are equivalent if, and only if, they have exactly the same solutions.

Inequations

Most critical findings in mathematics result from inequalities rather than equations. And understandibly so — it is often easier to determine whether particular expressions are greater or less than others, in comparison to concluding that they are equal. Whenever we replace the equal sign in an equation with one of the symbols {\leq} or ,{\geq,} we obtain an an inequation. Both equations and inequations are mathematical statements. As such, inequations, like equations, have truth sets — the set of all values that render the inequality true if any member of that set is substituted into a variable, or variables, of the inequality.

For example, suppose 2x3<8.{2x - 3 < 8.} The value x=5{x = 5} would be a member of the inequation's truth set, since 2(5)3=7<8.{2(5) - 3 = 7 < 8.} The value x=10,{x = 10,} however, would not be a member of that truth set, since 2(10)3=178.{2(10) - 3 = 17 \nless 8.} We can find the entire truth set for the inequation by solving for xx, using the same techniques for equations:

2x3<8=2x3+3<8+32x<8+32x<11x<112 2x - 3 < 8 = 2x - 3 + 3 < 8 + 3 \\ 2x < 8 + 3 \\ 2x < 11 \\ x < \dfrac{11}{2} \\

Although inequations can be solved by mostly applying the techniques for equations, there are some situations we should take care when handling. Particularly, we must be careful whenever we multiply or divide both sides of an inequation by the same nonzero number.

For example, suppose 1<3.{1 < 3.} If we multiplied both sides of the inequality by 10, we would obtain 10<30.{10 < 30.} Clearly this is true. But if were to multiply both sides by -5, we would obtain 5<15.{-5 < -15.} This is false. This observation returns 2 general rules:

lemma. If both sides of an inequation are multiplied or divided by a positive real number, then the inequation is preserved.

lemma. If both sides of an inequation are multiplied or divided by a negative real number, then the inequation is reversed.

More explicit rules:

lemma. If a<b,{a < b,} then a+c<b+c,{a + c < b + c,} and ac<bc.{a - c < b - c.}

lemma. If a<b{a < b} and cR+,{c \in \mathbb{R}^+,} then ac<bc,{ac < bc,} and > ac<bc.{\dfrac{a}{c} < \dfrac{b}{c}.}

lemma. a<b{a < b} and cR,{c \in \mathbb{R}^-,} then ac>bc,{ac > bc,} and > ac>bc.{\dfrac{a}{c} > \dfrac{b}{c}.}

lemma. If a<b{a < b} and b<c,{b < c,} then a<c.{a < c.}

Number Sets

Numbers have been around for a long time. Early humans used all sorts of ways to record their enumerations: Scratches on cave walls and bones; knots on strings; pebbles and shells; notches cut into twigs — all attempts to quantify, to make sense of the environment. It's as if the numbers revealed themselves magically.

But what is it really magic? Or was it purely out of necessity? The intuitive response is that our ancestors needed to quantify objects. Perhaps the earliest hunters needed to determine how much game was needed for the village. Or the chief having to decide a fair tribute. But this response neglects the fact that our forebears weren't just concerned with survival. They were also deeply superstitious.

Some of the earliest forms of counting date to religious ceremonies and mysticism. Many of these ceremonies required ordering participants and events, hinting at the possibility that the ordinal numbers — numbers used to denote position — came before the cardinal numbers — numbers used for quantity.

Indeed, there's a plausible argument there. Some of the earliest forms of the cardinal numbers were rudimentary and simple, in contrast to their ordinal counterparts. Cardinal numbers were limited to 2,{2,} and anything beyond that was a word for "many" or "lots." Moreover, the view that the ordinals preceded the cardinals would be in line with the recurring phenomenon of dividing the integers into "male" and "female," an idea that would eventually lead to "odd" and "even."

Regardless of whether the ordinal or the cardinal came first, the human need for numerating grew increasingly complex. And as complexity increased, so too did our sophistication. Sticks, stones, and bones were too ephemeral, so we developed number systems to more efficiently analyze our states of affairs. As far back as 3600 BC, the Ancient Egyptians used a numbering system where every 10{10} unique characters could be abstracted, or symbolized, by a new, unique character. The Egyptian system evidences one of the earliest forms of a base-10 system, or decimal system. Unlike the number system we're familiar with, the Ancient Egyptian number system introduced a new symbol for every 10{10} symbols. In other words, a symbol for 1,{1,} then for every ten 1s{1s}, a symbol for 10,{10,} then 100,{100,} then 1,000,{1,000,} then 10,000,{10,000,} then 100,000,{100,000,} and so on.

Centuries later, the Greeks concocted their own number system, where a unique character for 5{5} was introduced. The ancient Romans followed suit, utilizing the familiar capital letter numerals now relegated to formal headings and cryptic date tattoos. The numerals I,{I,} II,{II,} and III{III} are derived directly from using three fingers to symbolize 3;{3;} the numeral V{V} from the angle formed by all five fingers; and X{X} from the joining of two V.{V.} The numeral C{C} represented 100,{100,} and half of a C{C} — an L{L} — represented 50.{50.} The numeral M{M} represented 1,000.{1,000.} Split it in half and turn it on its side, we obtain something that looks like D,{D,} the numeral for 500.{500.}

The ancient Mayans too had a sophisticated and flourishing number system, where dots and tallies were used as numerals. The beauty in the Mayan system was its strong connection to addition. Numerals logically followed from arithmetic. In contrast to the Egyptians, Greeks, and Romans, the ancient Mayans introduced a new, unique character for every 20{20} symbols, an example of a base-20 system, or vigesimal system. Moreover, like the ancient Egyptians, the Mayans understood the need to represent zero and assigned it a unique symbol — well before zero was accepted in Europe.

Ancient China, whose characters made their way into the Korean peninsula and the Japanese isles, developed a number system from a mixture of pictographs and arithmetic. Unlike the Mayans and their connection to addition, the Chinese number system draws directly from multiplication and addition. By incorporating two arithmetic operators, the Chinese number system allowed for a smaller set of numerals, which in turn permitted efficient enumeration of large quantities.

The numerals we use today are the Hindu-Arabic numerals, derived from Hindu and Arabic number systems. Developed about 2,000{2,000} years ago in India, the numerals made their way into Europe, via the Spanish peninsula. The Moors (Spanish Muslims), in their search for trade, brought both the Hindu-Arabic numerals and Ancient Greek philosophy, preserved by Arabic scholars. In the 12th century, Leonardo Fibonacci, for whom the Fibonacci numbers are named, furthered the numerals spread into continental Europe. By the 15th century, the Roman numerals — until then the predominant symbols — were entirely replaced by the Hindu-Arabic numerals we know today.

All of this is to say that there is a distinction between a numeral and a number. The numeral is purely a symbol. We could have just as easily continued using Roman symbols to do mathematics, just as medieval Europeans did. Understanding this distinction is crucial for when we study numbers. The Chinese symbol 五 is different from the Hindu-Arabic symbol 5{5}, but they both communicate the same idea — the number five. Often, we'll find that numerals are inappropriate for a particular analysis because of the underlying idea they communicate — a discrete number. In those situations, we must let go of the numeral abstraction, and think in different terms, different symbols, and different relationships. To do so effectively, we study the properties of the real numbers.

The Natural Numbers

We define the natural numbers as the positive integers including zero (i.e., the nonnegative integers). These are what we refer to as the counting numbers in regular speech. Importantly, mathematicians differ on whether 0{0} is a natural number. This is more a matter of convenience than anything else, but in these materials, we follow the premise that 0{0} is a natural number. If there is a variation, we will explicitly state that 0{0} is not included in the set.

N={0,1,2,3,4,5,}\nat = \{ 0, 1, 2, 3, 4, 5, \ldots \}

Some texts define the natural numbers as the positive integers, and instead refer to the nonnegative integers as the whole numbers. For reasons beyond the scope of the current discussion, defining the natural numbers as including zero has benefits for later sections.

Integers

The integers are the numbers we use for counting (also called the cardinal numbers in other contexts), and their negatives. They are denoted with the special symbol, Z,{\uint,} from the German zehr, meaning "number".

Z={,-4,-3,-2,-1,0,1,2,3,4,}{ \uint = \{ \ldots , \texttt{-}4, \texttt{-}3, \texttt{-}2, \texttt{-}1, 0, 1, 2, 3, 4, \ldots \} }

Some elementary mathematical texts call the counting numbers the whole numbers, and go so far as to denote them with the symbol W.{\mathbb{W}.} This terminology is seldom used in higher level mathematics, so we'll stick to the more common integers. Nevertheless, the idea of a whole number is helpful for communicating the notion of an integer, so we'll momentarily use it.

But what is a whole number? We can think of the whole number as such: If a number has at least one non-zero digit after its decimal point, then it is not a "whole number" because we cannot count from or to zero with that number exactly 1 unit at a time on the number line. For example, 1.1{1.1} is not a whole number, because no matter how much we try, to get to zero, we have to subtract 0.1.{0.1.} This is not 1 unit, so it does not constitute a whole number. And since it does not constitute a whole number, it is not an integer.

The whole numbers greater than 0{0} are called the positive integers. We can think of these numbers as the numbers we use to count forwards (e.g., inserting a certain amount of objects into a space).

Z+={1,2,3,4,} \uint^{+} = \{ 1, 2, 3, 4, \ldots \}

Notice symbol for the set, a zehr with a superscripted plus sign. And the whole numbers less than 0{0} are called the negative integers. Here, we can think of these numbers as the numbers we use to count backwards (e.g., removing a certain number of objects from a space).

Z={,-4,-3,-2,-1} \uint^{-} = \{ \ldots , \texttt{-}4, \texttt{-}3, \texttt{-}2, \texttt{-}1 \}

The number 0{0} is neither negative nor positive. It has a special duality: It is both a non-positive and non-negative number. The non-positive integers are negative integers including zero:

Z0={,-4,-3,-2,-1,0} \Z_{\leq 0} = \{ \ldots , \texttt{-}4, \texttt{-}3, \texttt{-}2, \texttt{-}1, 0 \}

and the non-negative integers are the positive integers including zero:

Z0={0,1,2,3,4,}. \Z_{\geq 0} = \{ 0, 1, 2, 3, 4, \ldots \}.

Rational Numbers

When we divide one integer by another nonzero integer, we get a rational number. Specifically, if, and only if, a number can be written in the form ab\frac{a}{b}, where aa is an integer and bb is a nonzero integer, then the number is a rational number. For example, 4 is a rational number, since 4 can be written as 41\frac{4}{1}. 0.5 is a rational number, since it can be written as 12\frac{1}{2}.

Irrational Numbers

If a number cannot be written in the form ab\frac{a}{b}, where aa is an integer and bb is a nonzero integer, then the number is an irrational number. For example, the number π\pi is an irrational number, because it cannot be written in the form ab\frac{a}{b}.

Real Numbers

1,3.14,π,ϕ,e,4{1, 3.14, \pi, \phi, e, 4} and 0,{0,} all of these numbers are called real numbers. In the simplest terms, a real number is a number that can be written as a decimal. Thus, 7 is a real number, since 7=7.0.{7 = 7.0.} Similarly, while π{\pi} is an irrational number. It is a real number since it can be written as a decimal (3.14{3.14 \ldots}), albeit non-terminating.

Real numbers are written in decimal notation. This is the prevailing standard (with the exception of binary, favored by computers). The decimal system is useful because it allows us to quickly add numbers through the process of carrying over.

Complex Numbers

Outside of the real numbers, we have the imaginary numbers. These are the numbers that result from operations such as -1\sqrt{\texttt{-}1}. We will treat and analyze this numbers in a separate chapter.

Laws of Arithmetic

The following materials discuss arithmetic, the set of rules and conventions governing the manipulation of numbers. The word arithmetic is derived from the Greek arithmetike, meaning "art of counting," which is itself derived from the Greek arithmos, meaning "number."

There are four basic arithmetic operators:

SymbolOperation
a+ba + bAddition
aba - bSubtraction
a×ba \times bMultiplication
ab\dfrac{a}{b}Division
aba^bExponentiation

Above, we used the letters aa and bb. These are called variables. They are simply placeholders that we can plug numbers, other variables, or even other operations with variables into. Variables allow us to generalize, or abstract, statements. For example, 2+2=4{2 + 2 = 4}, and 3+8=11{3 + 8 = 11}. We make an abstraction of this pattern, or idea, by writing something like a+b=n.{a + b = n.} This communicates the idea, "The number n{n} is the sum of a{a} and b.{b.}"

There are many ways to read an equation, but they all express the same idea. The human mind, however, is a peculiar thing. Often, reading and writing the same equation in different ways can lead to drastically different insights. Consider all the different ways we can read a+b=n:{a + b = n:} The sum of a{a} and b{b} is n.{n.} The number n{n} is the sum of a{a} and b.{b.} The numbers a{a} and b{b} such that the sum is n.{n.} The number b{b} is the difference of n{n} and a.{a.}

All of this is to say that addition and subtraction are just fast and flexible ways to express counting forwards and backwards. For example, the expression 4+8{4 + 8} is just another way of saying "Count 8{8} positions from 4{4}." This leads to: 5,6,7,8,9,10,11,12.{5, 6, 7, 8, 9, 10, 11, 12.} This in turn is just another way of saying "Count 12{12} positions from 0{0}" or "Count 1{1} position from 11.{11.}" Notice that with these addition (and subtraction), we are just offsetting from one or the other of the operands. 1+1{1 + 1} is simply "1 offset 1," which is 2.{2.}

All of this evidences that addition and subtraction are abstractions of counting. Abstractions are useful because they allow us greater flexibility and accuracy with our statements. Suppose we want to say, "I'm looking for two numbers that add to 10." We can write: a+b=10a + b = 10. This flexibility and accuracy is a powerful tool in mathematics, because it gives us plenty of room to write statements that say what we can and cannot do — rules.

Understanding the basic rules of arithmetic at a very high level is priceless in mathematics. Part of the elegance and power of mathematics is that we can arrive at beautiful, and often shocking, results from a set of very basic rules.

Order of Operations

Arithmetic allows us to combine numbers, through the operations listed above, to form other numbers. The way we order operations matters in mathematics. Otherwise, we run into problems. For example, consider the following:

2+3×4 2 + 3 \times 4

What does this expression mean? Well, it could be: (1) Add 2 and 3, then (2) multiply the sum by 4. Or, it could mean: (1) Multiply 3 and 4, and (2) add 2 to the product. This ambiguity is unacceptable in mathematics. We need be clear and precise with our words. For these reason, mathematics follows a few conventions on how to write mathematical expressions:

  1. Evaluate expressions in parentheses first, working from the inner most to the outer most, applying the rules below.
  2. Perform all exponentiations.
  3. Perform all multiplications and divisions from left to right.
  4. Perform all additions and subtractions from left to right.

The conventions above are collectively called the order of operations. Again, these are conventions. They are not universal truths; they're simply what mathematicians have come to agree to about how mathematics is communicated and understood.

Addition

Addition can be thought of as moving to the right along the real number line by whatever number we are adding. So, for example, the expression 4+2{4 + 2} can be thought of as "Start at 4, and move 2 to the right." This give us the final destination of 6. Subtraction is really just addition, only instead of moving towards the right, we move towards the left. For example, 434 - 3 is really 4+(3)4 + (-3), returning 1. Visually:

A number line.

Identity Property of Real Addition

Pius, a bank robber, goes to Ultimate Safety Bank, and robs it. He leaves with 4,000 dollars. Pius then goes to another bank Safer Bank, 4,000 dollars and fake gun in hand, and tells the teller to put the money in the bag. The teller takes the bag, disappears below the counter, and pretends to place money in the bag; the teller, in fact, didn't place anything in the bag. She returns the bag, and Pius sprints out the building to his e-scooter, and zooms away. He gets home and opens the bag. How much money does Pius have?

Pius starts with 0, then takes 4,000 from Ultimate Safety, then goes to Safer Bank, and gets 0. When we add 0 to a number, we are effectively saying "no movement" along the number line. Thus, we have the following rule:

identity property of addition. Given aR,{a \in \reals,} it follows that a+0=0+a=a.{a + 0 = 0 + a = a.}

Real Subtraction

Knowing whether operands are negative or positive can allow us to immediately conclude certain propositions. First, the sum of two positive integers is a positive integer.

theorem. If a,bR,{a, b \in \reals,} then a+bR.{a + b \in \reals.}

From the rule property above, we have a further property: The sum of two negative numbers is a negative number.

theorem. If a,bR,{a, b \in \reals^{-},} then a+bR.{a + b \in \reals^{-}.}

The operation of subtraction is not exactly an independent operation. Really, subtraction is just addition with a negative. The "operation" of subtraction is really just a shorthand way for writing out addition of positive and negative numbers. For example, when we write a+(b),{a + (-b),}< we instead just write ab.{a - b.} It's just a faster way of writing the same expression.

The identity property of addition leads to the additive inverse. As aforementioned, subtraction is really just adding negative numbers. Considering the addition with zero rule, we have the following inference:

additive inverse. Given aR,{a \in \reals,} a+(a)=(a)+a=0{a + (-a) = (-a) + a = 0}

Given the equation a+(a)=0{a + (-a) = 0}, we say that a{-a} is the additive inverse of a.{a.} Whenever we see the equation a+a=0{-a + a = 0}, we should immediately become aware that we a{a} and a{-a} are on opposite sides of 0 on the real line. Furthermore, because a+a=0{-a + a = 0}, subtracting a{-a} from both sides, we have a=(a){-a = -(-a)}.

The additive inverse also allows us to draw another inference. Given the equation a+x=a,{a + x = a,} we should immediately know that x=0.{x = 0.} Similary, if we see the equation a+x=0,{a + x = 0,} then we should immediately know that x=a.{x = -a.}

Commutativity of Real Addition

The commutative law of addition states that:

a+b=b+a a + b = b + a

In other words, the order of the terms in an addition expression does not change the sum of those terms. We call the property above more generally as commutativity. One way to remember it: In addition, terms can "commute."" In other words, the terms can move around as they please, as long as they stay within addition (and as we'll see later, multiplication).

The idea is made apparent from a simple analogy. If we want sweet iced tea, we can: add the tea, then the ice, then the sugar; or we can add the sugar, then the ice, then the tea; or we can add the tea, then the sugar, then the ice, and so on and so forth. Either or, the result is the same — sweet iced tea.1

Associative Law of Addition

The associative law of addition states that the sum of the terms does not depend on the grouping of the terms. In other words, we can group the terms however way we'd like.

associative property of addition. If a,b,cR,{a, b, c \in \reals,} then a+b+c=(a+b)+c=a+(b+c).{a + b + c = (a + b) + c = a + (b + c).}

We call the property above more generally as associativity. We can remember it as: In addition, the terms are free to associate. In other words, the terms can gather together or separate as they please, as long as they stay within addition (and again, as we'll see later, multiplication).

From the definition above, if we have the expression a+b+c+d,{a + b + c + d,} we can generate numerous different possibilities:

a+b+c+da+b+c+da+(b+c+d)a+(b+(c+d))a+b+(c+d) a + b + c + d \\ a + b + c + d \\ a + (b + c + d) \\ a + (b + (c + d)) \\ a + b + (c + d)

All these possibilities are equivalent, but some may be easier to compute than others. The associative law of addition leads to another useful property:

theorem. If a+b=0a + b = 0, then b=ab = -a and a=ba = -b.

