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This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses division of fractions. By the end of the module students should be able to determine the reciprocal of a number and divide one fraction by another.

Section overview

  • Reciprocals
  • Dividing Fractions

Reciprocals

Reciprocals

Two numbers whose product is 1 are called reciprocals of each other.

Sample set a

The following pairs of numbers are reciprocals.

3 4 and 4 3 3 4 4 3 = 1

7 16 and 16 7 7 16 16 7 = 1

1 6 and 6 1 1 6 6 1 = 1

Notice that we can find the reciprocal of a nonzero number in fractional form by inverting it (exchanging positions of the numerator and denominator).

Practice set a

Find the reciprocal of each number.

3 10 size 12{ { {3} over {"10"} } } {}

10 3 size 12{ { {"10"} over {3} } } {}

2 3 size 12{ { {2} over {3} } } {}

3 2 size 12{ { {3} over {2} } } {}

7 8 size 12{ { {7} over {8} } } {}

8 7 size 12{ { {8} over {7} } } {}

1 5 size 12{ { {1} over {5} } } {}

5

2 2 7 size 12{2 { {2} over {7} } } {}

Write this number as an improper fraction first.

7 16 size 12{ { {7} over {"16"} } } {}

5 1 4 size 12{5 { {1} over {4} } } {}

4 21 size 12{ { {4} over {"21"} } } {}

10 3 16 size 12{"10" { {3} over {"16"} } } {}

16 163 size 12{ { {"16"} over {"163"} } } {}

Dividing fractions

Our concept of division is that it indicates how many times one quantity is con­tained in another quantity. For example, using the diagram we can see that there are 6 one-thirds in 2.

Two rectangles, each divided into three parts. The rectangles are connected to each other. There are 6 one-thirds in 2.

Since 2 contains six 1 3 size 12{ { {1} over {3} } } {} 's we express this as

Two divided by one-third is equal to six. Note also that two times three is equal to six, because one-third and three are reciprocals.

Using these observations, we can suggest the following method for dividing a number by a fraction.

Dividing one fraction by another fraction

To divide a first fraction by a second, nonzero fraction, multiply the first traction by the reciprocal of the second fraction.

Invert and multiply

This method is commonly referred to as "invert the divisor and multiply."

Sample set b

Perform the following divisions.

1 3 ÷ 3 4 size 12{ { {1} over {3} } div { {3} over {4} } } {} . The divisor is 3 4 size 12{ { {3} over {4} } } {} . Its reciprocal is 4 3 size 12{ { {4} over {3} } } {} . Multiply 1 3 size 12{ { {1} over {3} } } {} by 4 3 size 12{ { {4} over {3} } } {} .

1 3 4 3 = 1 4 3 3 = 4 9 size 12{ { {1} over {3} } cdot { {4} over {3} } = { {1 cdot 4} over {3 cdot 3} } = { {4} over {9} } } {}

1 3 ÷ 3 4 = 4 9 size 12{ { {1} over {3} } div { {3} over {4} } = { {4} over {9} } } {}

3 8 ÷ 5 4 size 12{ { {3} over {8} } div { {5} over {4} } } {} The divisor is 5 4 size 12{ { {5} over {4} } } {} . Its reciprocal is 4 5 size 12{ { {4} over {5} } } {} . Multiply 3 8 size 12{ { {3} over {8} } } {} by 4 5 size 12{ { {4} over {5} } } {} .

3 3 2 4 1 5 = 3 1 2 5 = 3 10 size 12{ { {3} over { { { {3}}} cSub { size 8{2} } } } cdot { { { { {4}}} cSup { size 8{1} } } over {5} } = { {3 cdot 1} over {2 cdot 5} } = { {3} over {"10"} } } {}

3 8 ÷ 5 4 = 3 10 size 12{ { {3} over {8} } div { {5} over {4} } = { {3} over {"10"} } } {}

5 6 ÷ 5 12 size 12{ { {5} over {6} } div { {5} over {"12"} } } {} . The divisor is 5 12 size 12{ { {5} over {"12"} } } {} . Its reciprocal is 12 5 size 12{ { {"12"} over {5} } } {} . Multiply 5 6 size 12{ { {5} over {6} } } {} by 12 5 size 12{ { {"12"} over {5} } } {} .

5 1 6 1 12 2 5 1 = 1 2 1 1 = 2 1 = 2 size 12{ { { { { {5}}} cSup { size 8{1} } } over { { { {6}}} cSub { size 8{1} } } } cdot { { {"12"} cSup { size 8{2} } } over { {5} cSub { size 8{1} } } } = { {1 cdot 2} over {1 cdot 1} } = { {2} over {1} } =2} {}

5 6 ÷ 5 12 = 2

2 2 9 ÷ 3 1 3 size 12{2 { {2} over {9} } div 3 { {1} over {3} } } {} . Convert each mixed number to an improper fraction.

