The first thing to note is that all fractions can be represented in many different ways. Thus 3 ⁄ 12, 4 ⁄ 16, 5 ⁄ 20 represent the same fraction which, in its lowest terms, is 1 ⁄ 4.
Example 1: Show that each of the above fractions is equal to ^{1}⁄_{4}.
Solution:
312 | = | 1 × 34 × 3 | = | 1 × 3/4 × 3/ | = | 14 |
416 | = | 1 × 44 × 4 | = | 1 × 4/4 × 4/ | = | 14 |
520 | = | 1 × 54 × 5 | = | 1 × 5/4 × 5/ | = | 14 |
The fractions in the left hand column each have a factor common to the numerator (top) and denominator (bottom) of that fraction, which is cancelled to give the fraction in its lowest terms.
In many problems it is necessary to carry out the reverse procedure, i.e. multiplying the numerator and denominator of a fraction by a common factor to obtain an equivalent fraction.
Example 2: Arrange each of the following fractions in the order of size: 3 ⁄ 4, 2 ⁄ 3, 5 ⁄ 6
Solution: To determine their relative order, each fraction must be written with the same denominator. The smallest such number in this case is 12.
34 | = | 3 × 34 × 3 | = | 912 |
23 | = | 2 × 42 × 4 | = | 812 |
56 | = | 5 × 26 × 2 | = | 1012 |
The order of size, staring with the smallest, is thus: 2 ⁄ 3, 3 ⁄ 4, 5 ⁄ 6.
Exercise 1: In each case below, arrange the fractions in increasing order of size.
Solution: The least common denominator of the fractions is 12, so
12 | = | 1 × 62 × 6 | = | 612 |
23 | = | 2 × 43 × 4 | = | 812 |
Since 7 ⁄ 12 already has 12 as a denominator, the required order of fractions is
Solution: The least common denominator of the fractions is 24, so
13 | = | 1 × 83 × 8 | = | 824 |
58 | = | 5 × 38 × 3 | = | 1524 |
34 | = | 3 × 64 × 6 | = | 1824 |
The fractions are thus increasing in the order they appear.
Solution: The least common denominator of the fractions is 36.
56 | = | 5 × 66 × 6 | = | 3036 |
89 | = | 8 × 49 × 4 | = | 3236 |
1112 | = | 11 × 312 × 3 | = | 3336 |
The fractions are thus increasing in the order they appear.
Solution: The least common denominator of the fractions is 30.
23 | = | 2 × 103 × 10 | = | 2030 |
35 | = | 3 × 65 × 6 | = | 1830 |
710 | = | 7 × 310 × 3 | = | 2130 |
The fractions, in increasing order of size, are:
Click on questions to reveal their solutions
Now try this short quiz.
Quiz 1: Of the following sets of fractions, only one is in ascending order of size. Which one?
Explanation: Putting each of them in the form of a fraction with denominator 24,
38 | = | 3 × 38 × 3 | = | 924 |
23 | = | 2 × 83 × 8 | = | 1624 |
34 | = | 3 × 64 × 6 | = | 1824 |
These are obviously in ascending order. Repeating this with the others will show that this is the only set in ascending order.
In this section we look at the addition (and subtraction) of fractions. If fractions are to be added then they must have the same denominators.
Example 3: Write the following sums of fractions as single fractions:
Solution: Taking all the fractions with denominator 24:
38 + 23 − 34 | = | 3 × 38 × 3 + 2 × 83 × 8 − 3 × 64 × 6 |
= | 924 + 1624 − 1824 | |
= | 9 + 16 − 1824 = 724 |
Solution: This time, taking all the fractions with denominator 12:
23 + 12 − 34 | = | 2 × 43 × 4 + 1 × 62 × 6 − 3 × 34 × 3 |
= | 812 + 612 − 912 | |
= | 8 + 6 − 912 = 512 |
The exercise below is designed to give you some practice at addition and subtraction of fractions.
Exercise 2: Evaluate the following, putting the final answer in its lowest terms.
Solution: The lowest common denominator is 24, so
13 + 18 | = | 1 × 83 × 8 + 1 × 38 × 3 |
= | 824 + 324 | |
= | 8+324 = 1124 |
Solution: Before proceeding, note that the second fraction is not in its lowest terms. Since 2 ⁄ 4 = (1 × 2) ⁄ (2 × 2) = 1 ⁄ 2,
56 + 24 | = | 56 + 12 |
= | 56 + 1 × 32 × 3 | |
= | 56 + 36 | |
= | 5+36 = 86 | |
= | 4 × 23 × 2 = 43 |
This fraction is called an improper fraction since the numerator is larger than the denominator. It is perfectly acceptable as a fraction but it may also be written as 1 ^{1}⁄_{3}.