The above property is the direct result of subtracting either aa or bb from both sides. Associativity and commutativity are what allow us to rewrite expressions in numerous different ways:

(ab)+c=(a+(b))+c=a+(b+c) by associativity=a+(cb) by commutativity=(a+c)b by associativity\begin{aligned} (a - b) + c &= (a + (-b)) + c \\ &= a + (-b + c) ~\text{by associativity} \\ &= a + (c - b) ~\text{by commutativity} \\ &= (a + c) - b ~\text{by associativity} \end{aligned}

Importantly, the associative property of addition does not apply to subtraction. For example, ab{a - b} does not imply that ba.{b - a.} ab{a - b} is equivalent to a+(b),{a + (-b),} but it is not equivalent to b+(a).{b + (-a).} The proposition ab=ba{a - b = b - a} is true iff a=b.{a = b.}

The Cancellation Law of Addition

The Cancellation Law for Addition Suppose we have the following relationship:

a+b=a+c a + b = a + c

Subtracting a{ a } from both sides, we have a+ba=a+ca{ a + b - a = a + c - a }. Simplifying, we have: b=c{ b = c }.

cancellation law of addition. If a+b=a+c{ a + b = a + c }, then b=c{ b = c }.

Closure

property. Let a,bR{a, b \in \reals} then a+bR{a + b \in \reals} and abR.{a \cdot b \in \reals.}

From the property above, we say that the reals are closed under addition and multiplication. In other words, if we add two reals numbers, then the sum is a real number. If we multiply two real numbers, then the product is real number. We can draw a few other useful properties.

Inequalities

Inequalities arguably appear more often than equalities in mathematics. This is because equality imposes an extraordinarily high burden of proof: When we write a=b,{a = b,} we are implying that a{a} and b{b} are the same in every possible way. Because of that burden, we often must speak in terms of relativity, using the symbols <,>,,,,,,.{ <, >, \leq, \geq, \nless, \ngtr, \nleq, \ngeq.} Let's discuss how each of these symbols are used and the propositions they can express.

First, when we write a<b,{a < b,} we are stating, a{a} is greater than b.{b.}_ The same proposition can be written as b<a.{b < a.}

Second, if a<0,{a < 0,} we say that a{a} is negative. If a>0,{a > 0,} we say that a{a} is positive.

Third, the inequality symbols operate similarly to the equal sign: There are two sides, and we can manipulate the expression accordingly — so long as the expression remains true. For example, consider the proposition a>b.{a > b.} We can manipulate this expression as such:

a>bab>bbab>0 \begin{aligned} a &> b \\ a - b &> b - b \\ a - b &> 0 \end{aligned}

We see that given the expression a>b,{a > b,} it follows that ab{a - b} is positive. This analysis leads to several laws. First, the Trichotomy law:

Trichotomy Law. For every aR,{a \in \reals,} one, and only one, of the following propositions is true: a=0{a = 0} aR+{a \in \reals^{+}} > aR+{-a \in \reals^{+}}

The rule above tells us that every real number a{a} can be only one of three things: zero, positive, or negative. Those are the only three possible states. Nothing else. Additionally, we have the laws of closure. We can think of these laws as defining the boundaries of addition and multiplication. First, the law of closure under addition:

Closure Under Addition. If a{a} and b{b} are in R+,{R^{+},} then a+b{a + b} is in R+.{R^{+}.}

And the law of closure under multiplication:

Closure Under Multiplication. If a{a} and b{b} are in R+,{R^{+},} then ab{a \cdot b} is in R+.{R^{+}.}

Complementing the laws above, we have the following definitions:

Definition. Given any two numbers a{a} and b,{b,} the following propositions hold:

  1. abR+a>b{a - b \in \reals^{+} \nc a > b}
  2. b>aa{b > a \nc a}
  3. (a>b)(a=b)ab{(a > b) \lor (a = b) \nc a \geq b}
  4. a>0aR+{a > 0 \iff a \in \reals^{+}}

As a consequence of the definitions above, we have:

definition. Given any two numbers a{a} and b,{b,} exactly one of the following propositions is true:

  • ab=0,{a - b = 0,}
  • ab>0,{a - b > 0,} xor
  • (ab)=ba>0{-(a - b) = b - a > 0}

Manipulating the above definitions, we have:

Definition. Given any two numbers a{a} and b,{b,} exactly one of the following propositions is true: a=b,{a = b,} a>b,{a > b,} xor b>a{b > a}

That is to say, given any numbers a{a} and b,{b,} one is less than or equal to the other.

Multiplication

Multiplication is simply repeated addition. Accordingly, we can think of multiplication as an even faster method of counting. For example, 22{2 \cdot 2} is just another way of saying "Move 2 positions 2 times forward, from the origin." This leads to 4. Similarly, 2-3{2 \cdot \texttt{-}3} is equivalent to, "Move 2 positions 3 timees backward, from the origin." This leads to -6.{\texttt{-}6.} This reveals that whenever we write the expression ab,{a \cdot b,} there's an implicit (ab)+0.{(a \cdot b) + 0.} Whenever we multiply integers, the following rule holds true:

Closure Property of Multiplication. The product of two integers is an integer.

Laws of Multiplication

Like addition, there are several laws of multiplication. These are equivalent to the laws of addition with just a few differences.

Commutative Law of Multiplication.

The commutative law of multiplication states that we the order in which we multiply two real numbers a{a} and b{b} doesn't affect the result. In other words, we can freely move terms around and arrive at the same product.2

Commutative Property of Multiplication. > ab=ba=(a)(b)=(b)(a){a \cdot b = b \cdot a = (a)(b) = (b)(a)}

Reciprocal Law of Multiplication

What is the result of 212?{2 \cdot \frac{1}{2}?} Clearly, it is 1.{1.} How about 313?{3 \cdot \frac{1}{3}?} It's also 1.{1.} Indeed, n1n{n \cdot \dfrac{1}{n}} will always return 1,{1,} where n{n} is a real number. We call this the reciprocal law of multiplication:

Reciprocal Law of Multiplication. Given any n0,{n \neq 0,} there exists a number n1=1n{n^{-1} = \dfrac{1}{n}} such that: nn1=n1n=1n \cdot n^{-1} = n^{-1} \cdot n = 1

While we haven't covered exponents yet, the notation n-1{n^{\texttt{-}1}} is the same as the expression 1n{\dfrac{1}{n}}. We call the number n-1{n^{\texttt{-}1}} the reciprocal or the multiplicative inverse of the number n.{n.} The law above tells us that multiplying a real number by its reciprocal always yields the number 1,{1,} as long as that real number isn't 0.{0.} Why not 0?{0?} Because if n=0,{n = 0,} then we would have the reciprocal 10.{\dfrac{1}{0}.} And for our purposes, division by 0{0} is always undefined.

Associative Law of Multiplication

The associative law of multiplication tells us that we can group factors however way we'd like, and still arrive at the same product.3

Associative Property of Multiplication. (ab)c=a(bc){ (ab)c = a(bc)}

The laws of commutativity and associativity apply whether a{ a }, b{ b }, and c{ c } are negative, positive, or 0.

Multiplication by Zero

A man goes to a café and asks the barista for a Bombón. The barista asks, "How many tablespoons of sugar would you like?" The man replies, "Zero."4

Question: How many table spoons of sugar would the man get? Very clearly zero! Zero of something is zero. We state this property of multiplication formally:

Zero Property of Multiplication. Given any real number a,{a,} > 0a=0{0 \cdot a = 0}

Identity Property of Multiplication

Ayuri is about to walk to a smoke shop to buy some flavored tobacco. She asks her roommate Elena, "Do you want any flavored tobacco?" looking for flavored tobacco. Elena responds, "I do need some flavored tobacco." "What would you like?" Ayuri asks. Elena answers, "Surprise me." At the smoke shop, Ayuri asks the shopkeeper what flavors of tobacco the shop has. The shopkeeper says, "There are 100 flavors — tobacco, mint, mango, lemon ... " enumerating each. After several minutes, Ayuri responds, "One of each!" Ayuri returns home with a heavy bag, greeted by her roommate Elena with, "Good grief, how many did you buy?" Ayuri stares, "100 I think.".

The computation above is governed by the identity law of multiplication. More formally:

Identity Law of Multiplication. Given aR,{a \in \reals,} then: 1a=a.{1 \cdot a = a.}

The above rules allow us to perform elaborate simplifications:

(2a)(5b)=2(a(5b)) by associativity=2(5a)b by commutativity=(25)ab by associativity=10ab\begin{aligned} (2a)(5b) &= 2(a(5b)) ~\text{by associativity}\\ &= 2(5a)b ~\text{by commutativity}\\ &= ( 2 \cdot 5 )ab ~\text{by associativity}\\ &= 10ab \end{aligned}

The above example represents a useful technique when handling expressions involving multiple factors: group all of the integers into a single multiplication expression, and all of the variables elsewhere.

example. Numbers consisting entirely of the digit "8" are added, resulting in the sum 1000. What are these numbers?

First, assume the following:

8+ 81000 \begin{matrix} \begin{alignedat}{8} \ldots 8 \\\ldots \\ + \space \ldots 8 \\ \hline 1000 \end{alignedat} \end{matrix}

What we know: Each number ends with an 8. Since the sum is 1000, the sum of the last digit, an 8, is 0.

example. How many 8s do we need to arrive at a sum where the last digit is 0? At least five 8s, since 8×5=408 \times 5 = 40. Thus, we have:

8888+ 81000 \begin{matrix} \begin{alignedat}{8} \ldots 8 \\ \ldots 8 \\ \ldots 8 \\ \ldots 8 \\ + \space \ldots 8 \\ \hline 1000 \end{alignedat} \end{matrix}

Now note that there is a 4 that must be carried to the next place. To keep the sum's digits 0, we need a sum of 8s and a 4 that returns the last digit 0. We know that 8+8+4=208 + 8 + 4 = 20. Further, we need one more 8 to get a 10:

88888+ 8881000 \begin{matrix} \begin{alignedat}{8} 8 \\ 8 \\ 8 \\ 88 \\ + \space 888 \\ \hline 1000 \end{alignedat} \end{matrix}

Thus, the numbers are 8, 8, 8, 88, and 888.

example. Suppose the following:

AAA+ BBBAAAC \begin{matrix} \begin{align*} AAA \\ + \space BBB \\ \hline AAAC \end{align*} \end{matrix}

The As{As} denote some digit, the Bs{Bs} denote another digit, and Cs{Cs} denotes a third digit. What are these digits? To start, we are looking for a number where A+B=nCA + B = nC, where nn and CC are digits. Furthermore, A+B+n=mAA + B + n = mA, where mm and AA are digits. Then, m+A+B=AAAm + A + B = AAA. Next, we note that A+BA + B must result in a two digit number. Second, we further note that AA appears in the thousands position of the result (AAACAAAC). Because AA appears in the thousands place, no other digit can be carried over into the thousands place. There is only one number for BB that can satisfy these conditions: 9.

111+ 9991110 \begin{align*} 111 \\ + \space 999 \\ \hline 1110 \end{align*}

example. Daniel claims that he can multiply any three-digit number by 1001 instantly. If his friend, Helga, says to him, "715" he gives the answer immediately. Compute his answer and explain Daniel's secret.

First, 715×1001=715715715 \times 1001 = 715715. We can see this with the multiplication algorithm:

1001×71550051001+7007715715 \begin{array}{cc} & & & 1 & 0 & 0 & 1 \\ \times & & & & 7 & 1 & 5 \\ \hline & & & 5 & 0 & 0 & 5 \\ & & 1 & 0 & 0 & 1 \\ + & 7 & 0 & 0 & 7 \\ \hline & 7 & 1 & 5 & 7 & 1 & 5 \end{array}

Under the multiplication algorithm, Daniel simply takes the three digits, ABC, and performs the following:

C00CB00BA00A C00C \\ B00B \\ A00A

Then, he adds the numbers, shifting one place to the left on each line:

C00CB00B+A00AABCABC \begin{array}{cc} & & & C & 0 & 0 & C \\ & & B & 0 & 0 & B \\ + & A & 0 & 0 & A \\ \hline & A & B & C & A & B & C \end{array}

The result then is ABCABCABCABC, a repetition of the digits ABCABC. Note further that there are two 0s, and the digits are repeated twice.

example. Compute 1010101010×571010101010 \times 57. The multiplication algorithm tells us that we are simply shifting the number 57 over and over, but, with an added 0, since the the first place does not shift:

010101010101575757575757 \begin{array}{cc} & 01 & 01 & 01 & 01 & 01 & 01 \\ & & & & & & 57 \\ \hline & & 57 & 57 & 57 & 57 & 57 \end{array}

Thus, the product is 5757575757.

example. Compute 10001×102030405010001 \times 1020304050. Solution We can compute this with the following:

1020304050×0100011020304050000000000000+102030405010204060804050 \begin{array}{cc} & & & 10 & 20 & 30 & 40 & 50 \\ \times & & & & & 01 & 00 & 01 \\ \hline & & & 10 & 20 & 30 & 40 & 50 \\ & & 00 & 00 & 00 & 00 & 00 & 00 \\ + & 10 & 20 & 30 & 40 & 50 \\ \hline & 10 & 20 & 40 & 60 & 80 & 40 & 50 \end{array}

Thus, the product is 10204060804050.

example. Compute 11111×111111111 \times 1111. Rewriting the expression with the multiplication algorithm:

11111×1111111111111111111+1111112344321 \begin{array}{cc} & & & & 1 & 1 & 1 & 1 & 1 \\ \times & & & & & 1 & 1 & 1 & 1 \\ \hline & & & & 1 & 1 & 1 & 1 & 1 \\ & & & 1 & 1 & 1 & 1 & 1 \\ & & 1 & 1 & 1 & 1 & 1 \\ + & 1 & 1 & 1 & 1 & 1 \\ \hline & 1 & 2 & 3 & 4 & 4 & 3 & 2 & 1 \end{array}

Distributivity

With the rules thus far, we can make even further inferences. We know that if a+b=0{ a + b = 0 }, then a=b{ a = -b } and b=a{ b = -a }. Thus, if a+b=0{ a + b = 0 }, then it must be true that (a+b)+(ab)=0{ (a + b) + (-a - b) = 0 }, since a+b=0{ a + b = 0 } and ab=0{ -a - b = 0 }. If (a+b)+(ab)=0{ (a + b) + (-a-b) = 0 }, then it follows that a+bab=0{ a + b - a -b = 0 }. Rearranging the equation, we have aa+bb=0{ a - a + b - b = 0 }. Rearranging again, we have: (a+b)=a+(b){ -(a + b) = -a + (-b) }. This yields the following inference:

(a+b)=ab-(a + b) = -a - b

Note that we should always be vigilant about distributing a minus sign to a sum involving negative numbers: (a)=a{ -(-a) = a }. To ensure that vigilance, we should make the following rules clear:

lemma. If a,bR{a,b \in \reals} and b{ b } are positive integers, then a+b{a+b} is a positive integer.

lemma. If a{a} and b{b} R{\reals^-}, then a+bR.{ a + b \in \reals^-.}

We can see that these rules are true with a simple proof. Let a=n{ a = -n } and b=m{ b = -m }, where m{ m } and n{ n } are positive. It follows then that a+b=nm=(n+m){ a + b = -n - m = -(n + m) }. Since n{ n } and m{ m } are positive, n+m{ n + m } is positive. Since n+m{ n + m } is positive, it follows that a+b{ a + b } is negative, because n+m{ n + m } is positive only if a+b{ a + b } is negative. This analysis hints at another law, the distributive property.

distributive property. If a,b,cR,{a, b, c \in \reals,} then: a(b+c)=ab+ac{a(b + c) = ab + ac}

With the distribute property, we have another way of proving that a0=0:{a \cdot 0 = 0:}

(a0)+(a0)=a(0+0)=a0\begin{aligned} (a \cdot 0) + (a \cdot 0) &= a \cdot (0 + 0) \\ &= a \cdot 0 \end{aligned}

The distributive property is also what leads to what at first appears to be a mysterious rule: The product of two negative numbers is a positive number. First, we start by proving that (a)(b)=(ab):{(-a)(b) = -(a \cdot b):}

(a)(b)+(a)(b)=((a)+a)(b)=(0)(b)=0\begin{aligned} (-a)(b) + (a)(b) &= ((-a) + a)(b) \\ &= (0)(b) \\ &= 0 \end{aligned}

Because (a)(b)=(ab),{(-a)(b) = -(a \cdot b),} we can make the following inferences. First, the factor (a)(b)=(ab).{(-a)(b) = -(ab).} Adding (ab){-(ab)} to both sides:

(a)(b)+((ab))=(a)(b)+(ab)=(a)((b)+b)=(a)0=0\begin{aligned} (-a)(-b) + (-(ab)) &= (-a)(-b) + (-ab) \\ &= (-a) \cdot ((-b) + b) \\ &= (-a) \cdot 0 \\ &= 0 \end{aligned}

Adding (a)(b){(a)(b)} to both sides:

(a)(b)=(a)(b)(-a)(-b) = (a)(b)

Multiplying 37254,{37 \cdot 25 \cdot 4,} we get:

37254=37(254)=37(100)=3700\begin{aligned} 37 \cdot 25 \cdot 4 &= 37 \cdot (25 \cdot 4) &= 37 \cdot (100) \\ &= 3700 \\ \end{aligned}

Anoter example: Multiply 125378{125 \cdot 37 \cdot 8}

125378=(25542)37=(10010)37=(1000)37=37000\begin{aligned} 125 \cdot 37 \cdot 8 = (25 \cdot 5 \cdot 4 \cdot 2) \cdot 37 \\ = (100 \cdot 10) \cdot 37 \\ = (1000) \cdot 37 \\ = 37000 \end{aligned}

Multivariable Relationships

Suppose we know the following relationship:

a+b=c a + b = c

Three number relationships are ripe for inferences. If we add b{ -b } to both sides of the equation, we obtain a+bb=cb{ a + b - b = c - b }. Simplifying, we have a+0=cb{ a + 0 = c - b }. We can then infer:

a=cbb=ca a = c - b \\ b = c - a

example. Add 357 + 17999 + 1 without paper and pencil.

357+17999+1=357+(17999+1)=357+(18000)=18357\begin{aligned} 357 + 17999 + 1 &= 357 + (17999 + 1)\\ &= 357 + (18000)\\ &= 18357 \end{aligned}

example. Add 899 + 1343 + 101 without paper and pencil.

