# Algebra/Real Numbers

## What Are Real Numbers?

Real numbers are, most likely, all the numbers you can think of! They consist of zero (0), the positive and negative integers (-3, -1, 2, 4), and all the fractional and decimal values in between (0.4, 3.1415927, 1/2). Real numbers are divided into rational and irrational numbers. Rational numbers are numbers that can be expressed as a ratio (that is, a division) of two integers (2/3, 0.6, 3, -4.7, 0.11111...). If a number has a terminating decimal, or a decimal that ends (3.6, 5.263) or repeats (1.33333....), it is rational. Irrational numbers have decimal parts that do not terminate or repeat (2.71828..., 3.14159...). There are several other different "sets" of rational numbers.

Natural numbers, also known as "counting numbers", are the first numbers you learn. The natural numbers include all the positive whole numbers (1, 24, 6, 2, 357). Note that zero is not included. Whole numbers are the natural numbers, plus zero. Integers are all positive and negative numbers without a decimal part (3, -1, 15, -42).

## Properties Of Real Numbers

We begin this section with a review of the fundamental properties of arithmetic. It may seem unusual to give so much emphasis to the few properties listed below, but there is a good reason. Roughly speaking, all of algebra follows from the 5 properties listed in the table below. In the table below, a, b and c can be any number unless stated otherwise. So let's take a look:

Commutative

 ${\displaystyle a+b=b+a}$ This doesn't work: ${\displaystyle a-b\neq b-a}$  This does: ${\displaystyle a+(-b)=(-b)+a}$
${\displaystyle a*b=b*a}$ This doesn't work: ${\displaystyle a/b\neq b/a}$  This does: ${\displaystyle a*1/b=1/b*a}$


Associative

 ${\displaystyle (a+b)+c=a+(b+c)}$ This doesn't work: ${\displaystyle (a-b)-c\neq a-(b-c)}$ This does: ${\displaystyle (a-b)-c=a-(b+c)=a+(-b-c)}$

 ${\displaystyle (a*b)*c=a*(b*c)}$ This doesn't work: ${\displaystyle (a/b)/c\neq a/(b/c)}$
This does: ${\displaystyle (a/b)/c=a*1/b*1/c=a/(b*c)}$


Identity

  ${\displaystyle a+0=a}$  ${\displaystyle a-0=a}$  ${\displaystyle a*1=a}$  ${\displaystyle a/1=a}$


Inverse

  ${\displaystyle a+-a=0}$  ${\displaystyle a-a=0}$  ${\displaystyle a*(1/a)=1}$   as long as a ≠ 0.  ${\displaystyle a/a=1}$   as long as a ≠ 0.


Distributive

  ${\displaystyle a*(b+c)=a*b+a*c}$  ${\displaystyle a*(b-c)=a*b-a*c}$  ${\displaystyle (a+b)/c=a/c+b/c}$
But wait: ${\displaystyle a/(b+c)\neq a/b+a/c}$


But what does all this mean? The commutative property is that you can exchange two numbers and still get the same answer. The associative property is that you can change the grouping (i.e., change the position of the parenthesis) and still get the same answer. The identity property is that there is a certain number that when operated with a number doesn't change it. The inverse property is something that results to the identity number. The distributive property means that you can distribute the operation. Out of all of those properties, the distributive property is the one you'll probably use the most, because it is the only one that mentions both addition and multiplication at the same time. To give an example: these properties even imply fundamental things such as: "multiplication is repeated addition". This book is not going to prove many things, but it would be useful for us to take a look at how this works.

We apply the distributive property for a = 7, b = 1 and c = 1.

 7 · 1 + 7 · 1 = 7 + 7


Though it may seem obvious, this is identity property for multiplication listed above. Now let's try to do the same thing with 7 · 3.

 7 · 3 = 7 · (1 + 1 + 1)


Just like before, this is just the fact that 3 = 1 + 1 + 1 together with substitution.

 7 · (1 + 1 + 1) = 7 · 1 + 7 · 1 + 7 · 1


Once again, we apply the distributive property. Note that we can apply it to expressions with more than two numbers being added in parentheses. The proof is below. While 7 · (1 + 1 + 1) = 7 · 1 + 7 · 1 + 7 · 1 is not covered by the distributive property alone, this problem is solved by grouping the last two 1s with parentheses. Rather than writing this as 7 · (1 + 1 + 1), we could write it as 7 · (1 + (1 + 1)), then used the distributive property with a = 7, b = 1 and c = (1 + 1). Then: 7 · (1 + (1 + 1)) = 7 · 1 + 7 · (1 + 1). Now we apply the distributive property just to the second (taking a = 7, b = 1, and c = 1. Then (looking just at the second term) we have 7 · (1 + 1) = 7 · 1 + 7 · 1. Finally we can substitute this expression for the second term back into the equation to get: 7 · (1 + 1 + 1) = 7 · 1 + 7 · 1 + 7 · 1.

