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We conclude that there is no answer to $4\xf70$ and so we say that division by 0 is undefined.
For any real number a , except 0, $\frac{a}{0}$ and $a\xf70$ are undefined.
Division by zero is undefined.
We summarize the properties of zero below.
Multiplication by Zero: For any real number a ,
$\begin{array}{cccc}\phantom{\rule{2em}{0ex}}a\xb70=0\phantom{\rule{1em}{0ex}}0\xb7a=0\hfill & & & \text{The product of any number and 0 is 0.}\hfill \end{array}$
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: ⓐ $\mathrm{-8}\xb70$ ⓑ $\frac{0}{\mathrm{-2}}$ ⓒ $\frac{\mathrm{-32}}{0}.$
ⓐ
$\begin{array}{cccc}& & & \hfill \mathrm{-8}\xb70\hfill \\ \text{The product of any real number and 0 is 0.}\hfill & & & \hfill 0\hfill \end{array}$
ⓑ
$\begin{array}{cccc}& & & \hfill \phantom{\rule{1.2em}{0ex}}\frac{0}{\mathrm{-2}}\hfill \\ \begin{array}{c}\text{Zero divided by any real number, except}\hfill \\ \text{itself, is 0.}\hfill \end{array}\hfill & & & \hfill \phantom{\rule{1.2em}{0ex}}0\hfill \end{array}$
ⓒ
$\begin{array}{cccc}& & & \hfill \phantom{\rule{5.2em}{0ex}}\frac{\mathrm{-32}}{0}\hfill \\ \text{Division by 0 is undefined.}\hfill & & & \hfill \phantom{\rule{5.2em}{0ex}}\text{Undefined}\hfill \end{array}$
Simplify: ⓐ $\mathrm{-14}\xb70$ ⓑ $\frac{0}{\mathrm{-6}}$ ⓒ $\frac{\mathrm{-2}}{0}.$
ⓐ 0 ⓑ 0 ⓒ undefined
Simplify: ⓐ $0\left(\mathrm{-17}\right)$ ⓑ $\frac{0}{\mathrm{-10}}$ ⓒ $\frac{\mathrm{-5}}{0}.$
ⓐ 0 ⓑ 0 ⓒ undefined
We will now practice using the properties of identities, inverses, and zero to simplify expressions.
Simplify: ⓐ $\frac{0}{n+5},$ where $n\ne \text{\u2212}5$ ⓑ $\frac{10-3p}{0},$ where $10-3p\ne 0.$
ⓐ
$\begin{array}{cccc}& & & \hfill \frac{0}{n+5}\hfill \\ \begin{array}{c}\text{Zero divided by any real number except}\hfill \\ \text{itself is 0.}\hfill \end{array}\hfill & & & \hfill 0\hfill \end{array}$
ⓑ
$\begin{array}{cccc}& & & \hfill \phantom{\rule{4em}{0ex}}\frac{10-3p}{0}\hfill \\ \text{Division by 0 is undefined.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\text{Undefined}\hfill \end{array}$
Simplify: $\mathrm{-84}n+\left(\mathrm{-73}n\right)+84n.$
$\begin{array}{cccc}& & & \hfill \mathrm{-84}n+\left(\mathrm{-73}n\right)+84n\hfill \\ \\ \\ \begin{array}{c}\text{Notice that the first and third terms are}\hfill \\ \text{opposites; use the commutative property of}\hfill \\ \text{addition to re-order the terms.}\hfill \end{array}\hfill & & & \hfill \mathrm{-84}n+84n+\left(\mathrm{-73}n\right)\hfill \\ \\ \\ \text{Add left to right.}\hfill & & & \hfill 0+\left(\mathrm{-73}\right)\hfill \\ \\ \\ \text{Add.}\hfill & & & \hfill \mathrm{-73}n\hfill \end{array}$
Simplify: $\mathrm{-27}a+\left(\mathrm{-48}a\right)+27a.$
$\mathrm{-48}a$
Simplify: $39x+\left(\mathrm{-92}x\right)+\left(\mathrm{-39}x\right).$
$\mathrm{-92}x$
Now we will see how recognizing reciprocals is helpful. Before multiplying left to right, look for reciprocals—their product is 1.
Simplify: $\frac{7}{15}\xb7\frac{8}{23}\xb7\frac{15}{7}.$
$\begin{array}{cccc}& & & \hfill \frac{7}{15}\xb7\frac{8}{23}\xb7\frac{15}{7}\hfill \\ \\ \\ \begin{array}{c}\text{Notice the first and third terms are}\hfill \\ \text{reciprocals, so use the commutative}\hfill \\ \text{property of multiplication to re-order the}\hfill \\ \text{factors.}\hfill \end{array}\hfill & & & \hfill \frac{7}{15}\xb7\frac{15}{7}\xb7\frac{8}{23}\hfill \\ \\ \\ \text{Multiply left to right.}\hfill & & & \hfill 1\xb7\frac{8}{23}\hfill \\ \\ \\ \text{Multiply.}\hfill & & & \hfill \frac{8}{23}\hfill \end{array}$
Simplify: $\frac{9}{16}\xb7\frac{5}{49}\xb7\frac{16}{9}.$
$\frac{5}{49}$
Simplify: $\frac{6}{17}\xb7\frac{11}{25}\xb7\frac{17}{6}.$
$\frac{11}{25}$
Simplify: ⓐ $\frac{0}{m+7},$ where $m\ne \text{\u2212}7$ ⓑ $\frac{18-6c}{0},$ 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: $\frac{3}{4}\xb7\frac{4}{3}\left(6x+12\right).$
$\begin{array}{cccc}& & & \hfill \frac{3}{4}\xb7\frac{4}{3}\left(6x+12\right)\hfill \\ \\ \\ \begin{array}{c}\text{There is nothing to do in the parentheses,}\hfill \\ \text{so multiply the two fractions first\u2014notice,}\hfill \\ \text{they are reciprocals.}\hfill \end{array}\hfill & & & \hfill 1\left(6x+12\right)\hfill \\ \\ \\ \begin{array}{c}\text{Simplify by recognizing the multiplicative}\hfill \\ \text{identity.}\hfill \end{array}\hfill & & & \hfill 6x+12\hfill \end{array}$
Simplify: $\frac{2}{5}\xb7\frac{5}{2}\left(20y+50\right).$
$20y+50$
Simplify: $\frac{3}{8}\xb7\frac{8}{3}\left(12z+16\right).$
$12z+16$
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 .
Back to our friends at the movies, we could find the total amount of money they need like this:
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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