# 1.9 Properties of real numbers  (Page 4/10)

 Page 4 / 10

We conclude that there is no answer to $4÷0$ and so we say that division by 0 is undefined.

## Division by zero

For any real number a , except 0, $\frac{a}{0}$ and $a÷0$ are undefined.

Division by zero is undefined.

We summarize the properties of zero below.

## Properties of zero

Multiplication by Zero: For any real number a ,

$\begin{array}{cccc}\phantom{\rule{2em}{0ex}}a·0=0\phantom{\rule{1em}{0ex}}0·a=0\hfill & & & \text{The product of any number and 0 is 0.}\hfill \end{array}$

Division of Zero, Division by Zero: For any real number $a,a\ne 0$

$\begin{array}{}\\ \\ \phantom{\rule{2em}{0ex}}\frac{0}{a}=0\hfill & & & \text{Zero divided by any real number, except itself is zero.}\hfill \\ \phantom{\rule{2em}{0ex}}\frac{a}{0}\phantom{\rule{0.2em}{0ex}}\text{is undefined}\hfill & & & \text{Division by zero is undefined.}\hfill \end{array}$

Simplify: $-8·0$ $\frac{0}{-2}$ $\frac{-32}{0}.$

## Solution

$\begin{array}{cccc}& & & \hfill -8·0\hfill \\ \text{The product of any real number and 0 is 0.}\hfill & & & \hfill 0\hfill \end{array}$

$\begin{array}{cccc}& & & \hfill \phantom{\rule{1.2em}{0ex}}\frac{0}{-2}\hfill \\ \begin{array}{c}\text{Zero divided by any real number, except}\hfill \\ \text{itself, is 0.}\hfill \end{array}\hfill & & & \hfill \phantom{\rule{1.2em}{0ex}}0\hfill \end{array}$

$\begin{array}{cccc}& & & \hfill \phantom{\rule{5.2em}{0ex}}\frac{-32}{0}\hfill \\ \text{Division by 0 is undefined.}\hfill & & & \hfill \phantom{\rule{5.2em}{0ex}}\text{Undefined}\hfill \end{array}$

Simplify: $-14·0$ $\frac{0}{-6}$ $\frac{-2}{0}.$

0 0 undefined

Simplify: $0\left(-17\right)$ $\frac{0}{-10}$ $\frac{-5}{0}.$

0 0 undefined

We will now practice using the properties of identities, inverses, and zero to simplify expressions.

Simplify: $\frac{0}{n+5},$ where $n\ne \text{−}5$ $\frac{10-3p}{0},$ where $10-3p\ne 0.$

## Solution

$\begin{array}{cccc}& & & \hfill \frac{0}{n+5}\hfill \\ \begin{array}{c}\text{Zero divided by any real number except}\hfill \\ \text{itself is 0.}\hfill \end{array}\hfill & & & \hfill 0\hfill \end{array}$

$\begin{array}{cccc}& & & \hfill \phantom{\rule{4em}{0ex}}\frac{10-3p}{0}\hfill \\ \text{Division by 0 is undefined.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\text{Undefined}\hfill \end{array}$

Simplify: $-84n+\left(-73n\right)+84n.$

## Solution

$\begin{array}{cccc}& & & \hfill -84n+\left(-73n\right)+84n\hfill \\ \\ \\ \begin{array}{c}\text{Notice that the first and third terms are}\hfill \\ \text{opposites; use the commutative property of}\hfill \\ \text{addition to re-order the terms.}\hfill \end{array}\hfill & & & \hfill -84n+84n+\left(-73n\right)\hfill \\ \\ \\ \text{Add left to right.}\hfill & & & \hfill 0+\left(-73\right)\hfill \\ \\ \\ \text{Add.}\hfill & & & \hfill -73n\hfill \end{array}$

Simplify: $-27a+\left(-48a\right)+27a.$

$-48a$

Simplify: $39x+\left(-92x\right)+\left(-39x\right).$

$-92x$

Now we will see how recognizing reciprocals is helpful. Before multiplying left to right, look for reciprocals—their product is 1.

