In the previous section we learned about properties of equivalent fractions. In this section we will see the comparison of unequal fractions.
Consider the following points:
• We are given some fractions
• All those given fractions have the same denominators. The numerators are different
■ Then, the fraction with the smallest numerator will be the smallest fraction
■ The fraction with the largest numerator will be the largest fraction
• Example: Given some fractions: 6⁄11, 3⁄11, 8⁄11, 4⁄11 ,9⁄11
• The smallest is 3⁄11, and the largest is 9⁄11
We have discussed the reason behind the above result here. Now we consider another such property:
Consider the following points:
• We are given some fractions
• All those given fractions have the same numerators. The denominators are different
■ Then, the fraction with the smallest denominator will be the largest fraction
■ The fraction with the largest denominator will be the smallest fraction
• Example: Given some fractions: 9⁄17, 9⁄12, 9⁄14, 9⁄21, 9⁄10
• The smallest is 9⁄21, and the largest is 9⁄10
We have discussed the reason behind the above result here.
Let us see some applications of the above two properties:
■ Compare 3⁄4 and 3⁄11
Ans: 3⁄11 < 3⁄4 (∵ when two fractions have the same numerator, that fraction with the largest denominator is the smallest)
■ Compare 7⁄5 and 7⁄9
Ans: 7⁄9 < 7⁄5 (∵ when two fractions have the same numerator, that fraction with the largest denominator is the smallest)
■ Compare 3⁄12 and 5⁄12
Ans: 3⁄12 < 5⁄12 (∵ when two fractions have the same denominator, that fraction with the smallest numerator is the smallest)
■ Compare 4⁄7 and 2⁄7
Ans: 2⁄7 < 4⁄7 (∵ when two fractions have the same denominator, that fraction with the smallest numerator is the smallest)
In some situations, the above two properties can be used together. Let us see an example:
■ Compare 3⁄7 and 4⁄5
In this case, neither the numerators nor the denominators are equal. So how do we proceed?
1. We have to compare 3⁄7 and 4⁄5
2. Set up a new fraction.
♦ It must have the same numerator as the first fraction 3⁄7
♦ It must have the same denominator as the second fraction 4⁄5
3. So the new fraction is: 3⁄5
4. Compare the new fraction with the first fraction. We get: 3⁄7 < 3⁄5. (∵ when two fractions have the same numerator, that fraction with the largest denominator is the smallest)
5. Compare the new fraction with the second fraction. We get 3⁄5 < 4⁄5. (∵ when two fractions have the same denominator, that fraction with the smallest numerator is the smallest)
6. So we have two results: 3⁄7 < 3⁄5 and 3⁄5 < 4⁄5
7. we can make an important conclusion from the above:
3⁄7 is less than 3⁄5. But this 3⁄5 is less than 4⁄5. So obviously, 3⁄7 will be less than 4⁄5
8. So the result of the comparison between 3⁄7 and 4⁄5 is: 3⁄7 < 4⁄5
■ The above steps can be summarised as follows:
• We have a fraction 3⁄7 in hand
• We are given a second fraction 4⁄5 for comparison
• If the second fraction has an increased numerator, and a decreased denominator, then that second fraction will be greater than the first fraction
Another example:
■ Compare 5⁄6 and 4⁄9
In this case, neither the numerators nor the denominators are equal. So how do we proceed?
1. We have to compare 5⁄6 and 4⁄9
2. Set up a new fraction.
♦ It must have the same numerator as the first fraction 5⁄6
♦ It must have the same denominator as the second fraction 4⁄9
3. So the new fraction is: 5⁄9
4. Compare the new fraction with the first fraction. We get: 5⁄9 < 5⁄6. (∵ when two fractions have the same numerator, that fraction with the largest denominator is the smallest)
5. Compare the new fraction with the second fraction. We get 4⁄9 < 5⁄9. (∵ when two fractions have the same denominator, that fraction with the smallest numerator is the smallest)
6. So we have two results: 5⁄9 < 5⁄6 and 4⁄9 < 5⁄9
7. we can make an important conclusion from the above:
5⁄6 is greater than 5⁄9. But this 5⁄9 is greater than 4⁄9. So obviously, 5⁄6 will be greater than 4⁄9
8. So the result of the comparison between 5⁄6 and 4⁄9 is: 4⁄9 < 5⁄6
■ The above steps can be summarised as follows:
• We have a fraction 5⁄6 in hand
• We are given a second fraction 4⁄9 for comparison
• If the second fraction has a decreased numerator, and an increased denominator, then that second fraction will be less than the first fraction.