899+1343+101=(899+1)+1343+(1011)=(900)+1343+(100)=(1000)+1343=2343\begin{aligned} 899 + 1343 + 101 &= (899 + 1) + 1343 + (101 - 1) \\ &= (900) + 1343 + (100) \\ &= (1000) + 1343 \\ &= 2343 \end{aligned}

example. (a+b)+(c+d)=(a+d)+(b+c){ (a + b) + (c + d) = (a + d) + (b + c) }

(a+b)+(c+d)=(a+d+b+c) by commutativity=(a+d)+(b+c) by associativity \begin{aligned} (a + b) + (c + d) &= (a + d + b + c) ~\text{by commutativity} \\ &= (a + d) + (b + c) ~\text{by associativity} \end{aligned}

example. Prove: (a+b)+(c+d)=(a+c)+(b+d){ (a + b) + (c + d) = (a + c) + (b + d) }

(a+b)+(c+d)=(a+b+c+d) by associativity(a+b)+(c+d)=(a+c+b+d) by commutativity(a+b)+(c+d)=(a+c)+(b+d) by associativity \phantom{(a + b) + (c + d)} = (a + b + c + d) ~\text{by associativity} \\ \phantom{(a + b) + (c + d)} = (a + c + b + d) ~\text{by commutativity} \\ \phantom{(a + b) + (c + d)} = (a + c) + (b + d) ~\text{by associativity}

example. Prove: (ab)+(cd)=(a+c)+(bd).{(a - b) + (c - d) = (a + c) + (-b - d).} Following the rule nm=n+(m),{n - m = n + (-m),} (ab)+(cd)=(a+(b))+(c+(d)){(a - b) + (c - d) = (a + (-b)) + (c + (-d))}. Then, by associativity, (a+(b))+(c+(d))=a+(b)+c+(d){(a + (-b)) + (c + (-d)) = a + (-b) + c + (-d)}. By commutativity, a+(b)+c+(d)=(a+c)+((b)+(d)).{a + (-b) + c + (-d) = (a + c) + ((-b) + (-d)).} Following the rule nm=n+(m){n - m = n + (-m)}, it follows that (a+c)+((b)+(d))=(a+c)+(bd).{(a + c) + ((-b) + (-d)) = (a + c) + (-b - d).} Therefore, (ab)+(cd)=(a+c)+(bd).{(a - b) + (c - d) = (a + c) + (-b - d).}

Expansions

There are some expansions that every student in mathematics should know. An expansion is just a way of rewriting — expanding — a shortened expression. The expansions are the following:

(a+b)2=a2+2ab+b2(ab)2=a22ab+b2(a+b)(ab)=a2b2\begin{aligned} (a + b)^2 &= a^2 + 2ab + b^2 \\ (a - b)^2 &= a^2 - 2ab + b^2 \\ (a + b)(a - b) &= a^2 - b^2 \end{aligned}

The Multiplication Algorithm

The law of distributivity is what leads to the multiplication algorithm we're all farmiliar with from elementary school. The multiplication algorithm allows us to compute large numbers without the need for memorization. For example, 17×38{17 \times 38}:

 17×38136+51646\begin{array}{cc} & \space & 1 & 7 & \\ & \times & 3 & 8 & \\ \hline & 1 & 3 & 6 & \\ + & 5 & 1 \\ \hline & 6 & 4 & 6 & \end{array}

Similarly:

 38×17266+38646\begin{array}{cc} & \space & 3 & 8 & \\ & \times & 1 & 7 & \\ \hline & 2 & 6 & 6 & \\ + & 3 & 8 \\ \hline & 6 & 4 & 6 & \end{array}

The algorithm above is really an embodiment of applying the distributive property:

17×38=17×(3×10+8)=(17×3×10)+(17×8)=510+136=646\begin{aligned} 17 \times 38 &= 17 \times (3 \times 10 + 8) \\ &= (17 \times 3 \times 10) + (17 \times 8) \\ &= 510 + 136 \\ &= 646 \end{aligned}

Exponentiation

We can get by with the operations of addition, subtraction, multiplication, and division alone. However, having just these operations would be particularly cumbersome for many mathematical statements. Consider for example, the following expression:

xxxxxxx+xxxxxxxxxx + xxx

This is difficult to parse, and delimiting with parentheses and operator symbols is no better:

(x)(x)(x)(x)(x)(x)(x)+(x)(x)(x)(x)(x)(x)(x)(x)(x)(x) + (x)(x)(x)
xxxxxxx+xxxx \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x + x \cdot x \cdot x

Even worse:5

xxxxxxx×xxx=xxxxxxxxxxxxxxxxx \times xxx = xxxxxxxxxx

Because of this difficulty, we have the special operation of exponentiation. We can think of this as repeated multiplication. And because multiplication is already a pretty fast way of counting, exponentiation is even faster.

xxxxxxx+xxx=x7+x3xxxxxxx + xxx = x^7 + x^3
xxxxxxx×xxx=x7×x3=x10xxxxxxx \times xxx = x^7 \times x^3 = x^{10}

From the multiplication example, we have the following rule:

product rule of exponents. For any aR,{a \in \reals,} given mN{m \in \nat} and nN:{n \in \nat:} aman=am+n{a^m a^n = a^{m + n}}

Now, consider the following:

aaaaaaaaa=a4a5\dfrac{a \cdot a \cdot a \cdot a}{a \cdot a \cdot a \cdot a \cdot a} = \dfrac{a^4}{a^5}

Really, this reduces to:

aaaaaaaaa=1a\dfrac{\cancel{a} \cdot \cancel{a} \cdot \cancel{a} \cdot \cancel{a}}{\cancel{a} \cdot \cancel{a} \cdot \cancel{a} \cdot \cancel{a} \cdot a} = \dfrac{1}{a}

The computation above provides us this rule:

the quotient rule of exponents. For any aR,{a \in \reals,} and mN,{m \in \nat,} > nN,{n \in \nat,} such that m>n,{m > n,} > aman=amn{\dfrac{a^m}{a^n} = a^{m - n}}

and this rule:

the negative rule of exponents. For any aR,{a \in \reals,} and nN,{n \in \nat,} > an=1an{a^{-n} = \dfrac{1}{a^n}}

From the Quotient Rule of Exponents, we can observe:

aaaaaa=a3a3 \dfrac{a \cdot a \cdot a}{a \cdot a \cdot a} = \dfrac{a^3}{a^3}
aaaaaa=a33\dfrac{\cancel{a} \cdot \cancel{a} \cdot \cancel{a}}{\cancel{a} \cdot \cancel{a} \cdot \cancel{a}} = a^{3-3}
aaaaaa=a0 \phantom{\dfrac{a \cdot a \cdot a}{a \cdot a \cdot a}} = a^0
aaaaaa=1 \phantom{\dfrac{a \cdot a \cdot a}{a \cdot a \cdot a}} = 1

Thus, we have the rule:

the zero exponent rule. For any {aR:a0},{\{a \in \reals : a \neq 0 \},} > a0=1.{a^0 = 1.}

The Zero Exponent Rule as applied to the number 0{0}00{0^0} — is a point of debate. Some mathematicians treat 0 as 00=1.{0^0 = 1.} In algebra and calculus, however, 00undefined.{0^0 \nc \text{undefined}.} In computer science, programming languages handle 00{0^0} in numerous different ways; some languages provide an explicit value (i.e., 1), others undefined, others return an error.

Each approach has valid premises. If you consider xy{x^y} as a function of two variables, then we have a problem with the definition 00=1.{0^0 = 1.} If we approach 0 along the line y=0,{y = 0,} then following the definition, limx0+=1.{\lim\limits_{x \to 0^{+}} = 1.} However, if we approach 0 along the line x=0,{x = 0,} then we get limy0y=limy0+0=0.{\lim\limits_{y \to 0^{y}} = \lim\limits_{y \to 0^{+}}0 = 0.} This would mean 00=0.{0^0 = 0.} Thus, continuous mathematics generally treats 00undefined.{0^0 \nc \text{undefined}.}

On the other hand, discrete mathematics treats 00=1.{0^0 = 1.} In set theory, AB{A^B} denotes the set of all functions from B{B} to A.{A.} If A{A} and B{B} denote the cardinalities (i.e., the "size" of the sets) then AB{A^B} is the size of the set of all functions from A{A} to B.{B.} If we use the value 0 for A{A} and B,{B,} then 0 denotes the empty set — 00{0^0} is the collection of all functions from the empty set to the empty set. This is the empty function. There is one, and only one, empty function. Accordingly, 00=1.{0^0 = 1.} Thus, it must be true that 00=1.{0^0 = 1.}

In these materials, we will limit the Zero Exponent Rule to a number x{x} where {xR:x0}.{\{ x \in \reals : x \neq 0 \}.} Next, consider the following example:

(b3)2(b^3)^2

This expression captures:

(bbb)(bbb)(b \cdot b \cdot b ) \cdot (b \cdot b \cdot b)

Following the associative property of multiplication:

(bbb)(bbb)=bbbbbb=b6\begin{aligned} (b \cdot b \cdot b ) \cdot (b \cdot b \cdot b) &= b \cdot b \cdot b \cdot b \cdot b \cdot b \\ & = b^6 \end{aligned}

Again, we have another rule:

the power rule of exponents. For any aR,{a \in \reals,} and mN,{m \in \nat,} > nN,{n \in \nat,} (am)n=amn{(a^m)^n = a^{m \cdot n}}

Using the commutative and associative properties of multiplication yields another inference:

(ab)3=(ab)(ab)(ab) expansion=aaabbb by associativity=(aaa)(bbb) by commutativity=a3b3 by the Definition of Exponent\begin{aligned} (ab)^3 &= (ab) \cdot (ab) \cdot (ab) ~\text{expansion} \\ &= a \cdot a \cdot a \cdot b \cdot b \cdot b ~\text{by associativity} \\ &= (a \cdot a \cdot a) \cdot (b \cdot b \cdot b) ~\text{by commutativity} \\ &= a^{3}b^{3} ~\text{by the Definition of Exponent} \end{aligned}

Thus, it follows that:

the power of a product rule of exponents. For all a,bR,{a, b \in \reals,} given a nN,{n \in \nat,} (ab)n=anbn{(ab)^n = a^n b^n}

This rule leads to another rule. Consider the expression (a2b2)3.{(a^{-2}b^{2})^3.} Expanding and reducing this expression with the negative rule of exponents, we get:

(a2b2)3=(b2a2)3 (a^{-2}b^{2})^3 = \left(\dfrac{b^2}{a^2}\right)^3

Applyting the definition of an exponent:

(a2b2)3=(b2a2)(b2a2)(b2a2)(a^{-2}b^{2})^3 = \left(\dfrac{b^2}{a^2}\right) \cdot \left(\dfrac{b^2}{a^2}\right) \cdot \left(\dfrac{b^2}{a^2}\right)

Then by commutativity:

(a2b2)3=b2b2b2a2a2a2(a^{-2}b^{2})^3 = \dfrac{b^2 \cdot b^2 \cdot b^2}{a^2 \cdot a^2 \cdot a^2}

And by the product rule of exponents:

(a2b2)3=b6a6 (a^{-2}b^{2})^3 = \dfrac{b^6}{a^6}

Finally, by the negative rule of exponents:

(a2b2)3=a6b6\phantom{(a^{-2}b^{2})^3} = a^{-6}b^{6}

This yields the rule:

the power of a quotient rule of exponents. For all a,bR{a, b \in \reals} and any nZ,{n \in \uint,} (ab)n=anbn\left( \dfrac{a}{b} \right)^n = \dfrac{a^n}{b^n}

Perfect Powers

With exponentiation, we arrive at several special properties. First, we have the notion of a perfect square. The set of all perfect square looks something like:

{0,1,4,9,16,25,36,49,64,81,100,121,144,}{\{ 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, \ldots \}}

Notice the pattern. These are all numbers that can be written in the form n2,{n^2,} where n{n} is some integer. for example, 9{9} is a perfect square because it can be written as 32.{3^2.} So too are 1{1} and 0.{0.} They can be written as 12{1^2} and 02,{0^2,} respectively. Formally:

definition. A perfect square is a positive integer n{n} such that: n=m2   (mZ+){n = m^2 \space \space \space (m \in \uint^{+})}

Alongside perfect squares, we have the perfect cube. The set of all perfect cubes looks like:

{0,1,8,27,54,125,216,343,512}{\{ 0, 1, 8, 27, 54, 125, 216, 343, 512 \}}

And the formal definition:

Definition. A perfect cube is a positive integer n{n} such that: n=m3   (mZ+){n = m^3 \space \space \space (m \in \uint^{+})}

Both perfect squares and perfect cubes are examples of perfect powers. In other words, we can have a "perfect tesseract"6

Definition. A positive integer n{n} is a perfect power of x{x} iff: n=mx   (mZ, xZ+){n = m^x \space \space \space (m \in \uint, \space x \in \uint^{+})}

Radicals

By introducing the exponentiation notation, it would be very helpful to have notation that quickly reverses the results. This notation is provided by the radical. The syntax:

Radical notation

The Square Root

Suppose we have a number a{a} where aR+.{a \in \reals^{+}.} The square root of a{a} is the number that, when multiplied by itself, i.e., a2,{a^2,} returns a.{a.} Now, that number may be positive or negative, because squaring two numbers, positive or negative, yields a positive number.

definition. Given aR+,{a \in \reals^{+},} a{\sqrt{a}} is the number n,{n,} where nR{n \in \reals} such that n2=a.{n^2 = a.}

From the square root definition, we have the following rule:

The Product Rule of Square Roots. Given aR0{a \in \R_{\geq 0}} and bR0,{b \in \R_{\geq 0},} ab=ab{\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}}

When simplifying radical expressions, there is a useful algorithm:

  1. Algorithm. Given a square root radical expression:
  2. Factor perfect squares from the radicand.
  3. Write the radical expression as a product of radical expressions.
  4. Simplify.

For example, consider implifying the following radical expression:

300\sqrt{300}

Factoring the perfect square:

1003 \sqrt{100 \cdot 3}

By the product rule:

1003 \sqrt{100} \cdot \sqrt{3}

Then, simplifying:

103 10 \sqrt{3}

Here's another example. Consider this square root:

162a2b4 \sqrt{162 a^2 b^4}

Factoring:

812a4ab4 \sqrt{81 \cdot 2 \cdot a^4 \cdot a \cdot b^4}

Then, by the product rule of square roots:

(81)(2)(a)(a4)(b4) (\sqrt{81}) (\sqrt{2}) (\sqrt{a}) (\sqrt{a^4}) (\sqrt{b^4})

And simplifying:

(9)(2)(a)(a2)(b2) (9) (\sqrt{2}) (\sqrt{a}) (a^2) (b^2)

By commutativity:

9a2b22a 9a^2 b^2 \sqrt{2a}

Another useful algorithm is combining multiple radical expressions into a single radical expression:

  1. Combine all the multiple radical expressions into a single expression.
  2. Simplify.

For example, consider the radical expression: (12)(3).{(\sqrt{12})(\sqrt{3}).} We can multiply the this expression's terms into a single expression:

(12)(3)=36 (\sqrt{12})(\sqrt{3}) = \sqrt{36}

The Quotient Rule of Square Roots

The Quotient Rule of Square Roots natural results from multiplication. First, suppose the following:

  1. a,b0.{a, b \geq 0.}
  2. c=ab.{c = \sqrt{\dfrac{a}{b}}.}
  3. x=a{x = \sqrt{a}} and y=b.{y = \sqrt{b}.}

This implies that x2=a{x^2 = a} and y2=b.{y^2 = b.} Next, suppose exists a nonnegative number n{n} such that n2=n.{\sqrt{n^2} = n.} It follows that, by substitution: (xy)2=x2y2=ab\left(\dfrac{x}{y}\right)^2 = \dfrac{x^2}{y^2} = \dfrac{a}{b}

Hence,

c=xy=ab=abc = \dfrac{x}{y} = \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}

by Assumptions 2 and 3. Thus, we have the following rule:

Property. Let a,b0.{a, b \geq 0.} Suppose n=ab,{n = \sqrt{\dfrac{a}{b}},} where nR0.{n \in \R_{\geq 0}.} Then: n=ab=abn = \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}

The rule above is particularly useful for rational expressions, which often contain radicands. Here's a useful algorithm for handling such expressions. Given a radical expression in the form of a quotient:

  1. Write the radical quotient has the quotient of two radical expressions.
  2. Simplify the numerator and denominator.

Adding and Subtracting Square Roots

We can add or subtract radical expressions only if (1) they have the same radicand and (2) they have the same nth{n^{\text{\scriptsize{th}}}} root. For example:

2+32=42\sqrt{2} + 3\sqrt{2} = 4 \sqrt{2}

Radican expressions can often be simplified by rewriting the relevant expression as a sum, followed by reductions.

2+18=2+29=2+(29)=2+32=42\begin{aligned} \sqrt{2} + \sqrt{18} &= \sqrt{2} + \sqrt{2 \cdot 9} \\ &= \sqrt{2} + (\sqrt{2} \cdot \sqrt{9}) \\ &= \sqrt{2} + 3 \sqrt{2} \\ &= 4 \sqrt{2} \end{aligned}

Here's a helpful algorithm. Given an expression that adds or subtract radical expressions,

  1. Simplify each radical expression to yield as many equal radicands as possible.
  2. Add or subtract expressions with equal radicands.

nth Root

The proposition a3=8,{a^3 = 8,} implies that 83=a.{\sqrt[3]{8} = a.} Thus, the phrase "the nth{n^{\text{\scriptsize{th}}}} root of a{a}" implies a number that, when raised to the the nth{n^{\text{\scriptsize{th}}}} power, equals a.{a.}

If a{a} is a real number with at least one nth{n^{\text{\scriptsize{th}}}} root, then the principal nth{n^{\text{\scriptsize{th}}}} root of a{a} is the number with the same sign as a{a} that, when raised to the nth{n^{\text{\scriptsize{th}}}} power, equals a.{a.}

Definition. Let aR,{a \in \reals,} where a{a} has at least one nth{n^{\text{\scriptsize{th}}}} root.

Then the principle nth{n^{\text{\scriptsize{th}}}} root of a,{a,} written as an,{\sqrt[n]{a},} is the number with the same sign as a,{a,} such that, when raised to the nth{n^{\text{\scriptsize{th}}}} power, equals a.{a.} We say that n{n} is the index of an.{\sqrt[n]{a}.}

Rational Exponents

Rather than using a radical symbol, we can express radical expressions with rational exponents. Needless to say, the index of such an exponent must be a positive integer:

a1n=ana^{\frac{1}{n}} = \sqrt[n]{a}

A useful fact: If the index n{n} is even, then a{a} cannot be negative.

We can also have rational exponents with numerators other than 1. If the numerator is a number other than 1, then we raise the base to a power and take an nth{n^{\text{\scriptsize{th}}}} root. The numerator tells us the power, and the denominator tells us the root:

amn=(an)m=amna^{\frac{m}{n}} = (\sqrt[n]{a})^{m} = \sqrt[n]{a^{m}}

Here's a useful procedure for handling rational exponents. Given an expression with a rational exponent:

  1. Determine the power by looking at the exponent's numerator. Determine the root by looking at the exponent's denominator.

  2. Using the base as the radicand, raise the radicand to the power and use the root as the index.

Ratios

The ratio of two quantities tells us the relative sizes of the quantities. For example, if we say, "There were 3 bachelors for every 5 bachelorettes," are expressing the ratio:

bachelorsbachelorettes=35=3:5 \dfrac{\text{bachelors}}{\text{bachelorettes}} = \dfrac{3}{5} = 3:5

The use of a colon is often helpful when expressing the relative sizes of the quantities of multiple discrete objects: The sentence, "For every 1 house on Main Street, there are 2 cats and 100 mice," may be expressed as: 1:2:100.{1:2:100.}

Example: The ratio of instructors to professors at the conference is 3:4. There are 84 total conference attendees, of which only instructors and professors attended.

