# 3.5 Multiplication and division of signed numbers

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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The basic operations with real numbers are presented in this chapter. The concept of absolute value is discussed both geometrically and symbolically. The geometric presentation offers a visual understanding of the meaning of |x|. The symbolic presentation includes a literal explanation of how to use the definition. Negative exponents are developed, using reciprocals and the rules of exponents the student has already learned. Scientific notation is also included, using unique and real-life examples.Objectives of this module: be able to multiply and divide signed numbers.

## Overview

• Multiplication of Signed Numbers
• Division of Signed Numbers

## Multiplication of signed numbers

Let us consider first the product of two positive numbers.

Multiply: $3\cdot 5$ .
$3\cdot 5$ means $5+5+5=15$ .

This suggests that

$\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)\cdot \left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)=\text{positive}\text{\hspace{0.17em}}\text{number}$ .

More briefly, $\left(+\right)\left(+\right)=+$ .

Now consider the product of a positive number and a negative number.

Multiply: $\left(3\right)\left(-5\right)$ .
$\left(3\right)\left(-5\right)$ means $\left(-5\right)+\left(-5\right)+\left(-5\right)=-15$ .

This suggests that

$\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)\cdot \left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)=\text{negative}\text{\hspace{0.17em}}\text{number}$

More briefly, $\left(+\right)\left(-\right)=-$ .

By the commutative property of multiplication, we get

$\left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)\cdot \left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)=\text{negative}\text{\hspace{0.17em}}\text{number}$

More briefly, $\left(-\right)\left(+\right)=-$ .

The sign of the product of two negative numbers can be determined using the following illustration: Multiply $-2$ by, respectively, $4,\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}0,\text{\hspace{0.17em}}-1,\text{\hspace{0.17em}}-2,\text{\hspace{0.17em}}-3,\text{\hspace{0.17em}}-4$ . Notice that when the multiplier decreases by 1, the product increases by 2.

$\begin{array}{ll}\begin{array}{l}4\left(-2\right)=-8\\ 3\left(-2\right)=-6\\ 2\left(-2\right)=-4\\ 1\left(-2\right)=-2\end{array}\right\}\to \hfill & \text{As}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{know},\text{\hspace{0.17em}}\left(+\right)\left(-\right)=-.\hfill \\ 0\left(-2\right)=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\to \hfill & \text{As}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{know},0\cdot \left(\text{any}\text{\hspace{0.17em}}\text{number}\right)=0.\hfill \end{array}$

$\begin{array}{ll}\begin{array}{l}-1\left(-2\right)=2\\ -2\left(-2\right)=4\\ -3\left(-2\right)=6\\ -4\left(-2\right)=8\end{array}\right\}\to \hfill & \text{This}\text{\hspace{0.17em}}\text{pattern}\text{\hspace{0.17em}}\text{suggests}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-\right)\left(-\right)=+.\hfill \end{array}$

We have the following rules for multiplying signed numbers.

## Rules for multiplying signed numbers

To multiply two real numbers that have

1. the same sign , multiply their absolute values. The product is positive.
$\begin{array}{c}\left(+\right)\left(+\right)=+\\ \left(-\right)\left(-\right)=+\end{array}$
2. opposite signs , multiply their absolute values. The product is negative.
$\begin{array}{c}\left(+\right)\left(-\right)=-\\ \left(-\right)\left(+\right)=-\end{array}$

## Sample set a

Find the following products.

$8\cdot 6$

$\begin{array}{ll}\hfill & \text{Multiply}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}\text{.}\hfill \\ \begin{array}{l}|8|=8\\ |6|=6\end{array}\right\}\hfill & 8\cdot 6=48\hfill \\ \hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{same}\text{\hspace{0.17em}}\text{sign,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{product}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{positive}\text{.}\hfill \\ 8\cdot 6=+48\hfill & \text{​}\text{​}\text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8\cdot 6=48\hfill \end{array}$

$\left(-8\right)\left(-6\right)$

$\begin{array}{ll}\hfill & \text{Multiply}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}\text{.}\hfill \\ \begin{array}{l}|-8|=8\\ |-6|=6\end{array}\right\}\hfill & 8\cdot 6=48\hfill \\ \hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{same}\text{\hspace{0.17em}}\text{sign,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{product}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{positive}\text{.}\hfill \\ \left(-8\right)\left(-6\right)=+48\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-8\right)\left(-6\right)=48\hfill \end{array}$