A general form of the results of the above two examples can be written as:
It is easy to remember the results shown above if we know this:
■ When numerator increases, the fraction increases and vice versa (denominator remains constant)
■ When denominator increases, the fraction decreases and vice versa (numerator remains constant)
We will now see some solved examples:
Find the larger of the two fractions in each of the following pairs:
■ 13/17, 14/15
Solution:
First fraction: 13/17
second fraction: 14/15. Numerator increased, denominator decreased
So Second fraction 14/15 is larger
■ 13/17, 11/18
solution:
First fraction: 13/17
second fraction: 11/18. Numerator decreased, denominator increased
So First fraction 13/17 is larger
■ 14/15, 11/18
solution:
First fraction: 14/15
second fraction: 11/18. Numerator decreased, denominator increased
So First fraction 14/15 is larger
We will now see an advanced case:
■ Compare 1⁄2 and 2⁄3
The methods that we have seen so far cannot be applied here. This is because:
• Neither numerators nor denominators are the same
• Numerator of the first fraction increased. The denominator of the second fraction also increased.
In such cases, we have to look for other methods. The most common method is to convert each of them into fractions with the same denominator. We have seen this method earlier here.
• Multiply both numerator and denominator of 1⁄2 by '3' (the other denominator). This will give 3⁄6
• Multiply both numerator and denominator of 2⁄3 by '2' (the other denominator). This will give 4⁄6
• Now we compare 3⁄6 and 4⁄6. Obviously, 4⁄6 is larger. That is., 3⁄6 < 4⁄6
• This is same as 1⁄2 < 2⁄3
Another example:
From the above two examples, it is clear that, once we convert each of the fractions into equivalent fractions of same denominators, we can ignore those 'same denominators'. All we have to do is to compare the 'new numerators'. We can write the steps algebraically as follows:
In the next section we will see a more advanced case.
Consider the following points:
• We are given some fractions
• All those given fractions have the same denominators. The numerators are different
■ Then, the fraction with the smallest numerator will be the smallest fraction
■ The fraction with the largest numerator will be the largest fraction
• Example: Given some fractions: 6⁄11, 3⁄11, 8⁄11, 4⁄11 ,9⁄11
• The smallest is 3⁄11, and the largest is 9⁄11
We have discussed the reason behind the above result here. Now we consider another such property:
Consider the following points:
• We are given some fractions
• All those given fractions have the same numerators. The denominators are different
■ Then, the fraction with the smallest denominator will be the largest fraction
■ The fraction with the largest denominator will be the smallest fraction
• Example: Given some fractions: 9⁄17, 9⁄12, 9⁄14, 9⁄21, 9⁄10
• The smallest is 9⁄21, and the largest is 9⁄10
We have discussed the reason behind the above result here.
Let us see some applications of the above two properties:
■ Compare 3⁄4 and 3⁄11
Ans: 3⁄11 < 3⁄4 (∵ when two fractions have the same numerator, that fraction with the largest denominator is the smallest)
■ Compare 7⁄5 and 7⁄9
Ans: 7⁄9 < 7⁄5 (∵ when two fractions have the same numerator, that fraction with the largest denominator is the smallest)
■ Compare 3⁄12 and 5⁄12
Ans: 3⁄12 < 5⁄12 (∵ when two fractions have the same denominator, that fraction with the smallest numerator is the smallest)
■ Compare 4⁄7 and 2⁄7
Ans: 2⁄7 < 4⁄7 (∵ when two fractions have the same denominator, that fraction with the smallest numerator is the smallest)
In some situations, the above two properties can be used together. Let us see an example:
■ Compare 3⁄7 and 4⁄5
In this case, neither the numerators nor the denominators are equal. So how do we proceed?
