Chapter 5.16 - Fractions between fractions

In the previous section we saw some properties of unequal fractions which are helpful for their comparisons. In this  section, we will see a more advanced case.

Consider the following example:
• We have two fractions ¹⁄₂ and ³⁄₄ in hand
• Out of the two, ¹⁄₂ is lesser. That is., ¹⁄₂ < ³⁄₄  (∵ 1 × 4   <   3 × 2)
• We are going to make a new fraction from the two fractions
• For that, we add the numerators and denominators
• So the new fraction is ⁽¹⁺³⁾⁄₍₂₊₄₎ = ⁴⁄₆ = ²⁄₃
• This new fraction ²⁄₃ is greater than ¹⁄₂. That is., ¹⁄₂ < ²⁄₃  (∵ 1 × 3   <   2 × 2)
• At the same time, this new fraction is less than ³⁄₄. That is., ²⁄₃ < ³⁄₄  (∵ 2 × 4   <   3 × 3) 
• So we can write: ¹⁄₂ < ²⁄₃ < ³⁄₄
• That means., the new fraction ²⁄₃ lies in between the original two fractions ¹⁄₂ and ³⁄₄
• We have seen how to represent fractions on a number line.
• The three fractions are marked on a number line in fig.5.34 below

Fig.5.34

We find that the new fraction ²⁄₃ lies in between the original two fractions ¹⁄₂ and ³⁄₄ 

Another example:
• We have two fractions ⁵⁄₇ and ⁶⁄₁₁ in hand
• Out of the two, ⁶⁄₁₁ is lesser. That is., ⁶⁄₁₁ < ⁵⁄₇  (∵ 6 × 7   <   5 × 11 ⇒ 42  <  55)
• We are going to make a new fraction from the two fractions
• For that, we add the numerators and denominators
• So the new fraction is ⁽⁶⁺⁵⁾⁄₍₁₁₊₇₎ = ¹¹⁄₁₈
• This new fraction ¹¹⁄₁₈ is greater than ⁶⁄₁₁. That is., ⁶⁄₁₁ < ¹¹⁄₁₈  (∵ 6 × 18   <   11 × 11 ⇒ 108 < 121)
• At the same time, this new fraction is less than ⁵⁄₇. That is., ¹¹⁄₁₈ < ⁵⁄₇  (∵ 11 × 7   <   5 × 18 ⇒ 77 < 90) 
• So we can write: ⁶⁄₁₁ < ¹¹⁄₁₈ < ⁵⁄₇
• That means., the new fraction ¹¹⁄₁₈ lies in between the original two fractions ⁶⁄₁₁ and ⁵⁄₇
• The three fractions are marked on a number line in fig.5.35

Fig.5.35

We find that the new fraction ¹¹⁄₁₈ lies in between the original two fractions ⁶⁄₁₁ and ⁴⁄₇ 

Another example:
• We have two fractions ⁸⁄₅ and ⁹⁄₇ in hand
• Out of the two, ⁹⁄₇ is lesser. That is., ⁹⁄₇ < ⁸⁄₅  (∵ 9 × 5   <   8 × 7 ⇒ 45  <  56)
• We are going to make a new fraction from the two fractions
• For that, we add the numerators and denominators
• So the new fraction is ⁽⁹⁺⁸⁾⁄₍₇₊₅₎ = ¹⁷⁄₁₂
• This new fraction ¹⁷⁄₁₂ is greater than ⁹⁄₇. That is., ⁹⁄₇ < ¹⁷⁄₁₂  (∵ 9 × 12   <   17 × 7 ⇒ 108 < 119)
• At the same time, this new fraction is less than ⁸⁄₅. That is., ¹⁷⁄₁₂ < ⁸⁄₅  (∵ 17 × 5   <   8 × 12 ⇒ 85 < 96) 
• So we can write: ⁹⁄₇ < ¹⁷⁄₁₂ < ⁸⁄₅
• That means., the new fraction ¹⁷⁄₁₂ lies in between the original two fractions ⁹⁄₇ and ⁸⁄₅
• The three fractions are marked on a number line in fig.5.36

Fig.5.36

We find that the new fraction ¹⁷⁄₁₂ lies in between the original two fractions ⁹⁄₇ and ⁸⁄₅ 