2 2 9 = 9 2 + 2 9 = 20 9 size 12{2 { {2} over {9} } = { {9 cdot 2+2} over {9} } = { {"20"} over {9} } } {} .

3 1 3 = 3 3 + 1 3 = 10 3 size 12{3 { {1} over {3} } = { {3 cdot 3+1} over {3} } = { {10} over {3} } } {} .

20 9 ÷ 10 3 size 12{ { {"20"} over {9} } div { {"10"} over {3} } } {} The divisor is 10 3 size 12{ { {"10"} over {3} } } {} . Its reciprocal is 3 10 size 12{ { {3} over {"10"} } } {} . Multiply 20 9 size 12{ { {"20"} over {9} } } {} by 3 10 size 12{ { {3} over {"10"} } } {} .

20 2 9 3 3 1 10 1 = 2 1 3 1 = 2 3 size 12{ { { { { {2}} { {0}}} cSup { size 8{2} } } over { { { {9}}} cSub { size 8{3} } } } cdot { { { { {3}}} cSup { size 8{1} } } over { { { {1}} { {0}}} cSub { size 8{1} } } } = { {2 cdot 1} over {3 cdot 1} } = { {2} over {3} } } {}

2 2 9 ÷ 3 1 3 = 2 3 size 12{2 { {2} over {9} } div 3 { {1} over {3} } = { {2} over {3} } } {}

12 11 ÷ 8 size 12{ { {"12"} over {"11"} } div 8} {} . First conveniently write 8 as 8 1 size 12{ { {8} over {1} } } {} .

12 11 ÷ 8 1 size 12{ { {"12"} over {"11"} } div { {8} over {1} } } {} The divisor is 8 1 size 12{ { {8} over {1} } } {} . Its reciprocal is 1 8 size 12{ { {1} over {8} } } {} . Multiply 12 11 size 12{ { {"12"} over {"11"} } } {} by 1 8 size 12{ { {1} over {8} } } {} .

12 3 11 1 8 2 = 3 1 11 2 = 3 22 size 12{ { { { { {1}} { {2}}} cSup { size 8{3} } } over {"11"} } cdot { {1} over { { { {8}}} cSub { size 8{2} } } } = { {3 cdot 1} over {"11" cdot 2} } = { {3} over {"22"} } } {}

12 11 ÷ 8 = 3 22 size 12{ { {"12"} over {"11"} } div 8= { {3} over {"22"} } } {}

7 8 ÷ 21 20 3 35 size 12{ { {7} over {8} } div { {"21"} over {"20"} } cdot { {3} over {"35"} } } {} . The divisor is 21 20 size 12{ { {"21"} over {"20"} } } {} . Its reciprocal is 20 21 size 12{ { {"20"} over {"21"} } } {} .

7 1 8 2 20 5 1 21 3 1 3 1 35 7 = 1 1 1 2 1 7 = 1 14 size 12{ { { {7} cSup { size 8{1} } } over { {8} cSub { size 8{2} } } } cdot { { {"20"} cSup { size 8{ {5} cSup { size 6{1} } } } } over { {"21"} cSub { {3} cSub { size 6{1} } } } } size 12{ cdot { { {3} cSup {1} } over { size 12{ {"35"} cSub {7} } } } } size 12{ {}= { {1 cdot 1 cdot 1} over {2 cdot 1 cdot 7} } = { {1} over {"14"} } }} {}

7 8 ÷ 21 20 3 25 = 1 14 size 12{ { {7} over {8} } div { {"21"} over {"20"} } cdot { {3} over {"25"} } = { {1} over {"14"} } } {}

How many 2 3 8 size 12{2 { {3} over {8} } } {} -inch-wide packages can be placed in a box 19 inches wide?

The problem is to determine how many two and three eighths are contained in 19, that is, what is 19 ÷ 2 3 8 size 12{"19" div 2 { {3} over {8} } } {} ?

2 3 8 = 19 8 size 12{2 { {3} over {8} } = { {"19"} over {8} } } {} Convert the divisor 2 3 8 size 12{2 { {3} over {8} } } {} to an improper fraction.

19 = 19 1 size 12{"19"= { {"19"} over {1} } } {} Write the dividend 19 as 19 1 size 12{ { {"19"} over {1} } } {} .

19 1 ÷ 19 8 size 12{ { {"19"} over {1} } div { {"19"} over {8} } } {} The divisor is 19 8 size 12{ { {"19"} over {8} } } {} . Its reciprocal is 8 19 size 12{ { {8} over {"19"} } } {} .