Suppose a{a} is the number of instructors, and b{b} is the number of professors. Write an equation using the ratio information. Remember, a ratio tells us nothing about the actual sizes of the sets. It only gives us relative sizes. Thus, we know that:

number of instructorsnumber of professors=ab=34 \dfrac{\text{number of instructors}}{\text{number of professors}} = \dfrac{a}{b} = \dfrac{3}{4}

Write an equation using the total number of conference attendees. We know that only instructors and professors attended. Hence:

a+b=84 a + b = 84

Question: How many professors were at the conference? We solve the system of equations. First, isolate the a{a} in the first equation:

ab=34a=3b4 \dfrac{a}{b} = \dfrac{3}{4} \\ a = \dfrac{3b}{4}

Then, substitute:

a+b=843b4+b=843b+4b4=847b4=847b=331b=48\begin{aligned} a + b = 84 \\ \dfrac{3b}{4} + b = 84 \\ \dfrac{3b + 4b}{4} = 84 \\ \dfrac{7b}{4} = 84 \\ 7b = 331 \\ b = 48 \end{aligned}

Example: The ratio of papayas to mangoes to rambutans in the fruit basket is 2:3:30. Only papayas, mangoes, and rambutans comprise the fruit basket.

What is the number of mangoes to the number of fruits in the fruit basket?

From the ratio, we know that there are at least 2 papayas, 3 mangoes, and 30 rambutans. Thus, there 2+3+30=35{2 + 3 + 30 = 35} total fruits. Hence:

mt=335 \dfrac{m}{t} = \dfrac{3}{35}

Suppose there are 385 fruits in the fruit basket. How many mangoes are there? The ratio of mangoes to fruits is 335.{\dfrac{3}{35}.} Thus, we have:

335=tm385 \dfrac{3}{35} = \dfrac{t_m}{385}

where tm{t_m} is the total amount of mangoes. Solving for tm{t_m}:

335=tm38535tm=3385tm=3113535tm=33 \dfrac{3}{35} = \dfrac{t_m}{385} \\ 35t_m = 3 \cdot 385 \\ t_m = \dfrac{3 \cdot 11 \cdot 35}{35}\\ \phantom{t_m} = 33

Exercise. Shylock has a jewel box filled with only emeralds and rubies. The ratio of emeralds to the total number of jewels in the box is 2:5.{2:5.} If Shylock adds 4 emeralds and removes 10 rubies, there will be twice as many emeralds in the box as rubies. How many jewels were in the box originally?

Let e{e} be the number of emeralds, r{r} the number of rubies, and t{t} the total number of jewels. We know that the box consists only of emeralds and rubies, so:

e+r=t e + r = t

Thus, the original ratio of emeralds to jewels can be expressed as:

ee+r=255e=2(e+r)5e=2e+2r3e=2r \dfrac{e}{e + r} = \dfrac{2}{5} 5e = 2(e + r) 5e = 2e + 2r 3e = 2r

Next, we are told that if Shylock adds 4 emeralds and removes 10 rubies, there are twice as many emeralds in the box as there are rubies. This statement expresses a ratio:

e+4r10=21=2 \dfrac{e + 4}{r - 10} = \dfrac{2}{1} = 2

Once more, we can express this ratio as a linear equation:

e+4r10=21=22(r10)=e+42r20=e+4 \dfrac{e + 4}{r - 10} = \dfrac{2}{1} = 2 \\ 2(r - 10) = e + 4 \\ 2r - 20 = e + 4

We know 2r=3e,{2r = 3e,} so we can substitute:

3e20=e+42e=24e=12 3e - 20 = e + 4 2e = 24 e = 12

Since e=12,{e = 12,} we know there are r=3e2=3(12)2=362=18{r = \dfrac{3e}{2} = \dfrac{3(12)}{2} = \dfrac{36}{2} = 18} rubies. Thus, there are e+r=12+18=30{e + r = 12 + 18 = 30} jewels in the box.

Exercise. 9 out of every 11 of Daniel's lawyers recommended against the merger. The rest said recommended it. 18 of Daniel's lawyers recommended the merger. How many lawyers does Daniel have?

Suppose a{a} is the number of lawyers against the merger, and t{t} is the total number of lawyers. Thus, we have the following ratio:

at=911 \dfrac{a}{t} = \dfrac{9}{11}

Since there are exactly 18 lawyers who recommend against the merger, there are t18{t - 18} lawyers who recommend the merger. Thus:

t18t=911 \dfrac{t - 18}{t} = \dfrac{9}{11}

Now we merely must solve for t:{t:}

t18t=91111(t18)=9t11t198=9t198=2t99=t \dfrac{t - 18}{t} = \dfrac{9}{11} \\ 11(t - 18) = 9t \\ 11t - 198 = 9t \\ -198 = -2t \\ 99 = t

Daniel has 99 lawyers — a hefty bill.

Exercise. The ratio of correct answers to all answers is 7/10. What is the ratio of incorrect answers to the number of correct answers?

Because there are at least 7 correct answers, the remaining 3 answers are incorrect answers. Thus, the ratio of incorrect answers to correct answers is 3/7.

Exercise. Nio is given a bag filled with pills. There are only red, blue, and green pills. The ratio of red pills to blue pills to green pills is 1:5:3.{1:5:3.} There are 27 green pills in the bag. How many total pills are in the bag?

We know there are 27 green pills. For every 3 of those pills, there is 1 red pill. Thus, there are 27/3=9{27 / 3 = 9} red pills. Then, for every 1 red pill, there are 5 blue pills. Thus, there are 95=45{9 \cdot 5 = 45} blue pills. The total number of pills then is:

9+45+27=81 9 + 45 + 27 = 81

pills. Alternatively, we know that with just the ratio, there are 1+5+3=9{1 + 5 + 3 = 9} total pills. Hence, the ratio of green pills to total pills is 3/9=1/3.{3/9 = 1/3.} Suppose t{t} is the total number of pills in the bag. Because there are 27 green pills total, we have:

27x=13x=273x=81 \dfrac{27}{x} = \dfrac{1}{3} \\ x = 27 \cdot 3 \\ \phantom{x} = 81

Exercise. Two-thirds of Thneedville's residents voted for executing Order 66. If 634 residents did not vote for executing Order 66, how many residents does Thneedville have?

Since 2/3 voted for, we know that 1/3 did not vote for. We know that 634 total did not vote for, so we have the ratio:

13=634t \dfrac{1}{3} = \dfrac{634}{t}

where t{t} is the total. Solving for t:{t:}

t=6343t=1902 t = 634 \cdot 3 \\ \phantom{t} = 1902

Thneedville has a total of 1,902 residents.

Exercise. Count Dracula changes height several times. The ratio of his original height to his second height is 24:5. The ratio of his second height to his third height is 1:12. The ratio of his original height to his fourth height is 16:1. The tallest of these four heights is 10 meters. What is Dracula's shortest height?

There are four heights total. We denote each with h1,{h_1,} h2,{h_2,} h3,{h_3,} and h4.{h_4.} We thus have the following ratios:

h1h2=245h2h3=112h1h4=161 \dfrac{h_1}{h_2} = \dfrac{24}{5} \\ \dfrac{h_2}{h_3} = \dfrac{1}{12} \\ \dfrac{h_1}{h_4} = \dfrac{16}{1}

Since h1:h2{h_1 : h_2} is 24:5,{24 : 5,} we know that h2:h1{h_2 : h_1} is 5:24.{5 : 24.} This means that h2{h_2} is 5/24{5/24} of h1.{h_1.}

h2=524h1 h_2 = \dfrac{5}{24}h_1

Then, since h2:h3{h_2:h_3} is 1:12,{1:12,} we know that h3{h_3} is 12 times the second height:

h3=12h2=12(524) h_3 = 12h_2 = 12 \left( \dfrac{5}{24} \right)

Since h1:h4=16:1,{h_1 : h_4 = 16:1,} the original height is 16 times the fourth height. Thus,

h4=x16 h_4 = \dfrac{x}{16}

From the analysis thus far, we have:

h1h2=5h124h3=5h12h4=h116 h_1 \\ h_2 = \dfrac{5h_1}{24} \\ h_3 = \dfrac{5h_1}{2} \\ h_4 = \dfrac{h_1}{16}

Of these values, h3{h_3} is the largest, since its constant multiplier is 5/2.{5/2.} Hence, we solve for x{x} with 10:

5x2=105x=20x=20/5x=4 \dfrac{5x}{2} = 10 \\ 5x = 20 \\ x = 20/5 \\ x = 4

The shortest height is h4,{h_4,} since its constant multiplier is 1/16.{1/16.} Evaluating:

h4=416=0.25 h_4 = \dfrac{4}{16} = 0.25

Hence, Count Dracula's shortest height is 0.25{0.25} meters.

Exercise. What is the ratio of x{x} to y{y} if:

10x3y13x2y=35 \dfrac{10x - 3y}{13x - 2y} = \dfrac{3}{5}

The objective is rewrite the equation to express a relation of xy.{\dfrac{x}{y}.} The key is cross multiplication:

10x3y13x2y=355(10x3y)=3(13x2y)50x15y=39x6y11x=9yxy=911 \dfrac{10x - 3y}{13x - 2y} = \dfrac{3}{5} \\ 5(10x - 3y) = 3(13x - 2y) \\ 50x - 15y = 39x - 6y \\ 11x = 9y \\ \dfrac{x}{y} = \dfrac{9}{11}

Exercise. If x/y=2/3{x/y = 2/3} and y/z=7/5,{y/z = 7/5,} then what is z/x?{z/x?}

Percentages

The word percent is likely an abbreviation for the pseudo-Latin per centum, meaning "per hundred." Denoted by the %,{\%,} a percent is a shorthand for n/100,{n/100,} where n{n} nR.{n \in \reals.} Thus, 1/4=25/100=25%.{1/4 = 25/100 = 25\%.}

Exercise. What percent of 240 is 48?

We solve for x{x}:

240(x/100)=48x/100=48/240x=100(48/240)=4800/240=480/24=20\begin{aligned} 240 (x/100) &= 48 \\ x/100 &= 48/240 \\ x &= 100(48/240) \\ &= 4800/240 \\ &= 480/24 \\ &= 20 \end{aligned}

Thus, 48{48} is 20%{20\%} of 240.{240.} Of course, we could have found the percentage with simple arithmetic: 48/240=0.2.{48/240 = 0.2.} Then, multiplying 0.2{0.2} by 100,{100,} we get 20%.{20\%.} We used algebra first because it provides the appropriate approach for handling more complex percentage problems.

Exercise. 36 is 120%{120\%} of what number?

We solve for x:{x:}

x(120/100)=36x(12/10)=3612x10=3612x=360x=360/12=30\begin{aligned} x(120/100) &= 36 \\ x(12/10) &= 36 \\ \dfrac{12x}{10} &= 36 \\ 12x &= 360 \\ x &= 360/12 \\ &= 30 \end{aligned}

When we increase a number x{x} by some percentage, we add the percent of x{x} to x{x} itself. For example, if we increase 100 by 1%{1\%}, we increase 100 by 1, since 1%{1\%} of 100 is 1 — 101. Likewise, if we increase 100 by 100%,{100\%,} we obtain 200, since 100%{100\%} of 100 is 100 (100+100=200.){(100 + 100 = 200.)}

Proportions

Suppose there is a ratio of two variable quantities. If this ratio is always the same, we say the two quantities are in direct proportion. For example, suppose x=5{x = 5} and y=9.{y = 9.} Now suppose that x{x} and y{y} are directly proportional. What is y{y} when x=14?{x = 14?} Since x{x} and y{y} are directly proportional:

xy=5959=14y9(14)=5y126=5yy=126/5=25.2\begin{aligned} \dfrac{x}{y} &= \dfrac{5}{9} \\ \dfrac{5}{9} &= \dfrac{14}{y} \\ 9(14) &= 5y \\ 126 &= 5y \\ y &= 126/5 = 25.2 \\ \end{aligned}

Another example: Suppose a{a} and b2{b^2} are directly proportional, a=5,{a = 5,} and b=9.{b = 9.} What is a{a} when b=12?{b = 12?} Since a{a} and b2{b^2} are directly proportional:

ab2=5(9)2=581581=a(12)281a=5(12)(12)(3)(3)(3)(3)a=5(3)(4)(3)(4)(3)(3)(3)(3)a=5(3)(4)(3)(4)9a=80a=80/9\begin{aligned} \dfrac{a}{b^2} &= \dfrac{5}{(9)^2} \\ &= \dfrac{5}{81} \\ \dfrac{5}{81} &= \dfrac{a}{(12)^2} \\ 81a &= 5(12)(12) \\ (3)(3)(3)(3) a &= 5(3)(4)(3)(4) \\ \cancel{(3)}\cancel{(3)}(3)(3) a &= 5\cancel{(3)}(4)\cancel{(3)}(4) \\ 9a &= 80 \\ a &= 80/9 \end{aligned}

Rational Expressions

The quotient of two polynomial expressions is called a rational expression. When handling rational expressions, we use the same rules applicable to fractions.

Simplifying Rational Expressions

Simplifying a rational expressions is no different from simplifying a fraction. The only differences is that we must often use the techniques of factorization, as well as be mindful of the laws governing variable manipulation. To simplify a rational expression:

  1. Factor the numerator and denominator;
  2. cancel any common factors

For example:

x29x2+4x+3=(x+3)(x3)(x+3)(x+1)=(x+3)(x3)(x+3)(x+1)=(x3)(x+1)\begin{aligned} \dfrac{x^2 - 9}{x^2 + 4x + 3} &= \dfrac{(x+3)(x-3)}{(x+3)(x+1)} \\ &= \dfrac{\cancel{(x+3)}(x-3)}{\cancel{(x+3)}(x+1)} &= \dfrac{(x-3)}{(x+1)} \end{aligned}

Multiplying Rational Expressions

We follow the same procedures for multiplying fractions when we multiply rational expressions: To multiply two rational expressions:

  1. Factor the numerator and denominator;
  2. multiply the numerators;
  3. multiply the denominators
  4. simplify.

For example, given:

(x2+4x53x+18)(2x1x+5) \left(\dfrac{x^2 + 4x - 5}{3x + 18}\right)\left(\dfrac{2x-1}{x+5}\right)

By factorization:

(x2+4x53x+18)(2x1x+5)=((x+5)(x1)3(x+6))(2x1x+5) \left(\dfrac{x^2 + 4x - 5}{3x + 18}\right)\left(\dfrac{2x-1}{x+5}\right) = \left(\dfrac{(x+5)(x-1)}{3(x+6)}\right)\left(\dfrac{2x-1}{x+5}\right)

Using the associative property of multiplication:

(x2+4x53x+18)(2x1x+5)=(x+5)(x1)(2x1)3(x+6)(x+5) \left(\dfrac{x^2 + 4x - 5}{3x + 18}\right)\left(\dfrac{2x-1}{x+5}\right) = \dfrac{(x+5)(x-1)(2x-1)}{3(x+6)(x+5)}

Cancelling like terms:

(x2+4x53x+18)(2x1x+5)=(x+5)(x1)(2x1)3(x+6)(x+5) \left(\dfrac{x^2 + 4x - 5}{3x + 18}\right)\left(\dfrac{2x-1}{x+5}\right) = \dfrac{\cancel{(x+5)}(x-1)(2x-1)}{3(x+6)\cancel{(x+5)}}

We can thus conclude:

(x2+4x53x+18)(2x1x+5)=(x1)(2x1)3(x+6) \left(\dfrac{x^2 + 4x - 5}{3x + 18}\right)\left(\dfrac{2x-1}{x+5}\right) = \dfrac{(x-1)(2x-1)}{3(x+6)}

Adding and Subtracting Rational Expressions

Once more, adding and subtracting rational expressions works the same way was adding and subtracting fractions: We may add and subtract the fractions iff the rational expressions have a common denominator. To ease the computation, the least common denominator — the smallest multiple that the denominators have in common — should be used whenever possible.

To add or subtract rational expressions:

  1. Factor the numerator and denominator.
  2. Find the least common denominator of the expressions — to find the least common denominator:
    1. Factor the expressions.
    2. Multiply all of the distinct factors.
  3. Multiply the expressions by a form of 1 that changes the denominators to the LCD.
  4. Add or subtract the numerators. Simplify.

For example, given:

5x+6y \dfrac{5}{x} + \dfrac{6}{y}

The least common denominator is (x)(y).{(x)(y).} Multiplying the expressions by a form of 1 to ensure the denominator is the least common denominator:

(5xyy)+(6yxx) \left(\dfrac{5}{x} \cdot \dfrac{y}{y}\right) + \left(\dfrac{6}{y} \cdot \dfrac{x}{x} \right)

Carrying out the multiplication:

(5xyy)+(6yxx)=5yxy+6xxy=5y+6xxy\begin{aligned} \left(\dfrac{5}{x} \cdot \dfrac{y}{y}\right) + \left(\dfrac{6}{y} \cdot \dfrac{x}{x} \right) &= \dfrac{5y}{xy} + \dfrac{6x}{xy} \\ &= \dfrac{5y + 6x}{xy} \end{aligned}

Simplifying Nested Rational Expressions

Often, the numerators and denominators of a rational expression will themselves contain rational expressions. They might look like:

a+bcde+fg \dfrac{a + \dfrac{b}{c}}{\dfrac{d}{e} + \dfrac{f}{g}}

We call such expressions nested rational expressions (the rational expressions in the numerators and denominators are colloquially called nesting rats). Nested rational expressions often strike fear in students, partly because of their size, partly because they are not often encountered. Nested rational expressions, however, are a good example of where Polish notation has an advantage over the more common infix notation (the notation above). The same expression above can be expressed as:

(/ [+ a (/ b c)][+ (/ d e)(/ f g)])\begin{aligned} (/ \space &[+ \space a \space (/ \space b \space c)] \\[1em] &[+ \space (/ \space d \space e) (/ \space f \space g)]) \end{aligned}

When encountering deeply nested rational expressions like the above, consider rewriting the expression in Polish notation and simplifying the expression from inside to outside. For most nested rational expressions, however, we use the following algorithm:

  1. Combine the expressions in the numerator into a single rational expression by adding or subtracting.
  2. Combine the expressions in the denominator into a single rational expression by adding or subtracting.
  3. Rewrite the overall expression as the numerator divided by the numerator.
  4. Rewrite the overall expression as a multplication.
  5. Multiply.
  6. Simplify.

For example,

y+1xxy. \dfrac{y + \dfrac{1}{x}}{\dfrac{x}{y}}.

Combine the expressions in the numerator into a single expression:

yxx+1xxy=yx+1xxy. \dfrac{\dfrac{yx}{x} + \dfrac{1}{x}}{\dfrac{x}{y}} = \dfrac{\dfrac{yx + 1}{x}}{\dfrac{x}{y}}.

Rewrite the expression as multiplication:

yxx+1xxy=yx+1xyx=y(xy+1)x2\begin{aligned} \dfrac{\dfrac{yx}{x} + \dfrac{1}{x}}{\dfrac{x}{y}} &= \dfrac{yx + 1}{x} \cdot \dfrac{y}{x} \\ &= \dfrac{y(xy + 1)}{x^2} \end{aligned}

Rationalizing Radical Denominators

Rational expressions can contain radicals in the denominator. These can be cumbersome to handle, and it is almost always best to simplify them.