One more example:
• We have two fractions ³⁄₅ and ¹⁵⁄₁₂ in hand
• Out of the two, ³⁄₅ is obviously lesser. That is., ³⁄₅ < ¹⁵⁄₁₂  (∵ ³⁄₅ is a proper fraction and ¹⁵⁄₁₂ is an improper fraction)
• We are going to make a new fraction from the two fractions
• For that, we add the numerators and denominators
• So the new fraction is ⁽³⁺¹⁵⁾⁄₍₅₊₁₂₎ = ¹⁸⁄₁₇
• This new fraction ¹⁸⁄₁₇ is greater than ³⁄₅. That is., ³⁄₅ < ¹⁸⁄₁₇  (∵ ³⁄₅ is a proper fraction and ¹⁸⁄₁₇ is an improper fraction)
• At the same time, this new fraction is less than ¹⁵⁄₁₂. That is., ¹⁸⁄₁₇ < ¹⁵⁄₁₂  (∵ 18 × 12   <   15 × 17 ⇒ 216 < 255) 
• So we can write: ³⁄₅ < ¹⁸⁄₁₇ < ¹⁵⁄₁₂
• That means., the new fraction ¹⁸⁄₁₇ lies in between the original two fractions ³⁄₅ and ¹⁵⁄₁₂
• The three fractions are marked on a number line in fig.5.37

Fig.5.37

We find that the new fraction ¹⁸⁄₁₇ lies in between the original two fractions ³⁄₅ and ¹⁵⁄₁₂ 

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Based on the above examples we can write: A new fraction ^(a+p)⁄_(b+q) obtained by adding numerators and denominators of two original fractions ^a⁄_b and ^p⁄_q will lie in between the two fractions. Not just 'in between' the two fractions. It follows a strict rule:
• We have two fractions ^a⁄_b and ^p⁄_q in hand
• One of them will be lesser than the other
• The new fraction will be greater than the 'lesser original fraction'
• The new fraction will be lesser than the 'greater original fraction'
■ So the 'lesser original fraction' will lie on the extreme left
■ The 'greater original fraction' will lie on the extreme right
■ The new fraction will lie in between the two

But we must show the proof for all the above:
If  ^a⁄_b <  ^p⁄_q, Prove that ^a⁄_b <  ^(a+p)⁄_(b+q) < ^p⁄_q
1. We have ^a⁄_b < ^p⁄_q. From this we get aq < pb
2. We have to prove that ^a⁄_b < ^(a+p)⁄_(b+q) < ^p⁄_q 
3. Consider the first two terms: ^a⁄_b < ^(a+p)⁄_(b+q)
4. If (3) is true, then a(b+q) < b(a+p) ⇒ (ab + aq) < (ba + bp)
5. ab is same as ba. That means we have 'one term same' on both sides in (4). They will cancel out each other
6. So the superiority or inferiority of the left side and right side in (4) is decided by aq and bp
8. It is given in (1) that aq < pb. So we find that left side of (4) is indeed inferior.
9. Hence (4) is proved, and consequently, (3) is established
10. Consider the last two terms in (2): ^(a+p)⁄_(b+q) < ^p⁄_q
11. If (10) is true, then (a+p)q < p(b+q) ⇒ (aq + pq) < (pb + pq)
12. We have 'one term same' on both sides in (11). The term is pq. They will cancel out each other
13. So the superiority or inferiority of the left side and right side in (11) is decided by aq and pb
14. It is given in (1) that aq < pb. So we find that left side of (11) is indeed inferior
15. Hence (11) is proved, and consequently, (10) is established
16. Taking (3) and (10) together we get: ^a⁄_b < ^(a+p)⁄_(b+q) < ^p⁄_q  

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So we learned the method to obtain a fraction between 'any two fractions'. Concentrate on the words: 'any two fractions'. The new fraction that we obtain by the above method can become one of the 'any two fractions'. For example, in fig.5.34 above, we obtained ²⁄₃ in between ¹⁄₂ and ³⁄₄. The fig. is shown again below:

• Now, ¹⁄₂ and ²⁄₃ can be considered as 'any two fractions'. 
• Let us add the numerators and denominators: ⁽¹⁺²⁾⁄₍₂₊₃₎ = ³⁄₅
• This ³⁄₅ will lie in between ¹⁄₂ and ²⁄₃ . This is shown in the fig.5.38 below:
[The reader is advised to check and confirm whether ³⁄₅ indeed lies in between ¹⁄₂ and ²⁄₃]