19 1 1 8 19 1 = 1 8 1 1 = 8 1 = 8 size 12{ { { {"19"} cSup { size 8{1} } } over {1} } cdot { {8} over { {"19"} cSub { size 8{1} } } } = { {1 cdot 8} over {1 cdot 1} } = { {8} over {1} } =8} {}

Thus, 8 packages will fit into the box.

Practice set b

Perform the following divisions.

1 2 ÷ 9 8 size 12{ { {1} over {2} } div { {9} over {8} } } {}

4 9 size 12{ { {4} over {9} } } {}

3 8 ÷ 9 24 size 12{ { {3} over {8} } div { {9} over {"24"} } } {}

1

7 15 ÷ 14 15 size 12{ { {7} over {"15"} } div { {"14"} over {"15"} } } {}

1 2 size 12{ { {1} over {2} } } {}

8 ÷ 8 15 size 12{8 div { {8} over {"15"} } } {}

15

6 1 4 ÷ 5 12 size 12{6 { {1} over {4} } div { {5} over {"12"} } } {}

15

3 1 3 ÷ 1 2 3 size 12{3 { {1} over {3} } div 1 { {2} over {3} } } {}

2

5 6 ÷ 2 3 8 25 size 12{ { {5} over {6} } div { {2} over {3} } cdot { {8} over {"25"} } } {}

2 5 size 12{ { {2} over {5} } } {}

A container will hold 106 ounces of grape juice. How many 6 5 8 size 12{6 { {5} over {8} } } {} -ounce glasses of grape juice can be served from this container?

16 glasses

Determine each of the following quotients and then write a rule for this type of division.

1 ÷ 2 3 size 12{1 div { {2} over {3} } } {}

3 2 size 12{ { {3} over {2} } } {}

1 ÷ 3 8 size 12{1 div { {3} over {8} } } {}

8 3 size 12{ { {8} over {3} } } {}

1 ÷ 3 4 size 12{1 div { {3} over {4} } } {}

4 3 size 12{ { {4} over {3} } } {}

1 ÷ 5 2 size 12{1 div { {5} over {2} } } {}

2 5 size 12{ { {2} over {5} } } {}

When dividing 1 by a fraction, the quotient is the .

is the reciprocal of the fraction.

Exercises

For the following problems, find the reciprocal of each number.

4 5 size 12{ { {4} over {5} } } {}

5 4 size 12{ { {5} over {4} } } {} or 1 1 4 size 12{1 { {1} over {4} } } {}

8 11 size 12{ { {8} over {"11"} } } {}

2 9 size 12{ { {2} over {9} } } {}

9 2 size 12{ { {9} over {2} } } {} or 4 1 2 size 12{4 { {1} over {2} } } {}

1 5 size 12{ { {1} over {5} } } {}

3 1 4 size 12{3 { {1} over {4} } } {}

4 13 size 12{ { {4} over {"13"} } } {}

8 1 4 size 12{8 { {1} over {4} } } {}

3 2 7 size 12{3 { {2} over {7} } } {}

7 23 size 12{ { {7} over {"23"} } } {}

5 3 4 size 12{5 { {3} over {4} } } {}

1

1

4

For the following problems, find each value.

3 8 ÷ 3 5 size 12{ { {3} over {8} } div { {3} over {5} } } {}

5 8 size 12{ { {5} over {8} } } {}

5 9 ÷ 5 6 size 12{ { {5} over {9} } div { {5} over {6} } } {}

9 16 ÷ 15 8 size 12{ { {9} over {"16"} } div { {"15"} over {8} } } {}

3 10 size 12{ { {3} over {"10"} } } {}

4 9 ÷ 6 15 size 12{ { {4} over {9} } div { {6} over {"15"} } } {}

25 49 ÷ 4 9 size 12{ { {"25"} over {"49"} } div { {4} over {9} } } {}

225 196 size 12{ { {"225"} over {"196"} } } {} or 1 29 196 size 12{1 { {"29"} over {"196"} } } {}

