Chapter 22 - Real Numbers

In the previous section we completed the discussion on Circles, arcs and sectors. In a previous chapter, we have seen different types of numbers. (see fig.16.1). In this chapter, we will see how and where 'Real numbers' fit into the list of numbers.

First let us recall how numbers are represented on a number line. In fig.22.1(a), a horizontal line segment is drawn between points A and B. We will fix this distance AB as 1 unit.

Fig.22.1

• So, if we take two line segments, each equal to AB, and put them end to end, we will get a distance of 2 units. This distance is the distance from zero to 2 on the number line shown in fig.d

• If we take 3 such segments and put them end to end, we will get a distance of 3 units. This distance is the distance from zero to 3 on the number line shown in fig.d. In this way, we can find the position of any number on the number line.

Now what about fractions? Consider fig.b. The segment AB is divided into 2 equal parts at C. So AC = BC = ¹⁄₂ unit. 
• If we take two line segments, one equal to AB, and the other equal to AC, and put them end to end, we will get a distance 1¹⁄₂ = ³⁄₂ units. If we measure out this end to end distance (from zero) on the number line, we can mark ³⁄₂. It is shown in fig.d

Another example: In fig.c, the segment AB is divided into 3 equal parts at D and E. So AD = DE = BE = ¹⁄₃ unit. 
• If we take three line segments, two of them equal to AB, and the third equal to AD, and put them end to end, we will get a distance 2¹⁄₃ = ⁷⁄₃ units.  If we measure out this end to end distance (from zero) on the number line, we can mark ⁷⁄₃. It is shown in fig.d

In the above examples, we marked points using a geometrical method. That is., we measured the distance from the 'standard 1 unit', or it's fractions, and then marked it on the number line. Can we do the marking with out actual measuring? Let us analyse:

• ³⁄₂ does not give us any problems because in decimal form, it is 1.5. It is a terminating decimal. 
• But ⁷⁄₃ is a different case. We discussed about such numbers here. It is a recurring decimal. It is written as ⁷⁄₃ = 2.333... Using this decimal form, we can represent ⁷⁄₃ on the number line by the following procedure: 

2.333... has infinite number of decimal places. The digit '3' repeats forever. If more number of decimal places are taken, we will get more accuracy. If we want such a greater accuracy, we will have to 'zoom in' on the region between 2 and 3. Because 2.333... lies in between 2 and 3. In the fig.22.2 below, ‘zoom level 1’ shows the portion between 2 and 3, at an enlarged scale.

Fig.22.2

In this zoom level, we have enough space to clearly mark ten subdivisions between 2 and 3. So the third subdivision will be 2.3 and the fourth subdivision will be 2.4. They are specially marked in red because 2.333... lies between them

Next, we zoom in on the region between this 2.3 and 2.4. This is shown as 'zoom level 2'. In this zoom level, we have enough space to clearly mark ten subdivisions between 2.3 and 2.4. So the third subdivision will be 2.33 and the fourth subdivision will be 2.34. They are specially marked in red because 2.333... lies between them

Next, we zoom in on the region between this 2.33 and 2.34. This is shown as 'zoom level 3'. In this zoom level, we have enough space to clearly mark ten subdivisions between 2.33 and 2.34. So the third subdivision will be 2.333 and the fourth subdivision will be 2.334. They are specially marked in red because 2.333... lies between them

So, if we take the left red subdivision in zoom level 3, it will represent 2.333. It is fairly accurate. We know that 0.003 is a small quantity. But we are able to mark it by 'zooming in'. It is like zooming in on Google Maps. At higher zoom levels, we are able to see smaller details. But in our case of number line, there is no limit. We can mark any quantity, however small it may be. All we need to do, is to zoom in to the required level. 
So we can conclude that:
• Any natural number can be represented on a number line
• Any fraction can be represented on a number line

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Now we try to do the reverse: We are given a number line with a mark  on it. We want to represent that mark as a number or a fraction. 
Consider the mark P in fig.22.3(a). It is exactly at 4

Fig.22.3

So we write: The mark P represents number 4. 
Consider mark Q. It is exactly at 6.5 
So we write: The mark Q represents number 6.5, which is equal to ¹³⁄₂
Consider mark R. We have seen that any recurring decimal can be represented on a number line. If it is given that R is 5.333... , can we represent it as a fraction? 
The answer is yes. The procedure is as follows:
• Let x = 5.333...
• 10x = 10 × 5.333... = 53.333... = 48 + 5.333... = 48 + x (∵ x = 2.333..)
⇒ 10x = 48 + x ⇒ 9x = 48 ⇒ x = ⁴⁸⁄₉ = 5³⁄₉ = 5¹⁄₃ = ¹⁶⁄₃

So, even if the given mark on the number line is a recurring decimal, we can convert it into a fraction. Let us see a few more examples:
■ Express 1.272727... as a fraction:
1. We can write: 1.272727... = 1.2̅7
2. A line is drawn above 2 and 7. That means the block '27' repeats for ever
3. Let x = 1.272727... Since two digits are repeating, we will multiply by 100
• 100x = 100 × 1.272727... = 127.272727... = 126 + 1.272727... = 126 + x (∵ x = 1.272727..)
⇒ 100x = 126 + x ⇒ 99x = 126 ⇒ x = ¹²⁶⁄₉₉ = 1³⁄₁₁