Fig.5.38

^• 2⁄₃ and ³⁄₄ can be considered as 'any two fractions'. 
• Let us add the numerators and denominators: ⁽²⁺³⁾⁄₍₃₊₄₎ = ⁵⁄₇
• This ⁵⁄₇ will lie in between ²⁄₃ and ³⁄₄ . This is also shown in the fig.5.38 above
• So altogether we get:
¹⁄₂ < ³⁄₅  < ²⁄₃ < ⁵⁄₇  < ³⁄₄  
³⁄₅ and ⁵⁄₇ are the two new fractions that we obtained
• We can continue like this for any number of times. For example, we can take ¹⁄₂ and ³⁄₅ as 'any two fractions'

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Now we will see some solved examples:

Solved example 5.37
(i) Find 3 fractions which are larger than ¹⁄₃ and smaller than ¹⁄₂
(ii) Find 3 fractions, all with denominator 24, which are larger than ¹⁄₃ and smaller than ¹⁄₂
(iii) Find 3 fractions, all with numerator 4, which are larger than ¹⁄₃ and smaller than ¹⁄₂
Solution:
(i) We know that ¹⁄₃ < ¹⁄₂
1. We can use the property: If  ^a⁄_b <  ^p⁄_q, Then ^a⁄_b <  ^(a+p)⁄_(b+q) < ^p⁄_q
2. Let us add the numerators and denominators: ⁽¹⁺¹⁾⁄₍₃₊₂₎ = ²⁄₅
3. This ²⁄₅ will lie in between ¹⁄₃ and ¹⁄₂ 
4. So we can write: ¹⁄₃ < ²⁄₅  < ¹⁄₂ 
5. Take ¹⁄₃ and ²⁄₅
6. Let us add the numerators and denominators: ⁽¹⁺²⁾⁄₍₃₊₅₎ = ³⁄₈
7. This ³⁄₈ will lie in between ¹⁄₃ and ²⁄₅
8. So we can write: ¹⁄₃ < ³⁄₈  < ²⁄₅ < ¹⁄₂  
9. Take ²⁄₅ and ¹⁄₂
10. Let us add the numerators and denominators: ⁽²⁺¹⁾⁄₍₅₊₂₎ = ³⁄₇
11. This ³⁄₇ will lie in between ²⁄₅ and ¹⁄₂
12. So we can write: ¹⁄₃ < ³⁄₈  < ²⁄₅ < ³⁄₇ < ¹⁄₂  
13. Thus we get 3 fractions: ³⁄₈, ²⁄₅ and ³⁄₇ between ¹⁄₃ and ¹⁄₂

(ii)  1. First we write ¹⁄₃ and ¹⁄₂ as fractions with denominator 24:
¹⁄₃ = ^(1×8)⁄_(3×8) = ⁸⁄₂₄ and  ¹⁄₂ = ^(1×12)⁄_(2×12) = ¹²⁄₂₄
2. So we get two fractions: ⁸⁄₂₄ and ¹²⁄₂₄
3. Now we use the property of fractions with the same denominators:
'When the denominators are the same, the fraction with the larger numerator will be the larger'
4. So the three required fractions are: ⁹⁄₂₄ , ¹⁰⁄₂₄ and ¹¹⁄₂₄
5. So we can write: ⁸⁄₂₄ < ⁹⁄₂₄  < ¹⁰⁄₂₄ < ¹¹⁄₂₄ < ¹²⁄₂₄   
6. This is same as:  ¹⁄₃ < ⁹⁄₂₄  < ¹⁰⁄₂₄ < ¹¹⁄₂₄ < ¹⁄₂    

(iii) 1. First we write ¹⁄₃ and ¹⁄₂ as fractions with numerator 4:
¹⁄₃ = ^(1×4)⁄_(3×4) = ⁴⁄₁₂ and  ¹⁄₂ = ^(1×4)⁄_(2×4) = ⁴⁄₈
2. So we get two fractions: ⁴⁄₁₂ and ⁴⁄₈
3. Now we use the property of fractions with the same numerators:
'When the numerators are the same, the fraction with the smaller denominator will be the larger'
4. So the three required fractions are: ⁴⁄₁₁ , ⁴⁄₁₀ and ⁴⁄₉
5. So we can write: ⁴⁄₁₂ < ⁴⁄₁₁  < ⁴⁄₁₀ < ⁴⁄₉ < ⁴⁄₈   
6. This is same as: ¹⁄₃ < ⁴⁄₁₁  < ⁴⁄₁₀ < ⁴⁄₉ < ¹⁄₂   

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In the next section we will see addition and subtraction with fractions.