To remove radicals from the denominators of rational expressions, we multiply the radical expression by the form of 1 that will eliminate the radical. If the denominator contains a single term, we multiply the rational expression by its reciprocal. I.e., if the rational-radical expression is 1bc,{\dfrac{1}{b \sqrt{c}},} we multiply the expression by cc.{\dfrac{\sqrt{c}}{\sqrt{c}}.}

Now, if the denominator of the rational-radical expression contains the sum of a rational and an irrational term, we multiply the numerator and the denominator by the conjugate of the denominator — the expression wherein the sign of the radial portion of the denominator is made opposite. For example, if the denominator is a+bc,{a + b \sqrt{c},} then the conjuage is abc.{a - b \sqrt{c}.}

Given a radical expression of the form ab,{\dfrac{a}{\sqrt{b}},}

  1. Multiply the numerator and the denominator by the radical in the denominator.
  2. Simplify.

Division

Of the core arithmetic operators, division is the most complex. In any discussion of division, we must first begin with two subsets of integers, the even integers:

{2,4,6,8,} \{ 2, 4, 6, 8, \ldots \}

and the odd integers:

{1,3,5,7,} \{1, 3, 5, 7, \ldots\}

Notice that both the even integers and the odd integers increment by 2. The difference, however, is that the even integers start at 2, and the odd integers start at 1. A number's property of odd-ness or even-ness is called the number's parity. Note that parity is a property held only by the positive integers. The negative integers are neither even nor odd. We can abstract this idea with algebra: An even integer is an integer that can be written as 2k,{2k,} where k{k} is a positive integer. Denoting the set of even integers with the symbol Zeven,{\Z_{\text{even}},} and the set of positive integers with the symbol Zodd,{\Z_{\text{odd}},} we can define an even integer more formally as:

Definition. Let xZ.{x \in \uint.} > xZevenx=2k:kZ+{x \in \Z_{\text{even}} \iff x = 2k : k \in \uint^{+}} I.e., an integer x{x} is even if, and only if, it can be written in the form 2k,{2k,} where k{k} is a positive integer.

For the odd integers, we write:

Definition. Let xZ.{x \in \uint.} > xZoddx=2k1:kZ+{x \in \Z_{\text{odd}} \iff x = 2k - 1 : k \in \uint^{+}}

I.e., an integer x{x} is odd if, and only if, it can be written in the form 2k1,{2k - 1,} where k{k} is a positive integer. Alternatively, we can define odd integers as the following:

Definition. Let xZ.{x \in \uint.} > xZoddx=2k+1:kN{x \in \Z_{\text{odd}} \iff x = 2k + 1 : k \in \nat}

In other words, an integer x{x} is odd if, and only if, it can be written in the form 2k+1,{2k + 1,} where k{k} is a natural number.

The only meaningful difference between the first and second definition is whether k{k} is a natural number or a positive integer. In other words, whether k{k} can be 0. If k{k} can be 0, then we must change the definition from 2k1{2k - 1} to 2k+1,{2k + 1,} because if k=0,{k = 0,} then 2k1=1.{2k - 1 = -1.} 1{-1} is neither odd nor even, because parity is a property that applies only to positive integers (0 is also neither odd nor even).

Knowing the parity of two integers allows us to quickly determine the parity of their sums:

lemma. Let a,bZ+.{a, b \in \uint^{+}.} Then:

  1. a,bZevena+bZeven{a, b \in \Z_{\text{even}} \nc a + b \in \Z_{\text{even}}}
  2. [(aZeven)(bZodd)]abZ_odd{[(a \in \Z_{\text{even}}) \land (b \in \Z_{\text{odd}})] \nc a - b \in \uint\_{\text{odd}}}
  3. [(aZodd)(bZeven)]abZ_odd{[(a \in \Z_{\text{odd}}) \land (b \in \Z_{\text{even}})] \nc a - b \in \uint\_{\text{odd}}}
  4. [(aZodd)(bZodd)]a+bZeven{[(a \in \Z_{\text{odd}}) \land (b \in \Z_{\text{odd}})] \nc a + b \in \Z_{\text{even}}}

That is, the sum of two even integers is an even integer; the sum of an odd integer and an even integer is an odd integer; and the sum of two odd integers is an even integer. From the rule of addition by 0, we also have the following propositions:

lemma. Let aZ+.{a \in \uint^{+}.} Then: aZevena+0Zeven{a \in \Z_{\text{even}} \nc a + 0 \in \Z_{\text{even}}} > aZodda+0Zodd{a \in \Z_{\text{odd}} \nc a + 0 \in \Z_{\text{odd}}}

Let's prove the first property. First, the hypothesis to be proved:

Let a,bZ+.{a, b \in \uint^{+}.} If aZeven{a \in \Z_{\text{even}}} and bZeven,{b \in \Z_{\text{even},}} then a+bZeven.{a + b \in \Z_{\text{even}}.}

Because a{a} and b{b} are even integers, then by definition, a{a} and b{b} can be written in the forms:

a=2n, b=2m a = 2n, ~ b = 2m

where n,mZ+{n, m \in \uint^{+}}. Thus, by substitution, the expression a+b{a + b} can be written as:

a+b=2n+2m a + b = 2n + 2m

By distributivity:

a+b=2(n+m) a + b = 2(n + m)

The sum of two positive integers is a positive integer. Accordingly, the sum of n{n} and m{m} is some positive integer k.{k.} As such, the expression 2(n+m){2(n + m)} can be written as:

a+b=2k a + b = 2k

Because a+b{a + b} is an expression of the form 2k,{2k,} then by definition, a+b{a + b} is even.

Now let's prove the second. Again, we'll use a direct proof. First, the hypothesis to proved:

Let a,bZ+.{a,b \in \uint^{+}.} If a{a} is even and b{b} is odd, then a+b{a + b} is odd.

Suppose a{a} is even and b{b} is odd. Then by definition, a{a} and b{b} are positive integers of the form:

a=2n,b=2m+1 a = 2n, b = 2m + 1

where n{n} is a positive integer, and m{m} is a natural number. It follows that:

a+b=2n+2m+1 a + b = 2n + 2m + 1

Then, by commutativity:

a+b=(2n+2m)+1 a + b = (2n + 2m) + 1

And by associativity:

a+b=2(n+m)+1 a + b = 2(n + m) + 1

Given that n{n} is a positive integer and m{m} is a non-negative integer (since m{m} is a natural number), n+m{n + m} is a natural number. Accordingly:

a+b=2k+1 a + b = 2k + 1

where k=n+m,{k = n + m,} which is some natural number. Hence, a+b{a + b} must be odd, by definition.

Parity provides yet another proposition:

Lemma. Let aZ+.{a \in \uint^{+}.} Then

  1. aZevena2Zeven{a \in \Z_{\text{even}} \nc a^2 \in \Z_{\text{even}}}
  2. aZodda2Zodd{a \in \Z_{\text{odd}} \nc a^2 \in \Z_{\text{odd}}}

The proposition above tells us that the square of an even integer is even, and the square of an odd integer is odd. Let's prove it.

The method here is by direct proof. There are two hypotheses to be proved:

  1. If a{a} is even, then a2{a^2} is even.
  2. If a{a} is odd, then a2{a^2} is odd.

We prove each hypothesis in turn, beginning with the first.

Suppose aZeven.{a \in \Z_{\text{even}}.} Since a{a} is an even integer, then by definition, a{a} is a positive integer of the form 2n,{2n,} where n{n} is a positive integer. By substitution:

a2=(2n)2 a^2 = (2n)^2

Applying the definition of an exponent:

a2=(2n)(2n) a^2 = (2n)(2n)

By the associative property of multiplication, we get:

a2=(2)(2nn) a^2 = (2)(2 \cdot n \cdot n)

And by the definition of an exponent:

a2=(2)(2n2) a^2 = (2)(2n^2)

Because the product of positive integer is a positive integer, the intger 2n2{2n^2} is a positive integer. Hence:

a2=2m a^2 = 2m

where m{m} is a positive integer. Because 2m{2m} is an even integer by definition, a2{a^2} is even.

Next, we consider the second hypothesis. Suppose aZodd.{a \in \Z_{\text{odd}}.} Given that a{a} is an odd integer, then by definition, a{a} is a positive integer of the form 2n+1,{2n + 1,} where n{n} is a positive integer. By substitution:

a2=(2n+1)2 a^2 = (2n + 1)^2

By expansion:

a2=4n2+4n+1 a^2 = 4n^2 + 4n + 1

By associativity:

a2=(4n2+4n)+1 a^2 = (4n^2 + 4n) + 1

By distributivity:

a2=2(2n2+2n)+1 a^2 = 2(2n^2 + 2n) + 1

Because n{n} is a positive integer, 2n2+2n2n^2 + 2n is a positive integer. Hence:

a2=2k+1 a^2 = 2k + 1

where k{k} is a positive integer. Because 2k+1{2k + 1} is an odd integer by definition, a2{a^2} is odd. The proposition we've just proven yields a corollary:

Corollary. Let aZ+.{a \in \uint^{+}.} Then:

  1. a2ZevenaZeven{a^2 \in \Z_{\text{even}} \nc a \in \Z_{\text{even}}}
  2. a2ZoddaZodd{a^2 \in \Z_{\text{odd}} \nc a \in \Z_{\text{odd}}}

As with many corollaries, the propositions above are implied by the previous theorem. If a2{a^2} is even, then a{a} must be even, for if a{a} is odd, then a2{a^2} is odd. Similarly, if a2{a^2} is odd, then a{a} must be odd, for if a{a} is even, then a2{a^2} is even. This is a straightforward example of logical contradiction.

Divisibility

It is from parity that we generalize the concept of divisibility. Suppose a{a} is a positive integer and b{b} is an integer. When we say that b{b} is divisible by a,{a,} we mean that a{a} divides b.{b.} We denote this proposition with the notation a  b.{a \space\vert\space b.} An integer b{b} is divisible by a{a} if, and only if, b=an,{b = an,} where n{n} is some integer.

Definition. Let aZ+{a \in \uint^{+}} and bZ.{b \in \uint.} Then:

a  bb=ana \space\vert\space b \iff b = an

where nZ.{n \in \uint.}

The definition above yields several useful propositions:

Proposition. Every integer is divisible by 1. Every positive integer is divisible by itself.

The first proposition is implied by the fact that every integer can be written as n=(1)(n).{n = (1)(n).} This in turn implies the second proposition.

Composites & Primes

Compounds and primes along the number line

From the definition above, there are two subsets within the natural numbers. Any natural number that can be grouped into a solid rectangle is called a composite number (also called compound numbers). More formally:

definition. A composite number k{k} is a natural number greater than 1, divisible by some positive integer n,{n,} where n1.{n \neq 1.}

The negation of this definition yields the prime numbers, numbers that cannot be grouped into a solid rectangle.

definition. A prime number m{m} is a natural number greater than 1, divisible only by 1 and itself.

Note that from the definitions above, we see that the numbers 0 and 1 have a special property: Both 0 and 1 are neither prime nor compound.

Division & Rational Numbers

The operation of division is closely tied to the rational numbers. A rational number is any number that can be written in the form:

mnorm/n\dfrac{m}{n} \text{or} m/n

where m{m} and n{n} are integers, and n0.{n \neq 0.} The last condition is crucial, because in this particular world (classical algebra), we cannot divide by 0. The rational numbers come in various forms: ,13,12,23,1.4,0.08,12.0,0.0,{\ldots, \dfrac{1}{3}, - \dfrac{1}{2}, \dfrac{2}{3}, 1.4, 0.08, 12.0, 0.0, \ldots}

The rule is, as long as a number can be written as m/n,{m/n,} where m,n{m, n} are integers and n0,{n \neq 0,} then the number is rational number.7

Otherwise, the number is an irrational number. Note what this definition means. Every integer satisfies the definition rational number definition, so we say that the rational numbers are a proper superset of the integers. In other words, every integer is a rational number, but not every rational number is an integer.

The Division Algorithm

Question: A six-digit number having 1 as its leftmost digit becomes three times bigger if we take this digit off and put it at the end of the number. What is this number?

Let the number be 1ABCDE1ABCDE. Then, ABCDE1=3×1ABCDEABCDE1 = 3 \times 1ABCDE. Rewrite this with the multiplication algorithm:

ABCDE×3ABCDE1\begin{array}{cc} & A & B & C & D & E \\ \times & & & & & 3 \\ \hline A & B & C & D & E & 1 \end{array}

Let's focus on EE. We need a digit EE where 3×E=n13 \times E = n1, where nn is a digit. This digit must be 7, for 7×3=217 \times 3 = 21. We know that EE is 7, so what might DD be? Well, we need a digit DD where (D×3)+2=7(D \times 3) + 2 = 7. Or, rearranging, D×3=5D \times 3 = 5. The only digit that would satisfy this condition is D=5D = 5 (since 3×5+2=173 \times 5 + 2 = 17.

We know that DD is 5, so now we look at CC. We need a digit CC such that (C×3)+1=5(C \times 3) + 1 = 5, or, rearranging, C×3=4C \times 3 = 4. There is only one digit that satisfies this condition, 8. Thus, C=8C = 8.

Now we look at BB. What digit satisfies the condition B×3=6B \times 3 = 6? It must be 2. Therefore, B=2B = 2.

Finally, we look at AA. What digit results in A×3=12A \times 3 = 12. It must be 4, since 3×4=123 \times 4 = 12. Hence, the digits are A=4A = 4, B=2B = 2, C=8C = 8, and D=5D = 5.

Question: Divide 123123123 by 123. Again, we use the division algorithm:

1001001123)123123123123123123012312312312301231230\begin{array}{r} 1001001 \\ 123{\overline{\smash{\big)}\,123123123}} \\ - 123 \phantom{123123} \\ \hline 0 123 \phantom{123} \\ - 123 \phantom{123} \\ \hline 0 123 \\ - 123 \\ \hline 0 \end{array}

We can confirm the result with the multiplication algorithm:

1001001×10011230030030030020020021+00100100112123123123\begin{array}{cc} \begin{align*} 1001001 \\ \times \phantom{1001} 123 \\ \hline \phantom{00} 3 003 003 \\ \phantom{00} 2 002 002 \phantom{1} \\ + \phantom{00} 1 001 001 \phantom{12} \\ \hline 123123123 \end{align*} \end{array}

Question: Can one predict the remainder when 1111{111 \ldots 1} (100 ones) is divided by 1111111{11 11 11 1}?

111111111 11 11 1 is 7 ones. Consider a simpler case:

1011)1111100010000001\begin{array}{r} 10 \\ 11 {\overline{\smash{\big)}\, 111}} \\ -11 \phantom{0} \\ \hline \phantom{0} 01 \\ - \phantom{00} 0 \\ \hline \phantom{000} 1 \end{array}

Question: Divide 100001000 \ldots 0 (20 zeros) by 7.

Consider a simpler case:

147)1007030284\begin{array}{r} 14 \\ 7 {\overline{\smash{\big)}\, 100}} \\ - 7 \phantom{0} \\ \hline 30 \\ - 28 \\ \hline 4 \end{array}

Let's keep adding 0s to see if there's a pattern:

1428577)100000070000030000028000020000140006000560040035050491\begin{array}{r} 142857 \\ 7 {\overline{\smash{\big)}\, 1 000 000}} \\ - 7 \phantom{000 00} \\ \hline 30 \phantom{000 0} \\ -28 \phantom{000 0} \\ \hline 20 \phantom{000} \\ - 14 \phantom{000} \\ \hline 60 \phantom{00} \\ - 56 \phantom{00} \\ \hline 40 \phantom{0} \\ - 35 \phantom{0} \\ \hline 50 \\ - 49 \\ \hline 1 \end{array}

We return to a remainder of 1. This tells us that if we add another 0, we're back at 10, which gives us 30, then 20, then 60, then 40, and so on.

This also tells us that after every six 0s, we return to 10. Thus, for twenty 0s, we make 3 "laps" of returning to 10, with two more "runs":

  • Lap 1 (six 0s): 142857 remainder 1
  • Lap 2 (twelve 0s): 142857 14285 remainder 7
  • Lap 3 (eighteen 0s): 142857 142857 remainder 1
  • Lap 4 (nineteen 0s): 142857 142857 1 remainder 4
  • Lap 5 (twenty 0s): 142857 142857 14 remainder 2

Hence, the quotient is: 14,285,714,285,714 + 2.

Absolute Value

Given a real number x,{x,} be a real number, the absolute value of xx is the distance between xx and the origin. We denote the absolute value of xx with the notation: x\lvert x \rvert For example, the numbers -8 and 8 are both 8 units from the origin. Thus, 8=8\lvert 8 \rvert = 8 and 8=8\lvert -8 \rvert = 8. We define the absolute value as follows:

absolute value. Given xR,{x \in \reals,}

x={x if x0x if x<0 \lvert x \rvert = \begin{cases} x &\text{ if } x \geq 0 \\ -x &\text{ if } x < 0 \end{cases}

For example, 8=(8)=8\lvert -8 \rvert = -(-8) = 8. From the definition of absolute value, we can make the following inferences:

xR{\forall x \in \reals}
x0\lvert x \rvert \geq 0
xxx \leq \lvert x \rvert and xx-x \leq \lvert x \rvert
x2=x2{\lvert x \rvert}^2 = x^2
x2=x\sqrt{x^2} = \lvert x \rvert
ab=ab\lvert ab \rvert = \lvert a \rvert \cdot \lvert b \rvert
ab=ab\lvert \dfrac{a}{b} \rvert = \dfrac{\lvert a \rvert }{\lvert b \rvert }
a+ba+b\lvert a + b \rvert \leq \lvert a \rvert + \lvert b \rvert

Removing Absolute Value Symbols

When we encounter expressions containing absolute values, it's often a good idea to rewrite the expression without the absolute value symbols. For example, consider the expression 2x2\lvert -2 - x^2 \rvert. Using our algebraic definition above, we can rewrite this expression:

2x2=(2x2)=2+x2\begin{aligned} \lvert -2 - x^2 \rvert &= -(-2 - x^2) \\ &= 2 + x^2 \end{aligned}

The example above reveals useful insight. Suppose that real numbers are points along the real number line. Let aa and bb be real numbers. Because aa and bb are points along a line, the distance between aa and bb is aba - b. And since the absolute value of a number is only concerned with the distance between a point and the origin, we can infer:

For real numbers aa and bb, the distance between aa and bb is, ab=ba.{\lvert a - b \rvert = \lvert b - a \rvert.}

Absolute Values and Inequalities

Whenever we see absolute values in inequalities, we should be on high alert. Suppose that aR,{a \in \mathbb{R},} uR,{u \in \mathbb{R},} and a>0.{a > 0.} The following rules hold:

u<aa<u<a{\lvert u \rvert < a \iff -a < u < a} u<a(u<a or u>a){\lvert u \rvert < a \iff (u < -a \text{ or } u > a) }

These rules follow from observing the number line. The statement a<u<a{-a < u < a} implies that uu is somewhere between a-a and aa, which is equivalent to saying that the distance from uu to 0 is less than aa (which, in mathematical notation, is simply u<a{\lvert u \rvert < a }).