$\left(-4\right)\left(7\right)$

$\begin{array}{ll}\hfill & \text{Multiply}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}\text{.}\hfill \\ \begin{array}{cc}|-4|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}& =4\\ |7|& =7\end{array}\right\}\hfill & 4\cdot 7=28\hfill \\ \hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{opposite}\text{\hspace{0.17em}}\text{signs,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{product}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{negative}\text{.}\hfill \\ \left(-4\right)\left(7\right)=-28\hfill & \hfill \end{array}$

$6\left(-3\right)$

$\begin{array}{ll}\hfill & \text{Multiply}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}\text{.}\hfill \\ \begin{array}{rr}\hfill |6|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}& \hfill =6\\ \hfill |-3|& \hfill =3\end{array}\right\}\hfill & 6\cdot 3=18\hfill \\ \hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{opposite}\text{\hspace{0.17em}}\text{signs,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{product}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{negative}\text{.}\hfill \\ 6\left(-3\right)=-18\hfill & \hfill \end{array}$

## Practice set a

Find the following products.

$3\left(-8\right)$

$-24$

$4\left(16\right)$

64

$\left(-6\right)\left(-5\right)$

30

$\left(-7\right)\left(-2\right)$

14

$\left(-1\right)\left(4\right)$

$-4$

$\left(-7\right)7$

$-49$

## Division of signed numbers

We can determine the sign pattern for division by relating division to multiplication. Division is defined in terms of multiplication in the following way.

If $b\cdot c=a$ , then $\frac{a}{b}=c,\text{\hspace{0.17em}}b\ne 0$ .

For example, since $3\cdot 4=12$ , it follows that $\frac{12}{3}=4$ .

Notice the pattern:

Since $\underset{b\cdot c=a}{\underbrace{3\cdot 4}}=12$ , it follows that $\underset{\frac{a}{b}=c}{\underbrace{\frac{12}{3}}}=4$

The sign pattern for division follows from the sign pattern for multiplication.

1. Since $\underset{b\cdot c=a}{\underbrace{\left(+\right)\left(+\right)}}=+$ , it follows that $\underset{\frac{a}{b}=c}{\underbrace{\frac{\left(+\right)}{\left(+\right)}}}=+$ , that is,

$\frac{\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)}{\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)}=\text{positive}\text{\hspace{0.17em}}\text{number}$

2. Since $\underset{b\cdot c=a}{\underbrace{\left(-\right)\left(-\right)}}=+$ , it follows that $\underset{\frac{a}{b}=c}{\underbrace{\frac{\left(+\right)}{\left(-\right)}}}=-$ , that is,

$\frac{\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)}{\left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)}=\text{negative}\text{\hspace{0.17em}}\text{number}$

3. Since $\underset{b\cdot c=a}{\underbrace{\left(+\right)\left(-\right)}}=-$ , it follows that $\underset{\frac{a}{b}=c}{\underbrace{\frac{\left(-\right)}{\left(+\right)}}}=-$ , that is,

$\frac{\left(\text{negative}\text{​}\text{\hspace{0.17em}}\text{number}\right)}{\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)}=\text{negative}\text{\hspace{0.17em}}\text{number}$

4. Since $\underset{b\cdot c=a}{\underbrace{\left(-\right)\left(+\right)}}=-$ , it follows that $\underset{\frac{a}{b}=c}{\underbrace{\frac{\left(-\right)}{\left(-\right)}}}=+$ , that is

$\frac{\left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)}{\left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)}=\text{positive}\text{\hspace{0.17em}}\text{number}$

We have the following rules for dividing signed numbers.

## Rules for dividing signed numbers

To divide two real numbers that have

1. the same sign , divide their absolute values. The quotient is positive.
$\begin{array}{cc}\frac{\left(+\right)}{\left(+\right)}=+& \frac{\left(-\right)}{\left(-\right)}=+\end{array}$
2. opposite signs , divide their absolute values. The quotient is negative.
$\begin{array}{cc}\frac{\left(-\right)}{\left(+\right)}=-& \frac{\left(+\right)}{\left(-\right)}=-\end{array}$

## Sample set b

Find the following quotients.