This looks like a lot of mindless parenthesis juggling, but the point is that the distributive property applies to arbitrarily long sums and products. It is also true that

a · (b + c + d + e) = a · b + a · c + a · d + a · e

 b + c = c + b This is the commutativity of addition applied to b + c a + (b + c) = a + (c + b) This follows from substitution a + (b + c) = (a + c) + b This is just using associativity on the right side of the line above.

Commutativity and associativity tell you that it doesn't matter in which order you add up a + b + c. You will get the same answer regardless of order. The rule holds even if there are more than three terms: There may be 4, 12, or several thousand. These properties would still tell us that it doesn't matter how we add things up.

The same properties for multiplication tell us it doesn't matter in what order we multiply things together. We are free to change the order to anything that we find easier. Does it ever really make things easier? Sure! For example if you were asked to calculate 4 · 3 · 5 · (1/4), then I would personally think it would be easier to calculate 4 · (1/4) · 3 · 5

The identity and inverse properties really capture what it means to say that "addition and subtraction are opposites" and "multiplication and division are opposites, as long as it isn't zero that we multiply by." We shall leave it as an exercise to the interested reader to think about why this is.

You can often simplify expressions using the Distributive Property. This is one of the reasons it is so important. For example, consider the expression 2(x − 7) + 14. What happens if we use the distributive property on the first term in this expression? Let's work it out. According to the Distributive Property

2(x − 7) = 2x − 2 · 7 = 2x − 14

Plugging this into the expression above we get 2(x − 7) + 14 = 2x − 14 + 14 = 2x. Clearly 2x is a lot easier to evaluate than 2(x − 7) + 14!

## Basic Laws In Algebra

There are several basic laws in algebra. Understanding these will help you to manipulate and solve equations, and to understand algebraic relationships.

### 1. Commutative Law

In general, the order of the items can be changed without affecting the results.

For addition, ${\displaystyle A+B=B+A}$ indicates that changing the order of the items added does not affect the sum.

For multiplication, ${\displaystyle XY=YX}$ indicating that the changing of the order of the items multiplied does not affect the product.

Note that the commutative law does not apply to subtraction or division.

### 2. Associative Law

In general, the grouping of the items can be changed without affecting the results. (Seems to be an extension of the commutative law).

For addition, ${\displaystyle A+(B+C)=(A+B)+C}$ indicates that changing the grouping of the items added does not affect the sum.

For multiplication, ${\displaystyle X(YZ)=(XY)Z}$ indicates that changing the grouping of the items multiplied does not affect the product.

As with the commutative law, the associative law does not apply to subtraction or division.

### 3. Distributive Law

Indicates that common terms can be factored, or that factors can be distributed. (A + B) X = (A X) + (B X) (The "X" terms on the right are combined into a factor on the left side; the factor "X" on the left is distributed on the right side).

Consider the substitution of X = (Y + Z) into the above equation yields (A + B) (Y + Z) = A (Y + Z) + B (Y + Z). Apply the distributive law to each term on the right yields A Y + A Z + B Y + B Z. We can skip the intermediate step if we multiply the terms identified by “F O I L” in the following expression (A + B) (Y + Z) =

Letter Description terms
F First terms A Y +
O Outside terms A Z +
I Inside terms B Y +
L Last terms B Z

### 4. Law of Identity

For addition and subtraction the law of identity indicates that the addition and subtraction of a given term or quantity results in the zero, 0, the identity element for addition and subtraction. Alternately, adding the identity element results in no change to the original value or quantity.

${\displaystyle A-A=0}$

Adding A to both sides of the first equation we get (A - A) + A = 0 + A. Re-arranging or substituting gives 0 + A = A

Note the special case(s) where A = A + 0 = A + 0 + 0

For multiplication and division the law of identity indicates that the multiplication and division of a given term or quantity results in "one," 1, the identity element for multiplication and division. Alternately, multipling or dividing by the identity element results in no change to the original value or quantity.

${\displaystyle 1={\frac {Y}{Y}}}$, or ${\displaystyle 1=({\frac {Y}{1}})({\frac {1}{Y}})}$

Note that dividing 1 by a term or quantity gives the reciprocal of the term or quantity. Multiplying by the reciprocal is the same as dividing by the term or quantity. In the above equation on the right (Y / 1), and (1 / Y) are reciprocals of each other
Note the special case where ${\displaystyle 1={\frac {1}{1}}}$, Multiplying this equation by “1” gives ${\displaystyle 1(1)=(1)({\frac {1}{1}})}$, and then dividing by one gives ${\displaystyle {\frac {1(1)}{1}}=(1)({\frac {1}{1}})=}$.
Simplify this by substititing the first special case equation to get ${\displaystyle 1=1(1)}$ , and ${\displaystyle 1=1(1)(1)}$, . . .