Simplify: $\frac{7}{15}·\frac{8}{23}·\frac{15}{7}.$

## Solution

$\begin{array}{cccc}& & & \hfill \frac{7}{15}·\frac{8}{23}·\frac{15}{7}\hfill \\ \\ \\ \begin{array}{c}\text{Notice the first and third terms are}\hfill \\ \text{reciprocals, so use the commutative}\hfill \\ \text{property of multiplication to re-order the}\hfill \\ \text{factors.}\hfill \end{array}\hfill & & & \hfill \frac{7}{15}·\frac{15}{7}·\frac{8}{23}\hfill \\ \\ \\ \text{Multiply left to right.}\hfill & & & \hfill 1·\frac{8}{23}\hfill \\ \\ \\ \text{Multiply.}\hfill & & & \hfill \frac{8}{23}\hfill \end{array}$

Simplify: $\frac{9}{16}·\frac{5}{49}·\frac{16}{9}.$

$\frac{5}{49}$

Simplify: $\frac{6}{17}·\frac{11}{25}·\frac{17}{6}.$

$\frac{11}{25}$

Simplify: $\frac{0}{m+7},$ where $m\ne \text{−}7$ $\frac{18-6c}{0},$ where $18-6c\ne 0.$

0 undefined

Simplify: $\frac{0}{d-4},\text{where}\phantom{\rule{0.2em}{0ex}}d\ne 4$ $\frac{15-4q}{0},\text{where}\phantom{\rule{0.2em}{0ex}}15-4q\ne 0.$

0 undefined

Simplify: $\frac{3}{4}·\frac{4}{3}\left(6x+12\right).$

## Solution

$\begin{array}{cccc}& & & \hfill \frac{3}{4}·\frac{4}{3}\left(6x+12\right)\hfill \\ \\ \\ \begin{array}{c}\text{There is nothing to do in the parentheses,}\hfill \\ \text{so multiply the two fractions first—notice,}\hfill \\ \text{they are reciprocals.}\hfill \end{array}\hfill & & & \hfill 1\left(6x+12\right)\hfill \\ \\ \\ \begin{array}{c}\text{Simplify by recognizing the multiplicative}\hfill \\ \text{identity.}\hfill \end{array}\hfill & & & \hfill 6x+12\hfill \end{array}$

Simplify: $\frac{2}{5}·\frac{5}{2}\left(20y+50\right).$

$20y+50$

Simplify: $\frac{3}{8}·\frac{8}{3}\left(12z+16\right).$

$12z+16$

## Simplify expressions using the distributive property

Suppose that three friends are going to the movies. They each need $9.25—that’s 9 dollars and 1 quarter—to pay for their tickets. How much money do they need all together? You can think about the dollars separately from the quarters. They need 3 times$9 so $27, and 3 times 1 quarter, so 75 cents. In total, they need$27.75. If you think about doing the math in this way, you are using the distributive property .

## Distributive property

$\begin{array}{cccc}\text{If}\phantom{\rule{0.2em}{0ex}}a,b,c\phantom{\rule{0.2em}{0ex}}\text{are real numbers, then}\hfill & & & a\left(b+c\right)=ab+ac\hfill \\ \\ \\ \hfill \text{Also,}& & & \left(b+c\right)a=ba+ca\hfill \\ & & & a\left(b-c\right)=ab-ac\hfill \\ & & & \left(b-c\right)a=ba-ca\hfill \end{array}$

Back to our friends at the movies, we could find the total amount of money they need like this:

$\begin{array}{}\\ \\ \phantom{\rule{1.3em}{0ex}}3\left(9.25\right)\hfill \\ \begin{array}{ccc}\hfill 3\left(9& +\hfill & 0.25\right)\hfill \\ \hfill 3\left(9\right)& +\hfill & 3\left(0.25\right)\hfill \\ \hfill 27& +\hfill & 0.75\hfill \end{array}\hfill \\ \\ \\ \phantom{\rule{1.6em}{0ex}}27.75\hfill \end{array}$

In algebra, we use the distributive property to remove parentheses as we simplify expressions.

For example, if we are asked to simplify the expression $3\left(x+4\right),$ the order of operations says to work in the parentheses first. But we cannot add x and 4, since they are not like terms. So we use the distributive property, as shown in [link] .

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