$\frac{-10}{2}$

$\begin{array}{ll}\begin{array}{ll}|-10|=\hfill & 10\hfill \\ \hfill |2|=& \hfill 2\end{array}\right\}\hfill & \text{Divide}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}.\hfill \\ \hfill & \frac{10}{2}=5\hfill \\ \frac{-10}{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-5\hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{opposite}\text{\hspace{0.17em}}\text{signs,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{quotient}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{negative}\text{.}\hfill \end{array}$

$\frac{-35}{-7}$

$\begin{array}{ll}\begin{array}{ll}|-35|=\hfill & 35\hfill \\ \hfill |-7|=& \hfill 7\end{array}\right\}\hfill & \text{Divide}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}.\hfill \\ \hfill & \frac{35}{7}=5\hfill \\ \frac{-35}{-7}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5\hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{same}\text{\hspace{0.17em}}\text{signs,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{quotient}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{positive}\text{.}\hfill \end{array}$

$\frac{18}{-9}$

$\begin{array}{ll}\begin{array}{ll}|18|=\hfill & 18\hfill \\ |-9|=\hfill & 9\hfill \end{array}\right\}\hfill & \text{Divide}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}.\hfill \\ \hfill & \frac{18}{9}=2\hfill \\ \frac{18}{-9}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-2\hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{opposite}\text{\hspace{0.17em}}\text{signs,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{quotient}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{negative}\text{.}\hfill \end{array}$

## Practice set b

Find the following quotients.

$\frac{-24}{-6}$

4

$\frac{30}{-5}$

$-6$

$\frac{-54}{27}$

$-2$

$\frac{51}{17}$

3

## Sample set c

Find the value of $\frac{-6\left(4-7\right)-2\left(8-9\right)}{-\left(4+1\right)+1}$ .

Using the order of operations and what we know about signed numbers, we get

$\begin{array}{lll}\frac{-6\left(4-7\right)-2\left(8-9\right)}{-\left(4+1\right)+1}\hfill & =\hfill & \frac{-6\left(-3\right)-2\left(-1\right)}{-\left(5\right)+1}\hfill \\ \hfill & =\hfill & \frac{18+2}{-5+1}\hfill \\ \hfill & =\hfill & \frac{20}{-4}\hfill \\ \hfill & =\hfill & -5\hfill \end{array}$

Find the value of $z=\frac{x-u}{s}$ if $x=57,\text{\hspace{0.17em}}u=51,\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}s=2$ .

Substituting these values we get

$z=\frac{57-51}{2}=\frac{6}{2}=3$

## Practice set c

Find the value of $\frac{-7\left(4-8\right)+2\left(1-11\right)}{-5\left(1-6\right)-17}$ .

1

Find the value of $P=\frac{n\left(n-3\right)}{2n}$ , if $n=5$ .

1

## Exercises

Find the value of each of the following expressions.