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Chapter 5.15 - Properties of Unequal fractions

In the previous section we saw the comparison of unequal fractions. Now we will see a more advanced case.
Consider the following example:
• We have a fraction ²⁄₃ in hand
• We are going to make a new fraction from this ²⁄₃
• For that, we increase both the numerator and denominator by 1
• So the resulting fraction is ⁽²⁺¹⁾⁄₍₃₊₁₎ = ³⁄₄
• Now compare the new fraction ³⁄₄ with the original fraction ²⁄₃

■ We find that the new fraction is greater than the original fraction
Another example:
• We have a fraction ⁵⁄₉ in hand
• We are going to make a new fraction from this ⁵⁄₉
• For that, we increase both the numerator and denominator by 1
• So the resulting fraction is ⁽⁵⁺¹⁾⁄₍₉₊₁₎ = ⁶⁄₁₀
• Now compare the new fraction ⁶⁄₁₀ with the original fraction ⁵⁄₉

■ We find that the new fraction is greater than the original fraction

We find that the new fraction is greater than the original fraction in both the above cases. But is this always true? Let us see another example:
• We have a fraction ⁴⁄₃ in hand
• We are going to make a new fraction from this ⁴⁄₃
• For that, we increase both the numerator and denominator by 1
• So the resulting fraction is ⁽⁴⁺¹⁾⁄₍₃₊₁₎ = ⁵⁄₄
• Now compare the new fraction ⁵⁄₄ with the original fraction ⁴⁄₃

In this example, we see that the new fraction is less than the original fraction. Why is this so?
• In the previous two examples, the numerator of the original fractions ²⁄₃ and ⁵⁄₉ is less than the denominator
• In the present example, the numerator of the original fraction ⁴⁄₃ is greater than the denominator
So we can write it algebraically:
■ If a < b, then ^a⁄_b <  ^(a+1)⁄_(b+1)
■ If a > b, then ^(a+1)⁄_(b+1) < ^a⁄_b

But we have to show proof for the above. It can be done as follows:
1. If a < b, then prove that  ^a⁄_b <  ^(a+1)⁄_(b+1)
2. We have seen earlier that: if ^a⁄_b <  ^p⁄_q then, aq < pb
3. So to prove (1), we need to prove that a(b+1) < b(a+1)
4. Expanding the left side of (3), we get: ab + a
5. Expanding the right side of (3) we get: ba + b
6. ab is same as ba. That means we have 'one term same' on both sides in (3). They will cancel out each other
7. So the superiority or inferiority of the left side and right side in (3) is decided by a and b
8. It is given in (1) that a < b. So we find that left side of (3) is indeed inferior.
9. Hence (3) is proved, and consequently, (1) is established

Similarly we can prove the second case also as follows:
1. If a > b, then prove that ^(a+1)⁄_(b+1) < ^a⁄_b
2. We have seen earlier that: if ^a⁄_b <  ^p⁄_q then, aq < pb
3. So to prove (1), we need to prove that (a+1)b < a(b+1)
4. Expanding the left side of (3), we get: ab + b
5. Expanding the right side of (3) we get: ba + a
6. ab is same as ba. That means we have 'one term same' on both sides in (3). They will cancel out each other
7. So the superiority or inferiority of the left side and right side in (3) is decided by a and b
8. It is given in (1) that a > b. So we find that left side of (3) is indeed inferior.
9. Hence (3) is proved, and consequently, (1) is established

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We will now see some solved examples

