# Algebra/Chapter 2/Real Numbers

2.7: Properties of Real Numbers

## 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:
| Property Name | Addition | Subtraction | Multiplication | Division |
| Commutative | $a+b=b+a$ | Doesn't work: $a-b\neq b-a$ This does: $a+(-b)=(-b)+a$ | $a*b=b*a$ | Doesn't work: $a/b\neq b/a$ This does: $a*1/b=1/b*a$ |
| Associative | $(a+b)+c=a+(b+c)$ | Doesn't work: $(a-b)-c\neq a-(b-c)$ This does: $(a-b)-c=a-(b+c)=a+(-b-c)$ | $(a*b)*c=a*(b*c)$ | Doesn't work: $(a/b)/c\neq a/(b/c)$ This does: $(a/b)/c=a*1/b*1/c=a/b*c$ |
| Identity | $a+0=a$ | $a-0=a$ | $a*1=a$ | $a/1=a$ |
| Inverse | $a+-a=0$ | $a-a=0$ | $a*(1/a)=1$ as long as a ≠ 0. | $a/a=1$ as long as a ≠ 0. |
| Distributive | $a*(b+c)=a*b+a*c$ | $a*(b-c)=a*b-a*c$ | $a*(b+c)=a*b+a*c$ | $(a+b)/c=a/c+b/c$ But wait: $a/(b+c)\neq a/b+a/c$ |
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
Or we could make it even longer! We could have as many terms in the sum as we like; as long as "a · " appears in front of each term on the right hand side we will have a true statement. We will use this fact without justification (that is, without proof). Let's remind ourselves what these properties tell us about arithmetic. Commutativity and Associativity together imply that it doesn't matter what order we add things up in. Let's see why. Associativity says that a + (b + c) = (a + b) + c. This should be thought of as a statement about the sum a + b + c. Why? Because usually addition is just defined between two things, so someone writes down something like a + b + c some people may first add b and c first then add in a, and other people might add a and b first and then add in c. This property says (using a formula) that it doesn't matter which way you do it. What about those people who add a and c together first? Well, that is where commutativity comes in. It tells us that we don't have add things up in exactly the order people write things down. You can switch things around and still get the same answer. Let's do one more example of using these properties to "juggle parentheses" to see that commutativity says you really can add a and c first and get the same answer.
| 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!

### Commutative properties of Division

Division is not commutative. That means usually a ÷ b is not equal to b ÷ a, and can be demonstrated simply by example.
${\frac {1}{2}}\neq {\frac {2}{1}}$
While division itself is not commutative, there are two special cases where the answer is the same if you reverse the order of operation. These cases occur when the answer (quotient) is 1 or when the answer is -1:
$a\div b=b\div a\iff {\mbox{(rewrite as fractions)}}$
${\frac {a}{b}}={\frac {b}{a}}\iff {\mbox{(multiply both sides by}}\ ab)$
$a^{2}=b^{2}\iff {\mbox{(take both square roots)}}$
$a={\sqrt {b^{2}}}\quad {\mbox{ or }}\quad a=-{\sqrt {b^{2}}}$
$a=b\quad {\mbox{ or }}\quad a=-b$
$a\div b=1\quad {\mbox{ or }}\quad a\div b=-1$

## 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, $A+B=B+A$ indicates that changing the order of the items added does not affect the sum.
For multiplication, $XY=YX$ 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, $A+(B+C)=(A+B)+C$ indicates that changing the grouping of the items added does not affect the sum.
For multiplication, $X(YZ)=(XY)Z$ 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.
$A-A=0$
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.
$1={\frac {Y}{Y}}$, or $1=({\frac {Y}{1}})({\frac {1}{Y}})$
    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 $1={\frac {1}{1}}$, Multiplying this equation by “1” gives $1(1)=(1)({\frac {1}{1}})$, and then dividing by one gives ${\frac {1(1)}{1}}=(1)({\frac {1}{1}})=$.
    Simplify this by substititing the first special case equation to get $1=1(1)$ , and $1=1(1)(1)$, . . .
By multiplying both sides of the first equation by “Y” we get $(Y)(1)=(Y)({\frac {Y}{Y}})$ , which simplifies and becomes (Y) = (1) Y.

## Closure

Closure is a property that is defined for a set of real numbers and an operation. This Wikipedia article gives a description of the closure property with examples from various areas in math. As an Algebra student being aware of the closure property can help you solve a problem. For instance a problem might state "The sum of two whole numbers is 24." With practice you will be able to see that the possible set of numbers will be either all odd (e.g. (1,23),(3,21), ... etc.) or all even (e.g. (2,22), (4,20), ... etc.). The problem might not explicitly state the idea of whole numbers. It might state that two sides of a square sum to 24. If you remember working a problem like this before you know that the sides of a square need to be equal and you divide by two. The author of the problem might want to be trickier and say that two sides of an equilateral triangle sum to 24 and then ask for the perimeter of the triangle. In this case you might want to write the equation $3x=p$ for the perimeter of an equilateral triangle. This might make it easier for you to see that again you just need to divide 24 by 2 to find the length of one side and plug it into the equation.