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 

---

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  

---

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'

---

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   

---

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

PREVIOUS      CONTENTS       NEXT

                        Copyright©2016 High school Maths lessons. blogspot.in - All Rights Reserved

Chapter 5.6 - Like Fractions and Comparing fractions

In the previous section we completed the discussion on the 'simplest form' of a fraction. In this section we will discuss about 'like fractions'.

The word 'like fractions' is in the plural form. Because we are saying 'fractions', not 'fraction'. So we are talking about more than one fraction here. If these fractions are to fall under the category of like fractions, their denominators must be the same. Consider the following fig.5.21:

Fig.5.21

Each of the three bars represents a fraction. All the three bars are divided into the same number of equal parts. So all the three fractions will have the same denominator. Thus they are 'like fractions'.

If the denominators are not the same, they are called unlike fractions. Example: 5/11 and 5/10 are unlike fractions.

Comparing fractions

Consider the situation: Mr. A wants to share a cake with his nephew. He decides to play a trick. He asks his nephew: “Do you want 1/3 of the cake or 2/5 ?” The nephew obviously wants the bigger part. But he cannot decide which is bigger. 1/3 or 2/5? He knows that:
• When the cake is divided into 3 equal parts, each piece will be larger than when it is divided into 5 equal parts.
• But uncle is offering '2 out the 5 smaller pieces'. Is it larger than '1 out of the bigger 3'? or smaller?
 Let us draw a fig. and find out:

Fig.5.22

From the fig.5.22 it is clear that 2/5 is larger than 1/3. So the nephew should ask for the 2/5 part. The problem is solved. But let us analyse it further:

What caused the nephew confusion? It is the 'methods of division'. 
• In the first option, the cake is to be divided into 5 equal parts. 
• In the second option, the cake is to be divided into 3 equal parts. 

The size of the equal parts will be different in each option. In addition to this, the numerators are different. So any body would be confused. 

The problem was solved by drawing a fig. But we will find a mathematical solution to such problems so that, we will not have to draw figures each time we find ourselves in such situations. We know that the 'unequal division' is the main reason for confusion. Can't we assume that they are divided equally? We can. For this we use the 'Equivalent fractions'.

• We have the first option 1/3. We must change it into a 'suitable' equivalent fraction. Let us call it EF1. How do we decide whether EF1 is suitable? We will see that soon.
• We have the second option 2/5. We must change it also into a 'suitable' equivalent fraction. Let us call it EF2.

The suitability of EF1 and EF2 is decided by one condition:
EF1 and EF2 should have the same denominator. So by this condition we are ensuring that the cake is divided into equal parts in both the two options. Let us now try to find these suitable EF1 and EF2:

We will write the possible equivalent fractions of 1/3 and 2/5 side by side in a tabular form as shown below:

From the table we can see that the 'equivalent fraction 5/15 for 1/3' and the 'equivalent fraction 6/15 for 2/5' have the same denominator 15. So we can make the selection of suitable equivalent fractions:

• 'EF1 for 1/3' is 5/15 and • 'EF2 for 2/5' is 6/15. So:
■ When uncle offers 1/3, he is offering 5 pieces out of 15
■ When he offers 2/5, he is offering 6 pieces out of 15.

6 is greater than 5. So the nephew should select the 2/5 option. Thus we can conclude that, whenever we want to compare unlike fractions, we must convert them into like fractions, by using the Equivalent fractions method.

Solved example 5.16
Pick out the larger fraction from among the following pairs
(a) 3/5, 2/3 (b) 2/5, 1/4
Solution:
(i) We will write a few equivalent fractions for 3/5 and 2/3:

From the above table, the suitable Equivalent fraction for 3/5 is 9/15, and that for 2/3 is 10/15. Because both have the same denominator 15, and so are like fractions. So
• 3/5 is 9 parts out of 15, and 
• 2/3 is 10 parts out of 15. 
Thus 2/3 is larger than 3/5
(ii) We will write a few equivalent fractions for 2/5 and 1/4:

From the above table, the suitable Equivalent fraction for 2/5 is 8/20, and that for 1/4 is 5/20. Because both have the same denominator 20, and so are like fractions. So
• 2/5 is 8 parts out of 20, and 
• 1/4 is 5 parts out of 20. 
Thus 2/5 is larger than 1/4

---

So now we can compare any given fractions and decide which one is the greatest among them. But the above method is lengthy: 

• We wrote several equivalent fractions for each given fraction. 

• From among them we selected the suitable fractions. 

There is a direct method by which we can calculate the 'suitable equivalent fractions':

In all the examples that we saw above, the suitability was decided by one condition:
• The denominators should be the same. 
If we examine those ‘same denominators’ we will see that each is a common multiple of the given denominators’. Let us see them again:
• In the solved example 5.16 (a) 15 is the common multiple of 5 and 3, (where 5 and 3 are the given denominators). That is., 15 is a multiple of both 5 and 3. 
• In (b), 20 is the common multiple of 5 and 4.

This gives us a method to calculate the denominator of the ‘suitable equivalent fraction’:
• Just find a common multiple of the given denominators. That will be our required denominator.
Once we find that denominator, we can get the ‘suitable equivalent fraction’. The following solved examples will demonstrate the method.