Solved example 5.36
Find the larger of each pair of fractions below with out pen and paper
(i) ³⁄₅ , ⁸⁄₁₃ (ii) ³⁄₅ ,⁶⁄₁₁  (iii) ⁷⁴⁄₇₅, ⁷⁵⁄₇₆  (iv) ¹⁰¹⁄₁₀₂ , ⁹⁸⁄₉₉
Solution:
(i) We can use the property: if aq < pb then, ^a⁄_b <  ^p⁄_q (Details here)
1. In the given problem, aq = 3 × 13 = 39, and pb = 8 × 5 = 40
2. 39 < 40. So aq < pb. 
3. Thus we get  ^a⁄_b <  ^p⁄_q  ⇒ ³⁄₅ < ⁸⁄₁₃
Once we understand the basic principles, we can do the above steps with out pen and paper
(ii) We can use the property: if aq < pb then, ^a⁄_b <  ^p⁄_q 
1. In the given problem, aq = 3 × 11 = 33, and pb = 6 × 5 = 30
2. 30 < 33. So pb < aq. 
3. Thus we get  ^p⁄_q <  ^a⁄_b  ⇒ ⁶⁄₁₁ < ³⁄₅
Once we understand the basic principles, we can do the above steps with out pen and paper
(iii) We can use the property that we saw earlier in this section: If a < b, then ^a⁄_b <  ^(a+1)⁄_(b+1)
1. We have ^a⁄_b = ⁷⁴⁄₇₅ and ^(a+1)⁄_(b+1) = ⁷⁵⁄₇₆ 
2. So we get ⁷⁴⁄₇₅ < ⁷⁵⁄₇₆
(iv) We can use the property that we saw earlier in this section: If a < b, then ^a⁄_b <  ^(a+1)⁄_(b+1)
1. Take ^a⁄_b = ⁹⁸⁄₉₉  So that,  ^(a+1)⁄_(b+1) = ⁹⁹⁄₁₀₀
2. From this we get ⁹⁸⁄₉₉ < ⁹⁹⁄₁₀₀
3. Take ^a⁄_b = ⁹⁹⁄₁₀₀  So that,  ^(a+1)⁄_(b+1) = ¹⁰⁰⁄₁₀₁
4. From this we get ⁹⁹⁄₁₀₀ < ¹⁰⁰⁄₁₀₁
5. Take ^a ⁄_b = ¹⁰⁰⁄₁₀₁  So that,  ^(a+1)⁄_(b+1) = ¹⁰¹⁄₁₀₂
6. From this we get ¹⁰⁰⁄₁₀₁ < ¹⁰¹⁄₁₀₂
7. Consider the results in (2), (4) and (6) together: ⁹⁸⁄₉₉ < ⁹⁹⁄₁₀₀ < ¹⁰⁰⁄₁₀₁  < ¹⁰¹⁄₁₀₂ 
8. From the first and last terms of the result in (7), we get ⁹⁸⁄₉₉ < ¹⁰¹⁄_{102
Once we understand the basic principles, we can do the above steps with out pen and paper}

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In the property that we discussed above, we increased the numerator and denominator by '1'. What if we increase by other natural numbers 2, 3, 4.. etc.,? Do we get the same result? We can check using algebra. We can put 'n' to represent any natural number.

1. If a < b, then prove that  ^a⁄_b <  ^(a+n)⁄_(b+n)
2. We have seen earlier that: if ^a⁄_b <  ^p⁄_q then, aq < pb
3. So to prove (1), we need to prove that a(b+n) < b(a+n)
4. Expanding the left side of (3), we get: ab + an
5. Expanding the right side of (3) we get: ba + bn
6. ab is same as ba. That means we have 'one term same' on both sides in (3). They will cancel out each other
7. So the superiority or inferiority of the left side and right side in (3) is decided by an and bn
8. Here also 'n' is common. It will cancel each other
9. So the superiority or inferiority of the left side and right side in (3) is decided by a and b
10. It is given in (1) that a < b. So we find that left side of (3) is indeed inferior
11. Hence (3) is proved, and consequently, (1) is established

Similarly we can prove the second case also as follows:
1. If a > b, then prove that ^(a+n)⁄_(b+n) < ^a⁄_b
2. We have seen earlier that: if ^a⁄_b <  ^p⁄_q then, aq < pb
3. So to prove (1), we need to prove that (a+n)b < a(b+n)
4. Expanding the left side of (3), we get: ab + nb
5. Expanding the right side of (3) we get: ba + na
6. ab is same as ba. That means we have 'one term same' on both sides in (3). They will cancel out each other
7. So the superiority or inferiority of the left side and right side in (3) is decided by nb and na
8. Here also 'n' is common. It will cancel each other
9. So the superiority or inferiority of the left side and right side in (3) is decided by b and a
10. It is given in (1) that a > b. So we find that left side of (3) is indeed inferior

11. Hence (3) is proved, and consequently, (1) is established

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In the next section we will see an even  more advanced case.

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