By multiplying both sides of the first equation by “Y” we get ${\displaystyle (Y)(1)=(Y)({\frac {Y}{Y}})}$ , which simplifies and becomes (Y) = (1) Y.

## Order and Absolute Value

### Practice Problems

Identify the following properties being expressed.

1. 4(3x + 4) = 12x + 16
2. 6 + 0 = 6
3. (2 + 7) + 5 = (2 + 5) + 7
4. (3/4)(4/3) = 1

Use the Distributive Property to simplify these expressions.

1. 2(14x - 26)
2. (2/3)(3x + 9)
3. 3(12x + 4y)
4. 2(5x - 6) + 3(3x + 2)

First set:

1. Distributive Property of Multiplication
4. Inverse Property of Multiplication (You multiply by the reciprocal. The reciprocal of 4 is 1/4, of -17 is -1/17, and of 2/5 is 5/2.)

Second set:

1. 28x - 52
2. 2x + 6
3. 36x + 12y
4. 10x - 12 + 9x + 6 = 19x - 6 (Combine like-terms, as we learned in Arithmetic!)

## Lesson Review

All numbers that we will be working with for the majority of Algebra are called Real Numbers. They consist of Rational and Irrational Numbers. Irrational Numbers are numbers that have infinite, non-repeating decimals, such as pi. Rational Numbers are all numbers that can be expressed as a fraction of integers, which include Natural Numbers, Whole Numbers, Integers, and Rational Numbers. For all Real Numbers, there are a few properties of addition and multiplication: Commutative, Associative, Identity, Inverse, and Distribution. The Distribution will come in handy for the rest of the course.

## Lesson Quiz

Identify the set or sets of numbers each number is in.

1. ${\displaystyle {\sqrt {7}}}$
2. ${\displaystyle -{\sqrt {36}}}$
3. ${\displaystyle (-4)^{2}}$

Identify the property being expressed.

1. ${\displaystyle 2\cdot \!3=3\cdot \!2}$
2. ${\displaystyle x\left({\frac {1}{x}}\right)=1}$
3. ${\displaystyle x+(-x)=0\,}$

Perform the Distributive Property of Multiplication to simplify each of these expressions.

1. ${\displaystyle 3(2x+7)\,}$
2. ${\displaystyle 15(6x-22)\,}$
3. ${\displaystyle 3(20x+42y)-2(7x+20y)\,}$

Challenge Questions.

1. When two rational numbers are multiplied, does it always result in a rational number? Why?
2. When two irrational numbers are multiplied, does it always result in an irrational number? Why?
3. When two irrational numbers are added, does it always result in an irrational number? Why?
4. When the square root of an irrational number is taken, does it have to be irrational? Why?
5. If ${\displaystyle x(x+1)}$ is irrational, does x have to be irrational? Why?

First set:

1. Irrational and Real
2. Integer, Rational, Real (Note that there is a negative outside the square root!)
3. Natural, Whole, Integer, Rational, and Real (Note that the negative comes before squaring, because it's in parentheses!)

Second set:

1. Commutative Property of Multiplication
2. Inverse Property of Multiplication

Third set:

1. 6x + 21
2. 90x - 330 (Remember that we could have simplified the expression to 30(3x - 11))
3. 46x + 86y (Don't forget to combine like-terms. Also, don't forget that you're subtracting 40y!)

Challenge Questions:

1. Yes, take two rational numbers ${\displaystyle {\frac {a}{b}}}$ and ${\displaystyle {\frac {c}{d}}}$ the product of those two numbers is ${\displaystyle {\frac {ac}{bd}}}$, which is still rational (remember that a rational number is one that can be expressed as a ratio of two numbers). This is an example of the closure property for rational numbers.
2. No, ${\displaystyle {\sqrt {2}}}$ is irrational, however, ${\displaystyle {\sqrt {2}}{\sqrt {2}}=2}$ is an integer.
3. No, consider two numbers ${\displaystyle 1+{\sqrt {2}}}$ and ${\displaystyle 1-{\sqrt {2}}}$ both of these numbers are irrational, however when you add them together, the irrational part "cancels" and you are left with a rational part. In other words, because you can always add a number and its negative to get 0 (which we consider to be rational), you can always get a rational number from an irrational number.
4. Yes, because if ${\displaystyle x}$ were irrational but ${\displaystyle {\sqrt {x}}}$ were equal to ${\displaystyle a/b}$ for some integers ${\displaystyle a}$ and ${\displaystyle b}$, then ${\displaystyle x}$ would be equal to the rational number ${\displaystyle a^{2}/b^{2}}$, contradicting the fact that ${\displaystyle x}$ is irrational.
5. Yes, the product of two rational numbers is rational (see the first challenge problem) similarly the sum of two rational numbers is also rational. If ${\displaystyle x}$ is rational then ${\displaystyle x+1}$ is rational, then ${\displaystyle x(x+1)}$ must also be rational. Therefore, ${\displaystyle x}$ must be irrational.