Solution: The lowest common denominator is 20, so
34 + 45 | = | 3 × 54 × 5 + 4 × 45 × 4 |
= | 1520 + 1620 | |
= | 15+1620 = 3120 |
This is another improper fraction which may be written 1 ^{11}⁄_{20}
Solution: The least common denominator of the two fractions is 12 so
43 − 14 | = | 4 × 43 × 4 − 1 × 34 × 3 |
= | 1612 − 312 | |
= | 16−312 = 1312 |
This is another improper fraction which is equal to 1 ^{1}⁄_{12}
Solution: The least common denominator of the two fractions is 6 so
56 − 32 | = | 56 − 3 × 32 × 3 |
= | 56 − 96 | |
= | 5−96 = − 46 | |
= | − 2 × 23 × 2 = − 23 |
where the common factor 2 has been cancelled to obtain the final answer.
Solution: The least common denominator of the two fractions is 30 so
56 − 310 | = | 5 × 56 × 5 − 3 × 310 × 3 |
= | 2530 − 930 | |
= | 25−930 = 1630 | |
= | 8 × 215 × 2 = 815 |
where the final answer is obtained after cancellation of the common factor 2.
Click on questions to reveal their solutions
To finish this section there follows two simple quizzes.
Quiz 2: Which of the following fractions is the result of evaluating the sum: 34 − 23 + 16 ?
Explanation: The least common denominator of the three fractions is 12, so
34 − 23 + 16 | = | 3 × 34 × 3 − 2 × 43 × 4 + 1 × 26 × 2 |
= | 912 − 812 + 212 | |
= | 9 − 8 + 212 = 11 − 812 | |
= | 312 = 1 × 34 × 3 = 14 |
Quiz 3: From the following fractions, choose the one which is mid-way between 2/3 and 4/5:
Explanation: The least common denominator of the two fractions is 15. Writing both fractions with this denominator gives:
23 | = | 2 × 53 × 5 | = | 1015 |
45 | = | 4 × 35 × 3 | = | 1215 |
The fraction mid-way between ^{10}⁄_{15} and ^{12}⁄_{ 15} is thus ^{11}⁄_{15}
.Multiplication of fractions is straightforward, as illustrated by the following:
Example 4: Evaluate the following products:
Solution:
Solution:
In both cases the numerators and denominators have been multiplied and the common factors have been cancelled to leave the fractions in their lowest forms.
Division of fractions is straightforward once we note that division by a fraction of the form ^{a}/_{b} is equivalent to multiplication by the fraction ^{b}/_{a}. This can be seen in the following example:
Example 5: Write the following as fractions in the usual way:
Solution: First note that:
Divide both sides of this equation by (^{3}⁄_{4}):
( 34 ) × ( 43 ) | = | 1 |
∴ ( 43 ) | = | 1 / ( 34 ) = 1 ÷ ( 34 ) |
Solution: The previous result may be used.
( 18 ) ÷ ( 34 ) = ( ^{1}⁄_{8} )( ^{3}⁄_{4} ) | = | ( ^{1}⁄_{8} ) × 11 × ( ^{3}⁄_{4} ) = ( ^{1}⁄_{8} )1 × 1( ^{3}⁄_{4} ) | |
= | ( 18 ) × ( 43 ) | - using the result from (a) | |
= | 12 × 13 = 16 |
- where a common factor of 4 has been cancelled.
Exercise 3: Evaluate each of the following in their lowest terms.
Solution:
( 13 + 18 ) × 211 | = | 1124 × 211 |
= | 224 = 112 |
- where the common factors of 11 and 2 have been cancelled.
Solution: Using the result of exercise 2(b) we have:
( 56 + 12 ) ÷ 13 | = | 43 ÷ 13 |
= | 43 × 31 | |
= | 41 = 4 |
after recalling that division by ^{1}⁄_{3} is the same as multiplication by ^{3}⁄_{1} = 3, and cancelling the common factor 3.
Solution: From exercise 2(c), the numerator is ^{31}⁄_{20}. The denominator is evaluated below.
14 − 15 | = | 520 − 420 = 120 |
Thus
( ^{3}⁄_{4} + ^{4}⁄_{5} )( ^{1}⁄_{4} − ^{1}⁄_{5} ) | = | ( ^{31}⁄_{20} )( ^{1}⁄_{20} ) |
= | ( 3120 ) × ( 201 ) | |
= | 31 |
- after cancellation of the common factor 20.
Solution: In this case, from exercise 2(c), the denominator is ^{13}⁄_{12} . Evaluating the numerator,
56 − 32 | = | 56 − 96 = − 23 |
= | − 46 = − 23 |
after cancelling the common factor 2. Thus:
( ^{5}⁄_{6} − ^{3}⁄_{2} )( ^{4}⁄_{3} − ^{1}⁄_{4} ) | = | − ( ^{2}⁄_{3} )( ^{13}⁄_{12} ) = − (23) × (1213) |
= | − (23) × (4 × 313) | |
= | − (21) × (4 × 113) = − 813 |
- after cancellation of the common factor 3.
Click on questions to reveal their solutions
Evaluate each of the following and choose the answer from those given below.