$\left(-2\right)\left(-8\right)$

16

$\left(-3\right)\left(-9\right)$

$\left(-4\right)\left(-8\right)$

32

$\left(-5\right)\left(-2\right)$

$\left(-6\right)\left(-9\right)$

54

$\left(-3\right)\left(-11\right)$

$\left(-8\right)\left(-4\right)$

32

$\left(-1\right)\left(-6\right)$

$\left(3\right)\left(-12\right)$

$-36$

$\left(4\right)\left(-18\right)$

$8\left(-4\right)$

$-32$

$5\left(-6\right)$

$9\left(-2\right)$

$-18$

$7\left(-8\right)$

$\left(-6\right)4$

$-24$

$\left(-7\right)6$

$\left(-10\right)9$

$-90$

$\left(-4\right)12$

$\left(10\right)\left(-6\right)$

$-60$

$\left(-6\right)\left(4\right)$

$\left(-2\right)\left(6\right)$

$-12$

$\left(-8\right)\left(7\right)$

$\frac{21}{7}$

3

$\frac{42}{6}$

$\frac{-39}{3}$

$-13$

$\frac{-20}{10}$

$\frac{-45}{-5}$

9

$\frac{-16}{-8}$

$\frac{25}{-5}$

$-5$

$\frac{36}{-4}$

$8-\left(-3\right)$

11

$14-\left(-20\right)$

$20-\left(-8\right)$

28

$-4-\left(-1\right)$

$0-4$

$-4$

$0-\left(-1\right)$

$-6+1-7$

$-12$

$15-12-20$

$1-6-7+8$

$-4$

$2+7-10+2$

$3\left(4-6\right)$

$-6$

$8\left(5-12\right)$

$-3\left(1-6\right)$

15

$-8\left(4-12\right)+2$

$-4\left(1-8\right)+3\left(10-3\right)$

49

$-9\left(0-2\right)+4\left(8-9\right)+0\left(-3\right)$

$6\left(-2-9\right)-6\left(2+9\right)+4\left(-1-1\right)$

$-140$

$\frac{3\left(4+1\right)-2\left(5\right)}{-2}$

$\frac{4\left(8+1\right)-3\left(-2\right)}{-4-2}$

$-7$

$\frac{-1\left(3+2\right)+5}{-1}$

$\frac{-3\left(4-2\right)+\left(-3\right)\left(-6\right)}{-4}$

$-3$

$-1\left(4+2\right)$

$-1\left(6-1\right)$

$-5$

$-\left(8+21\right)$

$-\left(8-21\right)$

13

$-\left(10-6\right)$

$-\left(5-2\right)$

$-3$

$-\left(7-11\right)$

$-\left(8-12\right)$

4

$-3\left[\left(-1+6\right)-\left(2-7\right)\right]$

$-2\left[\left(4-8\right)-\left(5-11\right)\right]$

$-4$

$-5\left[\left(-1+5\right)+\left(6-8\right)\right]$

$-\left[\left(4-9\right)+\left(-2-8\right)\right]$

15

$-3\left[-2\left(1-5\right)-3\left(-2+6\right)\right]$

$-2\left[-5\left(-10+11\right)-2\left(5-7\right)\right]$

2

$\begin{array}{cc}P=R-C.& \text{Find}\text{\hspace{0.17em}}P\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}R=2000\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}C=2500.\end{array}$

$\begin{array}{cc}z=\frac{x-u}{s}.& \text{Find}\text{\hspace{0.17em}}z\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x=23,\text{\hspace{0.17em}}u=25,\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}s=1.\end{array}$

$-2$

$\begin{array}{cc}z=\frac{x-u}{s}.& \text{Find}\text{\hspace{0.17em}}z\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x=410,\text{\hspace{0.17em}}u=430,\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}s=2.5.\end{array}$

$\begin{array}{cc}m=\frac{2s+1}{T}.& \text{Find}\text{\hspace{0.17em}}m\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}s=-8\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}T=5.\end{array}$

$-3$

$\begin{array}{cc}m=\frac{2s+1}{T}.& \text{Find}\text{\hspace{0.17em}}m\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}s=-10\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}T=-5.\end{array}$

$\begin{array}{cc}F=\left({p}_{1}-{p}_{2}\right){r}^{4}\cdot 9.& \text{Find}\text{\hspace{0.17em}}F\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}{p}_{1}=10,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{p}_{2}=8,r=3.\end{array}$

1458

$\begin{array}{cc}F=\left({p}_{1}-{p}_{2}\right){r}^{4}\cdot 9.& \text{Find}\text{\hspace{0.17em}}F\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}{p}_{1}=12,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{p}_{2}=7,r=2.\end{array}$

$\begin{array}{cc}P=n\left(n-1\right)\left(n-2\right).& \text{Find}\text{\hspace{0.17em}}P\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}n=-4.\end{array}$

$-120$

$\begin{array}{cc}P=n\left(n-1\right)\left(n-2\right)\left(n-3\right).& \text{Find}\text{\hspace{0.17em}}P\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}n=-5.\end{array}$

$\begin{array}{cc}P=\frac{n\left(n-2\right)\left(n-4\right)}{2n}.& \text{Find}\text{\hspace{0.17em}}P\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}n=-6.\end{array}$

40

## Exercises for review

( [link] ) What natural numbers can replace $x$ so that the statement $-4 is true?

( [link] ) Simplify $\frac{{\left(x+2y\right)}^{5}{\left(3x-1\right)}^{7}}{{\left(x+2y\right)}^{3}{\left(3x-1\right)}^{6}}$ .

${\left(x+2y\right)}^{2}\left(3x-1\right)$

( [link] ) Simplify ${\left({x}^{n}{y}^{3t}\right)}^{5}$ .

( [link] ) Find the sum. $-6+\left(-5\right)$ .

$-11$

( [link] ) Find the difference. $-2-\left(-8\right)$ .

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