15 4 ÷ 27 8 size 12{ { {"15"} over {4} } div { {"27"} over {8} } } {}

24 75 ÷ 8 15 size 12{ { {"24"} over {"75"} } div { {8} over {"15"} } } {}

3 5 size 12{ { {3} over {5} } } {}

5 7 ÷ 0 size 12{ { {5} over {7} } div 0} {}

7 8 ÷ 7 8 size 12{ { {7} over {8} } div { {7} over {8} } } {}

1

0 ÷ 3 5 size 12{0 div { {3} over {5} } } {}

4 11 ÷ 4 11 size 12{ { {4} over {"11"} } div { {4} over {"11"} } } {}

1

2 3 ÷ 2 3 size 12{ { {2} over {3} } div { {2} over {3} } } {}

7 10 ÷ 10 7 size 12{ { {7} over {"10"} } div { {"10"} over {7} } } {}

49 100 size 12{ { {"49"} over {"100"} } } {}

3 4 ÷ 6 size 12{ { {3} over {4} } div 6} {}

9 5 ÷ 3 size 12{ { {9} over {5} } div 3} {}

3 5 size 12{ { {3} over {5} } } {}

4 1 6 ÷ 3 1 3 size 12{4 { {1} over {6} } div 3 { {1} over {3} } } {}

7 1 7 ÷ 8 1 3 size 12{7 { {1} over {7} } div 8 { {1} over {3} } } {}

6 7 size 12{ { {6} over {7} } } {}

1 1 2 ÷ 1 1 5 size 12{1 { {1} over {2} } div 1 { {1} over {5} } } {}

3 2 5 ÷ 6 25 size 12{3 { {2} over {5} } div { {6} over {"25"} } } {}

85 6 size 12{ { {"85"} over {6} } } {} or 14 1 6 size 12{"14" { {1} over {6} } } {}

5 1 6 ÷ 31 6 size 12{5 { {1} over {6} } div { {"31"} over {6} } } {}

35 6 ÷ 3 3 4 size 12{ { {"35"} over {6} } div 3 { {3} over {4} } } {}

28 18 = 14 9 size 12{ { {"28"} over {"18"} } = { {"14"} over {9} } } {} or 1 5 9 size 12{1 { {5} over {9} } } {}

5 1 9 ÷ 1 18 size 12{5 { {1} over {9} } div { {1} over {"18"} } } {}

8 3 4 ÷ 7 8 size 12{8 { {3} over {4} } div { {7} over {8} } } {}

10

12 8 ÷ 1 1 2 size 12{ { {"12"} over {8} } div 1 { {1} over {2} } } {}

3 1 8 ÷ 15 16 size 12{3 { {1} over {8} } div { {"15"} over {"16"} } } {}

10 3 size 12{ { {"10"} over {3} } } {} or 3 1 3 size 12{3 { {1} over {3} } } {}

11 11 12 ÷ 9 5 8 size 12{"11" { {"11"} over {"12"} } div 9 { {5} over {8} } } {}

2 2 9 ÷ 11 2 3 size 12{2 { {2} over {9} } div "11" { {2} over {3} } } {}

4 21 size 12{ { {4} over {"21"} } } {}

16 3 ÷ 6 2 5 size 12{ { {"16"} over {3} } div 6 { {2} over {5} } } {}

4 3 25 ÷ 2 56 75 size 12{4 { {3} over {"25"} } div 2 { {"56"} over {"75"} } } {}

3 2 size 12{ { {3} over {2} } } {} or 1 1 2 size 12{1 { {1} over {2} } } {}

1 1000 ÷ 1 100 size 12{ { {1} over {"1000"} } div { {1} over {"100"} } } {}

3 8 ÷ 9 16 6 5 size 12{ { {3} over {8} } div { {9} over {"16"} } cdot { {6} over {5} } } {}

4 5 size 12{ { {4} over {5} } } {}

3 16 9 8 6 5 size 12{ { {3} over {"16"} } cdot { {9} over {8} } cdot { {6} over {5} } } {}

4 15 ÷ 2 25 9 10 size 12{ { {4} over {"15"} } div { {2} over {"25"} } cdot { {9} over {"10"} } } {}

3

21 30 1 1 4 ÷ 9 10 size 12{ { {"21"} over {"30"} } cdot 1 { {1} over {4} } div { {9} over {"10"} } } {}

8 1 3 36 75 ÷ 4 size 12{8 { {1} over {3} } cdot { {"36"} over {"75"} } div 4} {}

1

Exercises for review

( [link] ) What is the value of 5 in the number 504,216?

( [link] ) Find the product of 2,010 and 160.

321,600

( [link] ) Use the numbers 8 and 5 to illustrate the commutative property of multiplication.

( [link] ) Find the least common multiple of 6, 16, and 72.

144

( [link] ) Find 8 9 size 12{ { {8} over {9} } } {} of 6 3 4 size 12{6 { {3} over {4} } } {} .

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
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salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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Abhi
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kinnecy Reply
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ninjadapaul
20/(×-6^2)
Salomon
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ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Contemporary math applications. OpenStax CNX. Dec 15, 2014 Download for free at http://legacy.cnx.org/content/col11559/1.6
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