When inequalities contain expressions couched between absolute value symbols, the inequality is effectively expressing the expression's distance from 0. For example, suppse x3<1.{\lvert x - 3 \rvert < 1. } This statement tells us that the distance of x3{\lvert x - 3 \rvert } from 0 is less than 1. Because absolute values tell us the distance from 0, the statement x3<1,{\lvert x - 3 \rvert < 1,} allows us to infer:

1<x3<1 -1 < x - 3 < 1

Logarithms

Consider the number 16. How many times can we divide 16 by 2? Let's try it:

16/2=88/2=44/2=22/2=1\begin{aligned} 16 / 2 &= 8 \\ 8 / 2 &= 4 \\ 4 / 2 &= 2 \\ 2 / 2 &= 1 \end{aligned}

Above, we can see that we can divide the number 16 by 2 a total of 4 times. This corresponds to:

2222=24=16\begin{aligned} 2 \cdot 2 \cdot 2 \cdot 2 &= 2^4 \\ &= 16 \end{aligned}

The answer to "How many times can we divide a number n{n} by this constant x?{x?}" is the logarithm. More formally:

definition. Given b=ax,{b = a^x,} a1,{a \neq 1,} a>0,{a > 0,} and b>0,{b > 0,} > ax=bx=logab{a^x = b \iff x = \log_{a}b}

We say that x{x} is the logarithm in base a{a} of b.{b.} The logarithm provides a way of deducing, and rewriting, the exponent of a number. For example:

  • 23=8log28=3{2^3 = 8 \iff \log_{2}8 = 3}
  • 52=25log525=2{5^2 = 25 \iff \log_{5}25 = 2}
  • 103=1000log101000=3{10^3 = 1000 \iff \log_{10}1000 = 3}

Suppose that y=ax.{y = a^x.} From the definition above, it follows that a=logay.{a = \log_{a}y.} Then, it follows then that x=logaax{x = \log_{a}a^x} by substitution. We state this as a corollary:

Corollary. x=logaax{x = \log_{a}a^x}

If x=ay,{x = a^y,} then again from the definition, y=logax.{y = \log_{a}x.} It follows that x=alogax,{x = a^{\log_{a}x},} provided that x>0.{x > 0.} We again state this as a corollary:

Corollary. x=alogax,{x = a^{\log_{a}x},} where x>0{x > 0}

The definition and corollaries above directly lead to the laws of logarithms:

Log Laws. Where A,BN,{A, B \in \nat,} c>0,{c > 0,} and c1,{c \neq 1,}

logcA+logcB=logc(AB)logcAlogcB=logc(AB)nlogcA=logc(An)\begin{aligned} \log_{c}A + \log_{c}B &= \log_{c}(AB) \\[1em] \log_{c}A - \log_{c}B &= \log_{c}\left(\dfrac{A}{B}\right) \\[1em] n \log_{c}A &= \log_{c}(A^n) \end{aligned}

The Binomial Theorem

Recall that a binomial is a polynomial consisting of the sum of two monomials. In other words, a polynomial with two terms. For example, the expression:

axmbxn ax^m - bx^n

is a binomial, where a{a} and b{b} are numbers, m{m} and n{n} are distinct, nonnegative integers, and x{x} is a variable (or more formally, the indeterminate). Taking higher powers of binomials and expanding them is a process that quickly grows tedious. To see how, let's consider a few examples.

First, let's expand (a+b)0.{(a + b)^0.} This is straightforward; it's just 1:{1:}

(a+b)0=1 (a + b)^0 = 1

Similarly, (a+b)1{(a + b)^1} isn't so bad either:

(a+b)1=a+b (a + b)^1 = a + b

(a+b)2(a + b)^2 isn't terrible if we know our expansions by heart:

(a+b)2=a2+2ab+b2 (a + b)^2 = a^2 + 2ab + b^2

How about (a+b)3{(a + b)^3}? Well, we can carry it out:

(a+b)3=(a+b)(a+b)(a+b)=(a+b)(a2+2ab+b2)=a3+2a2b+ab2+a2b+2ab2+b3=a3+3a2b+3ab2+b3 \begin{aligned} (a + b)^3 &= (a + b)(a + b)(a + b) \\ &= (a + b)(a^2 + 2ab + b^2) \\ &= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3 \\ &= a^3 + 3a^2b + 3ab^2 + b^3 \end{aligned}

And (a+b)4:{(a + b)^4:}

(a+b)4=(a+b)(a3+3a2b+3ab2+b3)=a(a3+3a2b+3ab2+b3)+b(a3+3a2b+3ab2+b3)=a4+3a3b+3a2b2+ab3+a3b+3a2b2+3ab3+b4=a4+4a3b+6a2b2+4ab3+b4\begin{aligned} (a + b)^4 &= (a + b)(a^3 + 3a^2b + 3ab^2 + b^3) \\ &= a(a^3 + 3a^2b + 3ab^2 + b^3) + b(a^3 + 3a^2b + 3ab^2 + b^3)\\ &= a^4 + 3a^3b + 3a^2b^2 + ab^3 + a^3b + 3a^2b^2 + 3ab^3 + b^4 \\ &= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \end{aligned}

Notice how these expressions are getting more and more tedious. Laying out the expansions in a table:

expressionexpansionterm count
(a+b)0{(a + b)^0}11
(a+b)1{(a + b)^1}a+b{a + b}2
(a+b)2{(a + b)^2}a2+2ab+b2{a^2 + 2ab + b^2}3
(a+b)3{(a + b)^3}a3+3a2b+3ab2+b3{a^3 + 3a^2b + 3ab^2 + b^3}4
(a+b)4{(a + b)^4}a4+4a3b+6a2b2+4ab3+b4{a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4}5
(a+b)5{(a + b)^5}a5+5a4b+10a3b2+10a2b3+5ab4+b5{a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5}6

Needless to say, expanding binomials with higher powers by hand is tedious and prone to error. Attempting to expand something like (a+b)100{(a + b)^{100}} would take a very, very long time. And then after we're done, we must still go through the process of checking if our algebraic manipulation is correct.

Fortunately, there's a theorem to make our lives easier. But, before we examine this theorem, let's segue briefly into the world of counting.

Solving Quadratic Equations

There are four primary ways of finding the truth set of quadratic equations:

  1. using the square root method;
  2. factoring;
  3. completing the square; and
  4. using the quadratic equation

The next few sections explore how each of these methods are used.

Completing the Square

Given the quadratic equation x2+Ax=B{x^2 + Ax = B} ompleting the square can be summarized into a simple algorithm:

  1. Look at the cofficient of x.{x.}
  2. AA Divide the coefficient by 2: A2.{\dfrac{A}{2}.}
  3. Square the result: (A2)2{\left(\dfrac{A}{2}\right)^2}
  4. Add the number from step 1 to both sides of the equation:
    x2Ax+(A2)2=B+(A2)2 x^2 - Ax + \left(\dfrac{A}{2}\right)^2 = B + \left(\dfrac{A}{2}\right)^2
  5. Simplify the equation.
  6. Factor the result Solve for x.x.

For example, suppose we were confronted by the quadratic equation:

x22x=4. x^2 - 2x = 4.

Applying the method of completing the square, we look at the coefficient of xx first. We see that it's -2. We divide this number by 2 and square it:

((2)2)2=1. \left(\dfrac{(-2)}{2}\right)^2 = 1.

Then we add this number to both sides of the equation:

x22x+1=4+1. x^2 - 2x + 1 = 4 + 1.

Simplifying, we obtain:

x22x+1=5 x^2 - 2x + 1 = 5

We factor the left-hand side, and reach

(x1)(x1)=5. (x-1)(x-1) = 5.

Or, with a square,

(x1)2=5. (x-1)^2 = 5.

Taking the square root of both sides, we have

x1=±5. x - 1 = \pm \sqrt{5}.

(always be mindful of the the fact that squares result from both positive and negative numbers). Finally, we solve for x{x} by adding 1 to both sides:

x=1±5.x = 1 \pm \sqrt{5}.

Thus, the truth set of the equation x22x=4{x^2 - 2x = 4} is x{1+5,15}.{x \in \{ 1 + \sqrt{5}, 1 - \sqrt{5} \}.}

Why Does Completing the Square Work?

The method of completing the square follows from geometry. Suppose we had a square, ABCD{\square ABCD} with sides of length x{x}:

A square of arbitrary side length.

We know that the area of ABCD{\square ABCD} is x2.{x^2.} Now suppose that we were to add a rectangle with a width of bb and a length of xx to the square. This rectangle has an area of bx{bx}. When we add the rectangle, we obtain a total area of x2+bx:{x^2 + bx:}

Adding a rectangle of arbitrary width and length.

Now suppose we rearranged the resulting shape by (1) splitting the added rectangle vertically, (2) leaving one half at the original square's side, and (3) moving the other half to the square's base:

Rearranging the squares.

Notice that a gap is left. What might be the dimensions of the shape that would fill that gap? Well, we know that the total area of the rectange we created earlier is x2+bx{x^2 + bx}. We also know that the two halves we created must respectively have an area of x(b2).{x\left(\dfrac{b}{2}\right).} Thus, the shape we need to fill the gap must have an area of (b2)2{\left(\dfrac{b}{2}\right)^2}, or, simplifying the expression, b24.{\dfrac{b^2}{4}.}

To see the connection to the algebraic method, consider the shape above without filling the gap. That shape has an area of x2+(b2)x+(b2)x=x2+bx.{x^2 + \left(\dfrac{b}{2}\right)x + \left(\dfrac{b}{2}\right)x = x^2 + bx.} When we complete the square, we add the area of the "missing square": b24.{\dfrac{b^2}{4}.} Look familiar? Algebraically, the area of the incomplete shape is a quadratic equation.

If we had an equation of the form ax2+bx+c{ax^2 + bx + c}, where aa, bb, and cc are constants and a1,{a \neq 1,} that tells us that our original square has an area of ax2{ax^2}. That aa makes computations needlessly difficult, and we want our original shape to be a square of area x2{x^2}. Since aa is a constant, we can get rid of its place before xx by dividing all of the terms in the equation by a:a:

x2+(ba)x+ca=0. x^2 + \left(\dfrac{b}{a}\right)x + \dfrac{c}{a} = 0.

We can then move the term ca{\dfrac{c}{a}} to the right side of the equation, to give us our incomplete shape:

x2+(ba)x=ca x^2 + \left(\dfrac{b}{a}\right)x = - \dfrac{c}{a}

The term b/a{b/a} represents the side of our added rectangle. The side of our halved rectangles then, is b/a2=b2a{\frac{b/a}{2} = \frac{b}{2a}}. And since b2a{\frac{b}{2a}} represents the side of our halved rectangles, the missing square has an area of (b2a)(b2a)=b24a,{\left(\frac{b}{2a}\right)\left(\frac{b}{2a}\right) = \frac{b^2}{4a},} the term we add to both sides of x2+(ba)x=ca.{x^2 + \left(\frac{b}{a}\right)x = - \frac{c}{a}.}

The Quadratic Formula

The truth set for quadratic equations cannot be found by simple factorization. Instead, a more robust formula must be used: the quadratic formula.

From the method of completing the square, we can derive the quadratic formula, a formula that allows us to quickly deduce the truth set of a quadratic equation. To derive the formula, we begin wtih the general quadratic equation: ax2+bx+c=0,{ax^2 + bx + c = 0,} where a0.{a \neq 0.} We divide both sides by aa, and obtain:

x2+(ba)x+(ca)=0. x^2 + \left( \dfrac{b}{a} \right)x + \left( \dfrac{c}{a} \right) = 0.

Then we subtract ca{\dfrac{c}{a}} from both sides, and obtain:

x2+(ba)x=(ca) x^2 + \left( \dfrac{b}{a} \right)x = - \left( \dfrac{c}{a} \right)

Now we comlete the square, by adding b24a2{\dfrac{b^2}{4a^2}} to both sides:

x2+(ba)x+b24a2=b24a2ca(x+b2a)2=b24ac4a2x+b2a=±b24ac4a2=±b24ac2a=±b2rac2a\begin{aligned} x^2 + \left( \dfrac{b}{a} \right)x + \dfrac{b^2}{4a^2} &= \dfrac{b^2}{4a^2} - \dfrac{c}{a} \\ \left( x + \dfrac{b}{2a} \right)^2 &= \dfrac{b^2 - 4ac}{4a^2} \\ x + \dfrac{b}{2a} &= \pm \sqrt{\dfrac{b^2-4ac}{4a^2}}\\ &= \pm \dfrac{\sqrt{b^2 - 4ac}}{2 \lvert a \rvert } \\ &= \pm \dfrac{\sqrt{b^2 - rac}}{2a} \end{aligned}

Thus, the truth set of the quadratic equation a2+bx+c=0{a^2 + bx + c = 0} is:

x{b2a+b24ac2a, b2ab24ac2a} x \in \left\{-\dfrac{b}{2a} + \dfrac{\sqrt{b^2 - 4ac}}{2a}, \space - \dfrac{b}{2a} - \dfrac{\sqrt{b^2 - 4ac}}{2a} \right\}

The quadratic formula will always yield the truth set for quadratic equations of the form ax2+bx+c=0.{ax^2 + bx + c = 0.} In the event that the term b24acR{b^2 - 4ac \in \mathbb{R}^-} (the discriminant), however, the truth set of the equation consists of complex numbers, rather than real numbers.

Thus, given a quadratic equation of the form ax2+bx+c{ax^2 + bx +c}, where aa, bb, and cc are constants and a0{a \neq 0}, the truth set of the equation can be found with the equation:

x2+(ba)x+b24a=ca+b24a x^2 + \left(\dfrac{b}{a}\right)x + \dfrac{b^2}{4a} = - \dfrac{c}{a} + \dfrac{b^2}{4a}

Quadratic Formula. Given the quadratic equation ax2+bx+c=0{ ax^2 + bx + c = 0 }, where a0,{ a \neq 0, } the solutions are given by:

x=b±b24ac2a x = \dfrac{-b \pm \sqrt{b^2 - 4ac} }{2a}

The Discriminant

In the quadratic formula, the expression that appears below the radical sign, b24ac,{b^2 - 4ac,} is called the discriminant. If the discriminant evaluates to a positive real number, then the quadratic equation has 2 real solutions. If the discriminant evaluates to 0, then the quadratic equation has only 1 real solution. If the discriminant evaluates to a negative real number, then the quadratic equation has no real solutions — instead, it has a pair of complex conjugate solutions. Generalizing these results:

roots{1 distinct real rootiff b24ac>02 distinct real rootsiff b24ac=00 real rootsiff b24ac<0 \text{roots} \begin{cases} \text{1 distinct real root} &\text{iff } b^2 - 4ac > 0 \\ \text{2 distinct real roots} &\text{iff } b^2 - 4ac = 0 \\ \text{0 real roots} &\text{iff } b^2 - 4ac < 0 \end{cases}

The Product and Sum of Polynomial Roots

Suppose we were faced with the equation x2=53x.{x^2 = 5 - 3x.} What are the roots of this equation? First, let's rewrite the equation:

x2+3x5=0. x^2 + 3x - 5 = 0.

Then let's use the quadratic formula:

x=b±b24ac2a=3±324(1)(5)2(1)=3±9+202=3±292\begin{aligned} x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\ &=\dfrac{-3 \pm \sqrt{3^2-4(1)(-5)}}{2(1)} \\ &=\dfrac{-3 \pm \sqrt{9+20}}{2} \\ &=\dfrac{-3 \pm \sqrt{29}}{2} \end{aligned}

Thus, the truth set of x2=53x{x^2 = 5 - 3x} is:

x{3+292, 3292} x \in \left\{ \dfrac{-3 + \sqrt{29}}{2}, \space \dfrac{-3 - \sqrt{29}}{2}\right\}

What would happen if we multiplied the two roots?

(3292)(3+292)=(3)2(29)24=9294=204=5\begin{aligned} \left( \dfrac{-3 - \sqrt{29}}{2} \right) \left( \dfrac{-3 + \sqrt{29}}{2} \right) &= \dfrac{(-3)^2 - (\sqrt{29})^2}{4} \\ &= \dfrac{9 - 29}{4}\\ &= \dfrac{-20}{4} \\ &= -5 \end{aligned}

The product of the two roots is -5. What would happen if we added the two roots?

3+292+3292=329+(3)292=62=3\begin{aligned} \dfrac{-3 + \sqrt{29}}{2} + \dfrac{-3 - \sqrt{29}}{2} &= \dfrac{-3 - \sqrt{29} + (-3) - \sqrt{29}}{2} \\ &= \dfrac{-6}{2}\\ &= -3 \end{aligned}

The sum of the two roots is -3. Recall that the quadratic equation was x2+3x5=0.{x^2 + 3x - 5 = 0.} Is it pure coincidence that product of the two roots is -5, the cc term in the equation, and the sum of the two roots, -3, is the negative of the the bb term? Not a coincidence at all. In fact, it follows from a more general rule.

lemma. Let r1r_1 and r2r_2 be the roots of the quadratic equation x2+bx+c=0.{x^2 + bx + c = 0.} It follows that r1r2=c{r_1 r_2 = c} and r1+r2=b.{r_1 + r_2 = -b.} In other words, the product of the roots of x2+bx+c=0{x^2 + bx + c = 0} is the constant term in the equation, cc, and the sum of the roots is the negative of the coefficient of the xx term, bb (i.e., b-b).

The general rule follows from manipulating the quadratic formula. Suppose there exists a quadratic equation ax2+bx+c=0,{ax^2 + bx + c = 0,} where a=1.{a = 1.} From the quadratic formula, it follows that the roots of the equation are:

x=b±b24(1)c2(1)=b±b24c2.\begin{aligned} x &= \dfrac{-b \pm \sqrt{b^2 - 4(1)c}}{2(1)} \\[1em] &= \dfrac{-b \pm \sqrt{b^2 - 4c}}{2}. \end{aligned}

Thus, the roots are:

r1=b+b24c2r2=bb24c2 r_1 = \dfrac{-b + \sqrt{b^2 - 4c}}{2} \\[1em] r_2 = \dfrac{-b - \sqrt{b^2 - 4c}}{2}

The products of the roots:

r1r2=(b+b24c2)(bb24c2)r1r2=b2(b24c)4=4c4=c r_1 r_2 = \left(\dfrac{-b + \sqrt{b^2 - 4c}}{2}\right) \left( \dfrac{-b - \sqrt{b^2 - 4c}}{2} \right) \\[1em] \phantom{r_1 r_2} = \dfrac{b^2 - \left( b^2 - 4c \right)}{4} = \dfrac{4c}{4} = c

And the sum of the roots:

r1+r2=b+b24c2+bb24c2r1+r2=b+b24c+(b)b24c2=2b2=b r_1 + r_2 = \dfrac{-b + \sqrt{b^2 - 4c}}{2} + \dfrac{-b - \sqrt{b^2 - 4c}}{2} \\[1em] \phantom{r_1 + r_2} = \dfrac{-b + \sqrt{b^2 - 4c} + (-b) - \sqrt{b^2 - 4c}}{2} = \dfrac{-2b}{2} = -b

The Factorial

A common operation in counting is the factorial. It is defined as follows:

Definition: Factorial. Let n{n} be a positive integer. Then n!,{n!,} read n{n} factorial, is denoted:

n!=n(n1)(n2)321 n! = n \cdot (n - 1) \cdot (n-2) \cdot \ldots \cdot 3 \cdot 2 \cdot 1

Alongside the definition above is the following axiom:

Axiom: Zero Factorial. The factorial of zero is one: 0!=10! = 1

To illustrate, 6!{6!} is computed as:

6!=(6)(61)(62)(63)(64)(65)=(6)(5)(4)(3)(2)(1)=(6)(5)(2)(4)(3)=(6)(10)(12)=(6)(120)=720\begin{aligned} 6! &= (6)(6-1)(6-2)(6-3)(6-4)(6-5)\\[1em] &= (6)(5)(4)(3)(2)(1) \\[1em] &= (6)(5)(2)(4)(3) \\[1em] &= (6)(10)(12) \\[1em] &= (6)(120) \\[1em] &= 720 \end{aligned}

Binomial Coefficients

Suppose we had the following balls:

Our friend Leibniz comes up to us and says, Pick 1.{1.} How many possible choices do we have? Well, we would have 4{4} possible choices: A, B, C, and D. Now what if Leibniz said, Pick 2.{2.} How many choices do we have here? Well, we can count the number of possible choices: AB, AC, AD, BC, BD, CD. These are six possible choices. In fact, we can layout these possibilities:

Choose Possibilities Number of Possibilities

In mathematics, we have special syntax for expressing our thought process above. When Leibniz says, Pick 1, we express this as:

(41) \dbinom{4}{1}

This notation is read "Four choose one." There are four balls, and we are to pick one. When Leibniz says, Pick 2, we write this as:

(42) \dbinom{4}{2}

"Four choose two." And when Leibniz says, Pick 3, we have:

(43) \dbinom{4}{3}

"Four choose three." The notation (nk){\dbinom{n}{k}} is called a binomial coefficient, and it is read as n{n} choose k.{k.} Formally:

Definition: Binomial Coefficient. For nonnegative integers n{n} and k,{k,} where nk,{n \geq k,} the expression (nk),{\binom{n}{k},} also written as cCk,{_cC_k,} is called a binomial coefficient, and it is defined as:

(nk)=n!k!(nk)! \dbinom{n}{k} = \dfrac{n!}{k!(n - k)!}

Otherwise, where k>n,{k > n,} the binomial coefficient is defined as:

(nk)=0 \dbinom{n}{k} = 0

To understand where the binomial coefficient comes from, it's helpful to illustrate by example. Suppose we had the following ingredients for a cup of coffeee: coffee, milk, sugar. Now suppose we were asked, how many possible ways are there to make a cup of sweet coffee? Well, we can enumerate the number of ways with a table (where c{c} is for coffee, m{m} is for milk, and s{s} is for sugar):

pickpossible picksnumber of possibilities
01
1A, B, C, D4
2AB, AC, AD, BC, BD, CD6
3ABC, ABD, ACD, BCD4
4ABCD1

Reading the table above, we count six different orders. For example, (c,m,s){(c, m, s)} is coffee, then milk, then sugar, and (s,m,c){(s,m,c)} is sugar, then milk, then coffee. This result follows directly from the multiplication rule in counting. Since we have three different ingredients, the number of recipes is yielded by:

3×2×1=6 3 \times 2 \times 1 = 6

Generalizing the multiplication rule, given n{n} ingredients and k{k} choices, we have:

n×(n1)×(n2)×(nk+1)=n!(nk)! n \times (n - 1) \times (n - 2) \ldots \times (n - k + 1) = \dfrac{n!}{(n - k)!}

Importantly, the question of how many different recipes there are is distinct from asking: How many combinations of coffee, sugar, and milk are there? The former requires a distinct order, while the latter does not. The latter question — how many different combinations are there (disregarding order) — is what the binomial coefficient attempts to capture.

