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Algebra 1

1.02 Properties of real numbers

Lesson
Practice

Review: Properties of real numbers

Recall that the numbers that we use regularly for counting and measuring are called the real numbers. They include zero, the positive and negative whole numbers, the numbers that can be written as fractions, and every number in between.

Natural numbers

The counting numbers, starting from 1.

Whole numbers

The counting numbers, starting from 0.

Integers

A set of numbers that include positive whole numbers (natural numbers), their negative counterparts, and zero.

Rational numbers

The set of numbers that can be expressed in the form \dfrac{a}{b} where a and b are integers and b\neq0.

Irrational numbers

The set of numbers that cannot be written in the form \dfrac{a}{b} where a and b are integers.

A diagram showing the Real Number System divided into Rational and Irrational. Within Rational is the subset Integers; within Integers is the subset Whole; within Whole is the subset Natural. Examples of each type of number are shown. Speak to your teacher for more information.

The real numbers are so familiar to us that we hardly notice that they have special properties that involve the addition and multiplication operations.

For any real numbers a,\,b,\, and c, the following properties are always true.

AdditionMultiplication
Commutative property\ a+b = b+a\ a \cdot b = b \cdot a
Associative property\ \left(a + b\right) + c = a + \left(b + c\right)\ a \cdot \left(b \cdot c\right) = \left(a \cdot b\right) \cdot c
Inverse property\ a + \left(-a\right) = 0\ a \cdot \dfrac{1}{a} = 1
Identity property\ a + 0 = a\ a \cdot 1 = a
Distributive property a\left(b + c\right) = a \cdot b + a \cdot c

Notice the distributive property only applies to distributing the operation of multiplication. If the positions of the addition and multiplication signs are swapped so that we have a + \left(b \cdot c\right), a similar distributive rule is not true. This is one situation in which we must observe the correct order of operations when simplifying expressions.

Examples

Example 1

Verify the distributive property a\left(b + c\right) = a \cdot b + a \cdot c for the values a = 3,\, b = 13,\, and c = -1.

Worked Solution
Create a strategy

We need to substitute the values for a = 3,\, b = 13,\, and c = -1 into the distributive property equation a\left(b + c\right) = a \cdot b + a \cdot c. If both sides of the equation evaluate to the same number, then the property is true for those values.

Apply the idea

Evaluate the expressions.

\displaystyle a\left(b + c\right) \displaystyle =\displaystyle 3\left(13 + \left(-1\right)\right)Substitute a = 3,\, b= 13,\, and c=-1
\displaystyle =\displaystyle 3\left(12\right)Evaluate the subtraction inside parentheses
\displaystyle =\displaystyle 36Evaluate the multiplication
\displaystyle a \cdot b + a \cdot c \displaystyle =\displaystyle 3 \cdot 13 + 3 \cdot \left(-1\right)Substitute a = 3,\, b= 13,\, and c=-1
\displaystyle =\displaystyle 39 - 3Evaluate the multiplication
\displaystyle =\displaystyle 36Evaluate the subtraction

So, the expressions are the same for these numbers and the distributive property is true in this case.

Reflect and check

How might you apply the same strategy to verify the other properties of real numbers?

Example 2

Using the properties of real numbers, rewrite 3\left(x+4\right) = 9 in another way.

Worked Solution
Create a strategy

First identify the property we will be using. Given the addition and multiplication, the distributive property a\left(b + c\right) = a \cdot b + a \cdot c seems like a good fit.

Apply the idea

The distributive property tells us we can multiply each term in the parentheses individually by the 3 outside the parentheses as shown:3\left(x+4\right) = 3 \cdot x + 3 \cdot 4 = 3x + 12

Therefore, the expression 3\left(x+4\right) is equivalent to 3x + 12 by the distributive property. So we may rewrite the equation as 3x+12=9.

Reflect and check

We will use the distributive property many times to expand and simplify expressions. Can you think of any ways this may be helpful for us?

Example 3

If 5 \cdot \left(\dfrac{1}{x}\right) = 1, use the properties of real numbers to solve for x.

Worked Solution
Create a strategy

First identify the property we will be using. Given the multiplication by an inverse, the inverse property of multiplications seems like a good fit.

Apply the idea

The inverse property of multiplication tells us that for every real number a, there exists a number b such that a \cdot b = 1. The numbers a and b are called multiplicative inverses of one another. We usually write \dfrac{1}{a} instead of b.

That is, we can rewrite the property as a \cdot \dfrac{1}{a} = 1.

This looks a lot like our problem. Let a = 5 for this property. Then, we have 5 \cdot \dfrac{1}{5} = 1. This tells us x = 5 in our example by the inverse property of multiplication.

Reflect and check

See if you can challenge yourself to figure out what the value of x must be in this equation: x \cdot \left(\dfrac{1}{7}\right) = 1

Idea summary

The properties of real numbers can be used to simplify expressions and solve equations.

AdditionMultiplication
Commutative property\ a+b = b+a\ a \cdot b = b \cdot a
Associative property\ \left(a + b\right) + c = a + \left(b + c\right)\ a \cdot \left(b \cdot c\right) = \left(a \cdot b\right) \cdot c
Inverse property\ a + \left(-a\right) = 0\ a \cdot \dfrac{1}{a} = 1
Identity property\ a + 0 = a\ a \cdot 1 = a
Distributive property a\left(b + c\right) = a \cdot b + a \cdot c

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