Solved example 5.17
Pick out the larger fraction from among the following pairs:
(a) 7/15, 3/13 (b) 23/39, 14/34
Solution:
(a) 7/15, 3/13
■ Step 1: Find the suitable equivalent fraction for each given fractions
• Common multiple of 15 and 13.
There will be several common multiples for 15 and 13. Any one of them will serve our purpose. The easiest way to find one is to multiply them. Because the product of two numbers will obviously be a ‘common multiple’ of both.
So we have: common multiple of 15 and 13 = 15 ×13 = 195. This will be the denominator of both our ‘suitable equivalent fractions’
• Suitable equivalent fraction for 7/15:
    ♦ The denominator of the given fraction is 15.
   ♦ The denominator of the new equivalent fraction should be 195.
   ♦ We have seen above that, this 195 is obtained by multiplying 15 with 13
   ♦ So we have to multiply the numerator also by 13. The calculation steps and the result are given            below:

            ⁷ ⁄ ₁₅  =   ^{(7 × 13)} ⁄ _{(15 × 13)}  =  ⁹¹ ⁄ ₁₉₅  

   ♦ So the required equivalent fraction is  ⁹¹ ⁄ ₁₉₅ 
• Suitable equivalent fraction for 3/13:
    ♦ The denominator of the given fraction is 13.
   ♦ The denominator of the new equivalent fraction should be 195.
   ♦ We have seen above that, this 195 is obtained by multiplying 13 with 15
   ♦ So we have to multiply the numerator also by 15. The calculation steps and the result are given            below:

            ³ ⁄ ₁₃  =   ^{(3 × 15)} ⁄ _{(13 × 15)}  =  ⁴⁵ ⁄ ₁₉₅ 

   ♦ So the required equivalent fraction is  ⁴⁵ ⁄ ₁₉₅
It may be noted that, in this step, we are multiplying both numerator and denominator by the 'other denominator'. Based on this, we will soon see the easy steps to solve the problem.
■ Step 2: Write down the equivalent fractions
  ⁷ ⁄ ₁₅  =  ⁹¹ ⁄ ₁₉₅  and ³ ⁄ ₁₃  =  ⁴⁵ ⁄ ₁₉₅ 
■ Step 3: Compare the equivalent fractions
91 > 45. So 7/15 is greater than 3/13

(b) 23/39, 14/34
■ Step 1: Find the suitable equivalent fraction for each given fractions
• Common multiple of 39 and 34.
There will be several common multiples for 39 and 34. Any one of them will serve our purpose. The easiest way to find one is to multiply them. Because the product of two numbers will obviously be a ‘common multiple’ of both.
So we have: common multiple of 39 and 34 = 39 ×34 = 1326. This will be the denominator of both our ‘suitable equivalent fractions’
• Suitable equivalent fraction for 23/39:
    ♦ The denominator of the given fraction is 39.
   ♦ The denominator of the new equivalent fraction should be 1326.
   ♦ We have seen above that, this 1326 is obtained by multiplying 39 with 34
   ♦ So we have to multiply the numerator also by 34. The calculation steps and the result are given            below:

            ²³ ⁄ ₃₉  =   ^{(23 × 34)} ⁄ _{(39 × 34)}  =  ⁷⁸² ⁄ ₁₃₂₆  

   ♦ So the required equivalent fraction is ⁷⁸² ⁄ ₁₃₂₆
• Suitable equivalent fraction for 14/34:
    ♦ The denominator of the given fraction is 34.
   ♦ The denominator of the new equivalent fraction should be 1326.
   ♦ We have seen above that, this 1326 is obtained by multiplying 34 with 39
   ♦ So we have to multiply the numerator also by 39. The calculation steps and the result are given            below:

            ¹⁴ ⁄ ₃₄  =   ^{(14 × 39)} ⁄ _{(34 × 39)}  =  ⁵⁴⁶ ⁄ ₁₃₂₆  

   ♦ So the required equivalent fraction is ⁵⁴⁶ ⁄ ₁₃₂₆

It may be noted that, in this step, we are multiplying both numerator and denominator by the 'other denominator'. Based on this, we will soon see the easy steps to solve the problem.
■ Step 2: Write down the equivalent fractions

•  ²³ ⁄ ₃₉  =  ⁷⁸² ⁄ ₁₃₂₆  and ¹⁴ ⁄ ₃₄  =  ⁵⁴⁶ ⁄ ₁₃₂₆ 
■ Step 3: Compare the equivalent fractions
782 > 546. So 23/39 is greater than 14/34
Solved example 5.18
Compare the fractions in the following pairs:
(a) 5/11, 7/13  (b) 10/17, 8/15  (c) 6/23, 2/7
Solution:
(a) The steps are same as above. So this time we will write only those steps which are absolutely necessary, as shown below:

(b)

(c) 

---

In the above examples we found out the common multiple just by multiplying the two numbers. Some times it will be convenient to calculate the LCM. That is., the least common multiple. Out of the several available ‘common multiples’, The LCM will be the ‘least’ one, or the smallest one. In this way, we will be able to keep the ‘size of the numbers’ down.

---

Now we are in a position to compare any given fractions. If we are given a group of more than two fractions, write the 'suitable equivalent fraction' for each of them. And then compare. We will even be able to write the given fractions in ascending or descending order.

---

In some special cases, there will not be any need for 'a paper and pencil' to do the above calculations, to find which is the biggest among the given fractions. We will be able to solve them 'mentally'. We will discuss about them in the next section.

PREVIOUS      CONTENTS       NEXT

                        Copyright©2016 High school Maths lessons. blogspot.in - All Rights Reserved