Hence, using our sweet coffee example, when write (32),{\dbinom{3}{2},} we are expressing the number of combinations of two ingredients among three, where order does not matter. If we enumerated all of the possibilities out on a table:

Counting the number of possibilities, we have 6.{6.} But because we're disregarding order, the table above is incorrect because we've overcounted. Some of the possibilities are duplicated. As such, we must eliminate the duplicates:

ingredientcombinationcombination
c{c}(c,m){(c, m)}(c,s){(c, s)}
m{m}(m,c){(m, c)}(m,s){(m, s)}
s{s}(s,c){(s, c)}(s,m){(s, m)}

Notice that eliminating the overcount leaves us with three possibilities. Further notice that this is half of our original count. This is no coincidence. Looking back at our general formula for the multiplication rule, we can see that eliminating the overcount is a matter of dividing by a k!{k!} term (since we've overcounted by a factor of k!{k!}):

n×(n1)×(n2)×(nk+1)=n!(nk)!n×(n1)×(n2)×(nk+1)k!=n!(nk)!k!=n!k!(nk)!\begin{aligned} n \times (n - 1) \times (n - 2) \ldots \times (n - k + 1) &= \dfrac{n!}{(n - k)!} \\[2em] \dfrac{n \times (n - 1) \times (n - 2) \ldots \times (n - k + 1)}{k!} &= \dfrac{\dfrac{n!}{(n - k)!}}{k!} \\[2em] &= \dfrac{n!}{k!(n - k)!} \end{aligned}

And there it is — the binomial coefficient definition.

Pascal's Triangle

Do we notice any patterns in this triangle? There are several. The Binomial Theorem This is where the binomial theorem comes in handy:

Binomial Theorem. For any positive integer n,{n,} the following is true:

(a+b)n=k=0n(nk)ankbk=(n0)an+(n1)an1b+(n2)an2b2++(nn)bn\begin{aligned} (a + b)^n &= \sum\limits_{k = 0}^{n} \dbinom{n}{k} a^{n - k}b^k \\[2em] &= \dbinom{n}{0}a^n + \dbinom{n}{1}a^{n - 1}b + \dbinom{n}{2}a^{n-2}b^2 + \ldots + \dbinom{n}{n}b^n \end{aligned}

Invervals

An interval II is a special kind of set. It is a subset of R{\mathbb{R}} with the property: If sI,{s \in I,} tI,{t \in I,} and s<x<t,{s < x < t,} then xI.{x \in I.} In other words, an interval is simply a set of real numbers with two explicit members, called endpoints, with the added property that every number between the two endpoints is also an element of the set. Because there are infinitely many real numbers between two different real numbers, intervals with distinct endpoints have infinitely many members.

Intervals can be restricted to include or exclude its endpoints. If the interval excludes its endpoints, we say that the interval is an open interval. If the interval includes its endpoints, we say that the interval is an open interval.

Because of the special place intervals have in continuous mathematics, they have two different forms of notation: interval notation, and set notation. The table below lists some common intervals in both interval and set notation.

Interval NotationSet Notation
[a,b]{[a, b]}{xR:axb}{\{x \in \mathbb{R} : a \leq x \leq b\}}
(a,b){(a,b)}{xR:a<x<b}{\{x \in \mathbb{R} : a < x < b\}}
[a,b){[a, b)}{xR:ax<b}{\{ x \in \mathbb{R} : a \leq x < b \}}
(a,b]{(a, b]}{xR:a<xb}{\{ x \in \mathbb{R} : a < x \leq b \}}
(a,+){(a, + \infty)}{xR:x>a}{\{ x \in \mathbb{R} : x > a \}}
[a,+){[a, + \infty)}{xR:xa}{\{ x \in \mathbb{R} : x \geq a \}}
(,b){(- \infty, b)}{xR:x<b}{\{ x \in \mathbb{R} : x < b \}}
(,b]{(- \infty, b]}{xR:xb}{\{ x \in \mathbb{R} : x \leq b \}}
(,+){(- \infty, + \infty)}R{\mathbb{R}}

To graphically indicate that an interval includes its endpoints, we color in the the dot marking the endpoint. To indicate exclusion, we leave the mark unfilled.

Suppose the interval (a,b)R2,{(a, b) \in \mathbb{R}^2,} and a<b.{a < b.} Now suppose there is a number λ[0,1].{\lambda \in [0,1].} Since a<b{a < b}, we can infer ab<0.{a - b < 0.} We know that λa+(1λ)b=b+λ(ab){\lambda{a} + (1 - \lambda){b} = b + \lambda{(a-b)}} (simplifying this expression will yield 0=0{0 = 0}.) Since ab<0,{a - b < 0,} we can solve for bb: b=b+0(ab)b+1(ab)=a.{b = b + 0(a - b) \leq b + 1(a - b) = a.}

Now suppose there is a number x[a,b],{x \in [a,b],} and that x=λa+(1λ)b.{x = \lambda{a} + (1 - \lambda){b}.} Solving for λ,{\lambda,} we obtain λ=xbba.{\lambda = \dfrac{x - b}{b - a}.} Given that 0xbba,{0 \leq x - b \leq b - a,} it is evident that 0λ1.{0 \leq \lambda \leq 1.}

Thus, we can infer the following rule:

lemma. Given the open interval (a,b)R2{(a,b) \in \mathbb{R^2}} and a<b,{a < b,} every number of the form λa+(1λ)b,{\lambda{a} + (1 - \lambda)b,} > λ[0,1]{\lambda \in [0,1]} belongs to the interval [a,b].{[a,b].} Conversely, if x[a,b],{x \in [a,b],} then we can find a λ[0,1]{\lambda \in [0,1]} such that x=λa+(1λ)b.{x = \lambda{a} + (1 - \lambda){b}.}

Neighborhoods

Suppose aR.{a \in \mathbb{R}.} The set Na\mathcal{N}_a is a neighborhood of aa if: There exists an open inteval II centered at aa such that INa.{I \subseteq \mathcal{N}_a.} I.e., Na\mathcal{N} _a is a neighborhood of aa if there exists a δ>0{\delta > 0} such that the open interval (aδ,a+δ)Na.{(a - \delta, a + \delta) \subseteq \mathcal{N}_a.}

If Na\mathcal{N}_a is a neighborhood of aa, then the notation Na{a}{\mathcal{N}_a \setminus \{ a \}} signifies the deleted neighborhood of aa. This implies that Na{\mathcal{N}_a} is a neighborhood of aa if aa has neighbors left and right.

For example, consider the following intervals:

The first interval demarcates the neighborhood of aa. The second interval demarcates the sinistral neighborhood of aa. Finally, the third interval demarcates the dextral neighborhood of aa. More formally:

lemma. Let aR.{a \in \mathbb{R}.} The set VR{V \subseteq \mathbb{R}} is a dextral neighborhood (right-hand) of aa if there exists a δ>0{\delta > 0} such that [a,a+δ)V.{[a, a + \delta) \subseteq V.} The set VR{V' \subseteq \mathbb{R}} is a sinistral neighborhood (left-hand) of aa if there exists a δ>0{\delta' > 0} such that (aδ,a]V.{(a - \delta', a] \subseteq V'.}

In sum, neighborhoods are open intervals centered at a particular real number a.a. They can be large or small. The intervals (1000,1000){(-1000, 1000)}, (0.00001,0.00001){(-0.00001, 0.00001)}, and (π,π){(-\pi, \pi)} are all neighborhoods of the real number 0. Moreover, we can restrict our discussion to certain sections of the neighborhood: the sinistral (left-hand) and dextral (right-hand) neighborhoods. For example, the interval (100,0]{(-100, 0]} is the sinistral neighborhood of 0, and the interval [0,100){[0, 100)} is the dextral neighborhood of 0.

A neighborhood is free to include or exclude its reference point. Thus, where aR,{a \in \mathbb{R},} the δ-neighborhood{\delta \text{-neighborhood} } of aa is defined as Uδ(a)={xR:xa<δ}.{U_{\delta}(a) = \{ x \in \mathbb{R} : \lvert x - a \rvert < \delta \}.} This particular set is simply called the neighborhood of aa, and the variable Uδ(a){U_{\delta}(a)} is equivalent to Na{\mathcal{N}_a}. If, however, we were to exclude the reference point aa, then we obtain what is called a reduced neighborhood—the set Pδ(a)={xR:0<xa<δ}.{P_{\delta}(a) = \{ x \in \mathbb{R} : 0 < \lvert x - a \rvert < \delta \}.} Thus, the difference between a neighborhood of aa and a reduced neighborhood of aa is that the reduced neighborhood of aa excludes the reference point aa, whereas the neighborhood of aa includes aa.

The Cartesian Plane

Continuous mathematics depends heavily on graphs. One way to represent graphs is through the Cartesian plane:

The Cartesian plane consists of two perpendicular real number lines. The point where the lines intersect is 0, and is called the origin. The horizontal number line is usually called the x-axis, and the vertical number line the yy-axis. On the xx-axis, all points to the right of the origin represent the positive real numbers, and all points to the left of the origin represent the negative real numbers. On the yy-axis, all points above the origin reprepresent the positive real numbers, and all points below the origin represent the negative real numbers.

A point on the Cartesian plane is represented by an ordered pair, of the form x,y{ x, y }, where x{ x } is a point on the xx-axis, and y{ y } is a point on the yy-axis. For example:

Each of the points in the diagram above are Cartesian coordinates, or simply coordinates. The xx-coordinate is formally called the abscissa of the point, and the yy-coordinate the ordinate. When we refer to coordinates generally, or in the abstract, we write:

P(x,y) P(x,y)

The notation above is read "the point (x,y){ (x,y) }."

The Distance Formula

The Cartesian plane constructs a bridge between algebra and geometry. Many theorems and ideas from geometry can be applied in continuous mathematics. One such theorem is the Pythagorean theorem:

theorem. Given a right trangle, the lengths of the sides are related by the equation: a2+b2=c2,{a^2 + b^2 = c^2,} where a{a} and b{b} are the lengths of the sides forming the right triangle, and c{c} is the length of the hypotentuse.

Conversely, if a,{a,} b,{b,} and c{c} are the sides of a triangle, and all three lengths are related by an equation of the form a2+b2=c2,{a^2 + b^2 = c^2,} then the triangle is a right triangle, and c{c} is the hypotenuse.

We can visualize the Pythagorean Theorem with the following diagram:

Visualizing the Pythagorean Theorem

The Pythagorean Theorem leads directly to the Distance Formula, an equation that allows us to compute the distance between two points on the Cartesian plane:

Visualizing the distance formula.

Formula. Given two points P1(x1,y1){ P_1(x_1, y_1) } and P2(x2,y2){ P_2(x_2, y_2) }, the distance dd between P1{ P_1 } and P2{ P_2 } is given by the formula:

d2=x2x12+y2y12d=(x2x1)2+(y2y1)2\begin{aligned} d^2 &= {\lvert x_2 - x_1 \rvert}^2 + { \lvert y_2 - y_1 \rvert }^2 \\ d &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \end{aligned}

Functions

Functions. What are they? At the most basic level, a function is a rule assigning a number to another number. At a more abstract level, a function is a relation between mathematical objects. There are many relations in mathematics (equality, inequality, congruence, proximity, etc.). Functions are just one kind of relation. Most commonly, functions are relations between numbers, where one number shares a specific relationship, defined by a rule, to exactly one number. More explicitly: A function is a rule that assigns one number to another number. Do not confuse this idea of "one-to-one" with "one-to-unique." A relation can still be a function even if one or more numbers are related to the same number.

Functions can be represented in a variety of ways. We can represent functions algebraically, using an equation or formula: f(x)=lnx,{f(x) = \ln x,} g=h{g = \sqrt{h}}, or F=ma{F = ma}. We can represent functions verbally: "For every xx, there is a yy such that yy is xx squared." We can represent functions with tables, where one column lists the values of xx, and another lists their corresponding yy values. Finally, we can represent functions with graphs, using a coordinate system. Regardless of the way we represent them, functions all express the same idea: a relationship where one mathematical object maps to exactly one mathematical object, according to a rule.

A more formal definition:

definition. Let AA and BB be nonempty sets. A function from AA to BB is a rule of correspondence that assigns, or maps, to each element in AA exactly one element in B.{B.}

Consider the following relations:

Various relations

The relation uu maps the elements of AA to the elements of B.{B.} The relation vv maps the elements of CC to the elements of D.{D.} And the relation ww maps the elements of EE to the elements of F.{F.} The relation uu is a function, since it maps each element of AA to exactly one element of B.B. The same goes for relation vv — each element of CC is mapped to exactly one element of D,{D,} even if some elements of CC map to the same element of D.{D.} The relation w,{w,} however, is not a function, since at least one element of EE is mapped to more than one element of F.{F.}

We often refer to the inputs of a function with the variable xx, and the outputs of the function with the variable yy, or f(x)f(x). The input, xx, is also called the independent variable, and the output, f(x)f(x), is called the dependent variable.

A function will never give more than one output for one input. There are, of course, relationships where on number outputs more than one number. Those relationships, however, are not characterized by, or as, functions. We will explore those relationships later in due time.

definition Given two sets AA and BB, the set Cartesian product A×BA \times B the set of all ordered pairs (a,b)(a,b), where aAa \in A and bBb \in B. A subset of A×BA \times B is called a relation.

A function is a special kind of relation. It is a pairing of elements in AA with elements in BB, where each eleemnt in AA is paired with exactly one element in BB. Thus, a function ff from AA to BB is a rule or relation between AA and BB that assigns, or maps, each element aAa \in A to one element in bBb \in B.

If a function is a subset of A×BA \times B, then AA is the domain of the function, and BB is the range of the function.

Every function has a natural domain (also called the implied domain). The natural domain of a function is the set of all elements for which the function is defined. Whenever we analyze or discuss a function, we set a natural domain. That natural domain acts as the boundary of our analysis. For example, when we discuss the function f(x)=xf(x) = \sqrt{x}, we may specify that the natural domain of ff is R\mathbb{R}. Because the natural domain of ff is R\mathbb{R}, the domain of ff cannot include negative real numbers, since x\sqrt{x} is undefined for R\mathbb{R}^{-}. Because of this fact, we specify the domain of ff as:

dom(f)={xRx0} dom(f) = \{ x \in \mathbb{R} \mid x \geq 0 \}

Properties of Functions

Functions can be identified by several properties:

  1. A set of inputs, called the function's domain, denoted dom(f),{dom(f),} e.g., the domain of the function f(x)=x2{f(x) = x^2} is dom(f)=R.{dom(f) = \mathbb{R}.}
  2. A set of all possible inputs, called the function's natural domain, denoted D(f),{D(f),} e.g., the natural domain of the function c(d)=d{c(d) = \sqrt{d}} is D(c)=R{ D(c) = \mathbb{R}}, but the domain of the function is dom(d)={dRd0}.{dom(d) = \{ d \in \mathbb{R} \mid d \geq 0 \}.} In set notation, given a function f,{f,} dom(f)D(f).{dom(f) \subseteq D(f).}
  3. A set of outputs, called the function's range, or image, denoted ran(f){ran(f)} or Im(f),{Im(f),} e.g., the function y(w)=1w{y(w) = \dfrac{1}{w}} has the range of Im(y)=R.{Im(y) = \mathbb{R}.}
  4. A set of all possible outputs, called the function's codomain, or target, denoted target(f),{\text{target}(f),} e.g., the function h(z)=z+1{h(z) = z + 1} has a codomain of target(h)=R,{\text{target}(h) = \mathbb{R},} but if we restricted dom(h)={1,2},{dom(h) = \{ 1,2 \},} then Im(h)={2,3},{Im(h) = \{ 2,3 \},} but its codomain remains target(h)=R.{\text{target}(h) = \mathbb{R}.} In set notation, given a function f,{f,} Im(f)target(f);{Im(f) \subseteq \text{target}(f);}
  5. A name for a typical input, called a dummy variable, e.g., the function g(p)=1p{g(p) = \dfrac{1}{p} } has the dummy variable pp.
  6. A name for the function, e.g., the function m(j)=lnj{m(j) = \ln j} has the name m(j).{m(j).}
  7. An assignment rule or formula assigning to every element of the domain exactly one element of the codomain, e.g., the function r(t)=t2+4t+1{r(t) = t^2 + 4t + 1} has the assignment rule t2+4t+1.{t^2 + 4t + 1.}
  8. Some functions have graphs, denoted G(f).{G(f).} (We say "some" because a function need not concern numbers. A relation mapping common fruit names to their respective scientific names is just as much a function as s(t)=t1.){s(t) = \sqrt{t - 1}.)}

Arrow Notation

Functions can be represented in arrow notation:

Function mapping

For example, the function f(x)=x2{f(x) = x^2}, in arrow notation, is f:xx2,{f: x \mapsto x^2,} and its domain and codomain are represented as RR.{\mathbb{R} \mapsto \mathbb{R}.} The benefit of arrow notation is that it allows us to easily restrict a function's inputs and outputs, which in turn allows us to differentiate functions explicitly. This is a particularly useful asset when the functions share the same assignment rule:

Suppose the function f(x)=x3+3x+4.{f(x) = x^3 + 3x + 4.} The functions below all have the same assignment rule as f,{f,} but they are all different functions:

  • f:xx3+3x+4;{f: x \mapsto x^3 + 3x + 4;} f:RR.{f: \mathbb{R} \mapsto \mathbb{R}.}
  • g:xx3+3x+4;{g: x \mapsto x^3 + 3x + 4;} g:R[0,+).{g: \mathbb{R} \mapsto [0, + \infty).}
  • h:xx3+3x+4;{h: x \mapsto x^3 + 3x + 4;} h:[0,+)R.{h: [0, + \infty) \mapsto \mathbb{R}.}
  • i:xx3+3x+4;{i: x \mapsto x^3 + 3x + 4;} i:[0,+)[0,+).{i: [0, + \infty) \mapsto [0, + \infty).}

The function ff is neither injective nor surjective, the function bb is injective but not surjective, the function cc is surjective but not injective, and the function dd is a bijection.