1. We have to compare 3⁄7 and 4⁄5
2. Set up a new fraction.
♦ It must have the same numerator as the first fraction 3⁄7
♦ It must have the same denominator as the second fraction 4⁄5
3. So the new fraction is: 3⁄5
4. Compare the new fraction with the first fraction. We get: 3⁄7 < 3⁄5. (∵ when two fractions have the same numerator, that fraction with the largest denominator is the smallest)
5. Compare the new fraction with the second fraction. We get 3⁄5 < 4⁄5. (∵ when two fractions have the same denominator, that fraction with the smallest numerator is the smallest)
6. So we have two results: 3⁄7 < 3⁄5 and 3⁄5 < 4⁄5
7. we can make an important conclusion from the above:
3⁄7 is less than 3⁄5. But this 3⁄5 is less than 4⁄5. So obviously, 3⁄7 will be less than 4⁄5
8. So the result of the comparison between 3⁄7 and 4⁄5 is: 3⁄7 < 4⁄5
■ The above steps can be summarised as follows:
• We have a fraction 3⁄7 in hand
• We are given a second fraction 4⁄5 for comparison
• If the second fraction has an increased numerator, and a decreased denominator, then that second fraction will be greater than the first fraction
Another example:
■ Compare 5⁄6 and 4⁄9
In this case, neither the numerators nor the denominators are equal. So how do we proceed?
1. We have to compare 5⁄6 and 4⁄9
2. Set up a new fraction.
♦ It must have the same numerator as the first fraction 5⁄6
♦ It must have the same denominator as the second fraction 4⁄9
3. So the new fraction is: 5⁄9
4. Compare the new fraction with the first fraction. We get: 5⁄9 < 5⁄6. (∵ when two fractions have the same numerator, that fraction with the largest denominator is the smallest)
5. Compare the new fraction with the second fraction. We get 4⁄9 < 5⁄9. (∵ when two fractions have the same denominator, that fraction with the smallest numerator is the smallest)
6. So we have two results: 5⁄9 < 5⁄6 and 4⁄9 < 5⁄9
7. we can make an important conclusion from the above:
5⁄6 is greater than 5⁄9. But this 5⁄9 is greater than 4⁄9. So obviously, 5⁄6 will be greater than 4⁄9
8. So the result of the comparison between 5⁄6 and 4⁄9 is: 4⁄9 < 5⁄6
■ The above steps can be summarised as follows:
• We have a fraction 5⁄6 in hand
• We are given a second fraction 4⁄9 for comparison
• If the second fraction has a decreased numerator, and an increased denominator, then that second fraction will be less than the first fraction.
A general form of the results of the above two examples can be written as:
It is easy to remember the results shown above if we know this:
■ When numerator increases, the fraction increases and vice versa (denominator remains constant)
■ When denominator increases, the fraction decreases and vice versa (numerator remains constant)
We will now see some solved examples:
Find the larger of the two fractions in each of the following pairs:
■ 13/17, 14/15
Solution:
First fraction: 13/17
second fraction: 14/15. Numerator increased, denominator decreased
So Second fraction 14/15 is larger
■ 13/17, 11/18
solution:
First fraction: 13/17
second fraction: 11/18. Numerator decreased, denominator increased
So First fraction 13/17 is larger
■ 14/15, 11/18
solution:
First fraction: 14/15
second fraction: 11/18. Numerator decreased, denominator increased
So First fraction 14/15 is larger
We will now see an advanced case:
■ Compare 1⁄2 and 2⁄3
The methods that we have seen so far cannot be applied here. This is because:
• Neither numerators nor denominators are the same
• Numerator of the first fraction increased. The denominator of the second fraction also increased.
In such cases, we have to look for other methods. The most common method is to convert each of them into fractions with the same denominator. We have seen this method earlier here.
• Multiply both numerator and denominator of 1⁄2 by '3' (the other denominator). This will give 3⁄6
• Multiply both numerator and denominator of 2⁄3 by '2' (the other denominator). This will give 4⁄6
• Now we compare 3⁄6 and 4⁄6. Obviously, 4⁄6 is larger. That is., 3⁄6 < 4⁄6
• This is same as 1⁄2 < 2⁄3
Another example:
From the above two examples, it is clear that, once we convert each of the fractions into equivalent fractions of same denominators, we can ignore those 'same denominators'. All we have to do is to compare the 'new numerators'. We can write the steps algebraically as follows:
In the next section we will see a more advanced case.
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