Note that in the last function, the function name ii was used. This is generally considered poor writing in mathematics, as the letter ii is almost always exclusively reserved for the imaginary number ii (setting aside the unorthodox practices of electrical engineers using jj).

Function Graphs

A function with a domain and range of real numbers can be graphed. When graphing functions, we use two axes—a horizontal axis for the function's domain (i.e., inputs) and vertical axis for function's range (i.e., outputs). We call the input axis the x-axis{\text{x-axis}} and the output axis the y-axis.{\text{y-axis.}}

Definition. Given a function f,{f,} the graph of f{f} on the xy{x-y} plane consists of those points (x,y){(x, y)} such that x{x} is the domain of f{f} and y=f(x).{y = f(x).}

The Vertical Line Test

The vertical line test allows us to determine if a particular graph represents a function. This is a particularly helpful result when we aren't provided any equation for a possible function, or if we are uncertain about what sets the function maps to and from.

A graph in the xy{x-y} plane represents a function of xx iff any vertical line intersects the graph in at most one point.

For example, the graph below is not a function, since the blue vertical line passes through the plot at more than one point:

Graphing Transformations

Graphs of functions serve as a bridge between geometry and algebra. Crossing from geometry to algebra are the definitions of reflection, translation, and scaling. All of these defines are broadly referred to as transformations—if we reflect, translate, or scale a function's graph, we are transforming the function's graph.

Vertical Translation

When we vertically translate a graph, we either shift the graph up or down the y-axis. For example, the following are vertical translations of the graph f(x)=x2.{f(x) = x^2.}

Observe that when add a positive constant to the function's expression, we shift the graph up the yy-axis. And when we add a negative constant (i.e., subtraction) to the function's expression, we shift the grown down the yy-axis. This logically follows from the fact that we are essentially adding a constant value to each element in the function's range. In other words, for every output, add or substract some constant—shift the graph up or down respectively.

definition. Let f(x){f(x)} be a function and c{c} a constant, where cR.{c \in \reals.} Suppose y=f(x).{y = f(x).} Substituting y{y} with yc,{y - c}, we obtain yc=f(x),{y - c = f(x),} which is equivalent to y=f(x)+c.{y = f(x) + c.} This is a vertical translation of f(x).{f(x).}

If cR+,{c \in \reals^{+},} then f(x){f(x)} transforms to f(x)+c,{f(x) + c,} which shifts G(f){G(f)} by c{c} units up the yy-axis.

If cR,{c \in \reals^{-},} then f(x){f(x)} transforms to f(x)+c=f(x)c,{f(x) + c = f(x) - c,} which shifts G(f){G(f)} by c{c} units down the yy-axis.

If c=0,{c = 0,} then there is no transformation, and a fortiori, no vertical translation.

Horizontal Translation

When we horizontally translate a graph, we either shift the graph left or right along the xx-axis. In essence, we are are adding or subtracting some constant from each element in the function's domain: For every input, add or substract some constant value. Horizontal translations, however, are a common source of error.

Consider again the graph of y=x2.{y = x^2.} To shift the graph to the right, we add some negative constant c{c} to every input. Thus, y=(xc)2{y = (x - c)^2} indicates a shift c{c} units to the right along the xx-axis. Why is it a shift right? Because to get back to the original function y{y}, we have to add c{c} units. It is the same logic behind why scrolling down with a mouse wheel moves this page up, or why a gear moving clockwise turns a second gear counter-clockwise. For example, given y=x2,{y = x^2,} the parabola's vertex is (0,0).{(0,0).} However, given y=(x1)2,{y = (x-1)^2,} the parabola's vertex is now (1,0);{(1,0);} a shift to the right.

In contrast, given y=(x+c)2,{y = (x + c)^2,} we have a horizontal shift to the left. Again, consider y=x2.{y = x^2.} The parabola is (0,0).{(0, 0).} When y=(x+1)2,{y = (x + 1)^2,} the vertex is now (1,0);{(-1, 0);} a shift to the left. Below are horizontal translations:

In sum: If we subtract a constant, we shift right; add a constant, we shift left.

Definition. Let f(x){f(x)} be a function and c{c} a constant, where cR.{c \in \reals.} Suppose y=f(x).{y = f(x).} Substituting x{x} with xc,{x - c}, we obtain y=f(xc).{y = f(x - c).} This is a horizontal translation of f(x).{f(x).}

If cR+,{c \in \reals^{+},} then f(x){f(x)} transforms to f(xc),{f(x - c),} which shifts G(f){G(f)} by c{c} units to the right along the yy-axis.

If cR,{c \in \reals^{-},} then f(x){f(x)} transforms to f(x+c){f(x + c)} which shifts G(f){G(f)} by c{c} units to the left along the yy-axis.

If c=0,{c = 0,} then there is no transformation, and a fortiori, no horizontal translation.

Slope

The graph of an equation is the set of all points with coordinates satisfying the equation. For example, suppose we had the equation y=x2+1{y = x^2 + 1}. Does the point (3,10){(3,10)} lie on the equation's graph? Yes. Plugging (3,10){(3,10)} into the equation y=x2+1{y = x^2 + 1}, we get 10=(3)2+1=9+10{10 = (3)^2 + 1 = 9 + 10}, which is true. Thus, the point 3,10{3,10} does indeed lie on the graph.

At very small intervals, graphs can be viewed as lines. Lines can either slant or remain straight, but in both cases, the line has a slope — a number that measures the line's direction.

Definition. Given two points, (x1,y1){(x_1, y_1)} and (x2,y2){(x_2, y_2)}, the slope of the line passing through the two points is the number mm:

m=y2y1x2x1 m = \dfrac{y_2 - y_1}{x_2 - x_1}

For example, suppose we had two points, (2,1){(-2,-1)} and (3,5){(3,5)}. On the Cartesian plane, these two points appear as:

The line between these two points is:

m=513(2)=45 m = \dfrac{5-1}{3-(-2)} = \dfrac{4}{5}

Thus, the line between (2,1){(-2,1)} and (3,5){(3,5)} has a slope of 45{\frac{4}{5}}. The slope equation can also be written as:

m=ΔyΔx m = \dfrac{\Delta y}{\Delta x}

The variable Δy\Delta y represents the change in yy-coordinates (y2y1{y_2 - y_1}), and the variable Δx\Delta x represents the change in xx-coordinates (x2x1{x_2 - x_1}).

Comparing Slopes

The slope mm of a line is a real number (mR{m \in \mathbb{R}}). Thus, mm can be a member of only one of three subsets:

  1. mR{m \in \mathbb{R}^-}
  2. m=0{m = 0}
  3. mR+{m \in \mathbb{R}^+}

If mm is a negative real number — mR{m \in \mathbb{R}^-} — then the line slants downwards to the right. If mm is a positive real number — mR+{m \in \mathbb{R}^+} — then the line slants upwards to the right. If m=0{m = 0}, then the line is a perfectly straight horizontal line.

Given the Cartesian plane, one might wonder, aren't there two cases for straight lines: (a) a vertical straight line or (b) a horizontal straight line? Yes, there are indeed two cases, straight horizontal lines and straight vertical lines. But, the slope is not defined for a vertical line. If the two points share the same ordinate (the y-coordinate{y \text{-coordinate}}), then the line is a horizontal line (e.g., (a,b),(c,b){(a,b), (c,b)}). If, however, the two points share the same abscissa (the x-coordinate{x \text{-coordinate}}), then the line is a vertical line (e.g., (a,b),(a,c){(a,b), (a,c)}), but the slope is not defined. This is because calculating the slope for such a line would lead to an undefined result:

m=bcaa=bc0 m = \dfrac{b - c}{a - a} = \dfrac{b - c}{0}

Does the magnitude, or size, of mm yield any inferences? Yes. m{\lvert m \rvert } represents the "steepness" of the line. The larger m{\lvert m \rvert } is, the steeper the line slants (i.e., tending towards a straight vertical line), and the smaller m{\lvert m \rvert } is, the less the line's steepness (i.e., tending towards a straight horizontal line).

The Point-Slope Formula

The slope formula can be rearranged in another form, yielding the point-slope formula.

Formula. Given a point P1(x1,y1)P_1{(x_1, y_1)}, the equation of the line passing through P1{P_1} is given by the equation:

yy1=m(xx1) y - y_1 = m(x - x_1)

Because of the point-slope formula, it follows that the equation of a horizontal line is simply the ordinate of P1{P_1}. For example, suppose P1=(4,2){P_1 = (4, -2)}. A horizontal line is a line where m=0{m = 0}, so it follows that the line's equation is:

y(2)=0(x4)y=2 \begin{aligned} y - (-2) &= 0(x - 4) y &= -2 \end{aligned}

We know that a horizontal line will always have a slope of m=0{m = 0}. For vertical lines, only the yy-coordinate varies, and the xx-coordinate remains the same. Thus, we can state more general rules for horizontal and vertical lines:

Rule. Given a point P1(a,b){P_1(a,b)}, the equation of a line passing through P1{P_1} is y=b{y = b}.

Rule. Given a point P2(a,b){P_2(a,b)}, the equation of a line passing through P2{P_2} is x=a{x = a}.

Slope-Intercept Formula

If we know that a line passes through the point (0,b){(0,b)}, then we can derive yet another equation for a line: the slope-intercept form. Since the line passes through (0,b){(0,b)} (i.e., the line crosses the yy-axis), using the point-slope formula, we can conclude that yb=m(x0){y - b = m(x-0)}. Simplifying and adding bb to both sides of the equation, we obtain the following rule.

Formula. Given a line with slope mm and a yy-intercept bb, the equation of the line is given by the equation:

y=mx+b y = mx + b

If we know that a point satisfies this equation, then it must be the case that it lies on the line with the equation. Furthermore, whenever we see an equation of this form, we should be immediately aware that the equation graphically represents a line.

Parallel and Perpendicular Lines

In geometry, we can easily notice that parallel lines slant in the same direction (i.e., they have the same "steepness" to their slant):

Because perpendicular lines have the same slant, in algebraic terms, the lines have the same slope. Thus, given line AB{\overleftrightarrow{\rm AB}} (where AA and BB are two arbitrary points on the line) with the slope m1m_1 and the line CD{\overleftrightarrow{\rm CD}} (where CC and DD are two arbitary points on the line) with the slope m2m_2, the lines AB{\overleftrightarrow{\rm AB}} and CD{\overleftrightarrow{\rm CD}} are parallel if, and only if, m1=m2{m_1 = m_2}. In other words, two lines are parallel, if and only if, they have the same slope.

In contrast, perpendicular lines slant in the exact opposite direction, forming 90 degree angles:

Given that two lines are perpendicular if and only if they form 90 degree angles, it follows that for a line to be parallel to another line, the line's slope must be the negative reciprocal of the other line. Thus, given a line EF{\overleftrightarrow{\rm EF}} (where EE and FF are arbitrary points on the line) with the slope m1m_1 and a line GH{\overleftrightarrow{\rm GH}} (where GG and HH are arbirary points on the line) with the slope m2m_2, lines EF{\overleftrightarrow{\rm EF}} and GH{\overleftrightarrow{\rm GH}} are perpendicular if, and only if, m1=1m2{m_1 = - \dfrac{1}{m_2}}.

To see why a perpendicular line's slope must be the negative reciprocal of the other line, consider the following: First, suppose we have three lines, 1\ell_1, 2\ell_2, and the vertical line x=1{x = 1}. Now suppose that 12{\ell_1 \perp \ell_2}, 1{\ell_1} has a slope m1{m_1}, and 2{\ell_2} has a slope m2{m_2}. Finally, suppose that 1{\ell_1} and 2{\ell_2} intersect at the origin, (0,0){(0,0)}. The lines can be visually represented as:

Perpendicular lines

Since 12{\ell_1 \perp \ell_2}, we know that that the vertical line x=1{x = 1} intersects 1{\ell_1} at (1,m1){(1, m_1)} and 2{\ell_2} at (1,m2).{(1,m_2).} This is because both lines have a Δx=1.{\Delta x = 1.}

Where 1{\ell_1} and 2{\ell_2} intersect, a right angle, ABC{\angle ABC}, is formed, since perpendicular lines form right angles. Thus, it follows that there is a right triangle ABC,{\triangle ABC,} since the resulting triangle from ABC{\angle ABC} forms a right triangle.

Now, consider the distance between the origin, B(0,0){B(0,0)}, and the point A(1,m1){A(1,m_1)}. From the distance formula, the length of BA{\overline{\rm BA}} (the distance between BB and AA) is BA=1+m12.{\overline{\rm BA} = \sqrt{1 + {m_1}^2}.} Similarly, BC=1+m22.{\overline{\rm BC} = \sqrt{1 + {m_2}^2}.} Thus, AC=(11)2+(m1m2)2=m1m2.{\overline{\rm AC} = \sqrt{(1-1)^2 + (m_1-m_2)^2} = m_1 - m_2.}

From this analysis, we now know all of the lengths of the sides of ABC{\triangle ABC}:

  1. BA=1+m12{\overline{\rm BA} = \sqrt{1+{m_1}^2}}
  2. BC=1+m22{\overline{\rm BC} = \sqrt{1+{m_2}^2}}
  3. AC=m1m2{\overline{\rm AC} = m_1 - m_2}

Because we know the lengths for all of the sides of ABC,{\triangle ABC,} we can express ABC{\triangle ABC} as an equation with the Pythagorean Theorem:

(1+m12)2+(1+m22)2=(m1m2)2 (\sqrt{1+{m_1}^2})^2 + (\sqrt{1+{m_2}^2})^2 = (m_1 - m_2)^2

With this form, we can isolate m1{m_1} and m2{m_2} to separate sides of the equation:

1+m12+1+m22=(m1m2)22+m12+m22=m122m1m2+m222=2m1m21=m1m2m1=1m2 \begin{aligned} 1 + {m_1}^2 + 1 + {m_2}^2 &= (m_1 - m_2)^2 \\ 2 + {m_1}^2 + {m_2}^2 &= {m_1}^2 - 2{m_1}{m_2} + {m_2}^2 \\ 2 &= -2{m_1}{m_2} \\ -1 &= {m_1}{m_2} \\ m_1 &= - \dfrac{1}{m_2} \\ \end{aligned}

The analysis above applies regardless of where the point of intersection is, since perpendicular lines will always form right triangles. Accordingly, it follows that given a line 1{\ell_1} with the slope m1{m_1} and a line 2{\ell_2} with the slope m2{m_2}, for 12{\ell_1 \perp \ell_2} to be true, it must be the case that m1=1m2{m_1 = - \dfrac{1}{m_2}}.

Footnotes

  1. Granted, greater minds may differ on how sweet iced tea should be made.

  2. A helpful mnemonic for the commutative law: "Factors freely flow."

  3. A helpful mnemonic for the associative law: "Factors freely faction."

  4. As an aside, the coffee example above is a play on the philosopher Slavoj Zizek's joke:

    A man comes into a restaurant. He sits down at the table and he says, “Waiter, bring me a cup of coffee without cream.” Five minutes later, the waiter comes back and says, “I'm sorry, sir, we have no cream. Can it be without milk?”

    Zizek's joke is an attempt at revitilizing Hegelian logic through an alternative reading. The idea is that the existence of a negation can be just as significant and worthy of consideration as the negation itself. In other words, when we encounter the proposition "coffee without cream", it's just as important to consider the possibility that what's being expressed is not "coffee without cream", but rather "coffee without".

    This idea has its roots in Kant's essay An Attempt to Introduce the Concept of Negative Magnitudes of Philosophy. There, Kant presented a thought experiment: Two equal forces start at zero. One moves five feet to the right, the other moves five feet to the left. Is the zero resulting from their movement different from the zero they started at? Kant argued yes, they are different. The "zero" resulting from the two forces cancelling one another is different from the zero before any force moved. The +5{+5} and the 5{-5} have in themselves a 0,{0,} in that +5{+5} means "Move 5 to the right from 0{0}" and 5{-5} means "Move 5 to the left from 0.{0.}" That inherent 0{0} never went anywhere. It's always been there.

    Hegel took this several steps further by arguing that the zero we get from 5+(5){5 + (-5)} is not only different, but that the process of negation alters the very thing we're negating. 5+(5){5 + (-5)} results in a non-being 0,{0,} from a process that starts with a being, the origin 0.{0.} This is the act of negation, and its result is different from zero, which represents nothingness. In Hegel's view, nothingness is always there. It requires no action by another to "be". In contrast, negation — the act of (5)+(5){(5) + (-5)} — requires an action by another.

    Hegel's arguments on negation (and logic in general), no doubt had an impact on mathematical logic. In particular, Hegel's formulation of the absolute (a concept that follows from his work on negations) is contemporaneous with the concept of absolute value, a topic to be discussed in later sections.

    The zero property of multiplication may seem straightforward, but it leads a particularly useful fact to keep in mind. The equation ab=ac{a \cdot b = a \cdot c} does not immediately imply that b=c.{b = c.} Why? Because if a=0,{a = 0,} then either b{b} or c{c} could be any real number and bc.{b \neq c.} Never assume that ab=ac{ab = ac} implies b=c{b = c} without first verifying, or stating explicitly, that a0.{a \neq 0.}

    Additionally, if we know that ab=0,{a \cdot b = 0,} then we have three possibilities: (1) a=0,{a = 0,} (2) b=0,{b = 0,} or (3) a=0{a = 0} and b=0.{b = 0.}

  5. As an aside, the eighth power of a number is called zenzizenzizenzic. Coined by the 16th century Welsh physician Robert Recorde, the root of the word is "zenzic," The German spelling of the Latin "censo," meaning "squared." At the time, there was no easy way to denote powers of numbers, so various forms of "zenzic" were used: zenzic for squared, zenzizenzic for the square of a square, and zenzizenzizenzic for the square of a square of a square.

  6. A "perfect tesseract" is not a formal term. The names "square" and "cube" come directly from geometry, so the next logical name would be a tesseract — perfect squares of squares. A more general definition for the perfect power:

  7. In computer science, the rational numbers are called floats or floating point numbers (the decimal point "floats" as needed).