■ Express 0.2353535... as a fraction:
1. We can write: 0.2353535... = 0.23̅5
2. A line is drawn above 3 and 5. That means the block '35' repeats for ever
3. Let x = 0.2353535... Since two digits are repeating, we will multiply by 100
• 100x = 100 × 0.2353535... = 23.535353... =  23.3 + 0.2353535... = 23.3 + x (∵ x = 0.2353535...)
⇒ 100x = 23.3 + x ⇒ 99x = 23.3 ⇒ x = ²³³⁄₉₉₀

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■ Based on the above discussion, we can write the following:
1. Any natural number can be represented on a number line
2. Any fraction can be represented on a number line
    ♦ The fractions  like ¹⁄₂, ¹⁄₅, ³⁄₅ etc., that give terminating decimals
    ♦ The fractions like ¹⁄₃, ¹⁄₇, ³⁄₅ etc., that give non-terminating, recurring decimals
■ Reverse of the above is also possible: 
If a point is marked on a number line, it can be represented by any one of the two categories given below:
1. A natural number
2. A fraction
    ♦ The fraction may be one  like ¹⁄₂, ¹⁄₅, ³⁄₅ etc., that give terminating decimals
    ♦ The fraction may be one like ¹⁄₃, ¹⁄₇, ³⁄₅ etc., that give non-terminating, recurring decimals

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■ Any number that can be expressed as a fraction ^p⁄_q where q is not equal to zero is called a rational number. 

• Simple natural numbers like 2, 5 etc., can be written as ²⁄₁ and ⁵⁄₁. So they are rational numbers. 
• Decimals like 0.2, 0.35, 0.5 etc., can be easily written in the form of fractions (with non-zero denominators). So they are rational numbers.
• Non-terminating recurring decimals like 0.333..., 1.2̅7, 0.23̅5, etc., can be written in the form of fractions (with non-zero denominators). So they are rational numbers.

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In this section we saw the details about rational numbers. In the next section we will see irrational numbers.

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Chapter 6.10 - Representing Recurring decimals

In the previous section we saw how ¹⁄₃ is expressed as decimals. In this section, we will see another example.
Let us try to write ¹⁄₆ in decimal form. Just like in the case of ¹⁄₆, there is no natural number, which when multiplied with 6, will give any 'power of 10'. So we will use the other method:
1. We know that ¹⁄₆ = ^(1×10) ⁄_(6×10). 
2. Let us rearrange the right side: ¹⁄₆ = ¹⁄₁₀ × ¹⁰⁄₆ 
3. In the above result, we can write ¹⁰⁄₆ as (1 + ⁴⁄₆ )
4. So (2) becomes ¹⁄₆ = ¹⁄₁₀ × (1 + ⁴⁄₆ ). So we get:
5. ¹⁄₆ = ¹⁄₁₀ + ⁴⁄₆₀
6. Look at the above result carefully. We have two fractions on the right side: ¹⁄₁₀ and ⁴⁄₆₀ 
    ♦ Out of these two, ¹⁄₁₀ is in an 'ideal form'. Because it has a 'power of 10' in the denominator. So it can be readily converted into a decimal form
    ♦ But the other fraction ⁴⁄₆₀ is causing a problem. It cannot be readily converted into a decimal form
    ♦ If this ⁴⁄₆₀ is very very small, then we can ignore it. In that case, (5) will become ¹⁄₆ = ¹⁄₁₀
    ♦ But unfortunately, ⁴⁄₆₀ is not very small, and we cannot ignore it. 
• After reaching (5), if we write ¹⁄₆ = 0.1, we are ignoring ⁴⁄₆₀
• That is not a good thing to do because, ⁴⁄₆₀ is not a small quantity, that can be 'just ignored'
7. We arrived at (5) by writing ¹⁄₆ as ^(1×10) ⁄_(6×10) in (1).  Now let us write it in a modified form: 

8. We know that ¹⁄₆ = ^(1×100) ⁄_(6×100). 
9. Let us rearrange the right side: ¹⁄₆ = ¹⁄₁₀₀ × ¹⁰⁰⁄₆ 
10. In the above result, we can write ¹⁰⁰⁄₆ as (16 + ⁴⁄₆ )
11. So (9) becomes ¹⁄₆ = ¹⁄₁₀₀ × (16 + ⁴⁄₆ ). So we get:
12. ¹⁄₆ = ¹⁶⁄₁₀₀ + ⁴⁄₆₀₀
13. Look at the above result carefully. We have two fractions on the right side: ¹⁶⁄₁₀₀ and ⁴⁄₆₀₀
    ♦ Out of these two, ¹⁶⁄₁₀₀ is in an 'ideal form'. Because it has a 'power of 10' in the denominator. So it can be readily converted into a decimal form
    ♦ But the other fraction ⁴⁄₆₀₀ is causing a problem. It cannot be readily converted into a decimal form
    ♦ If this ⁴⁄₆₀₀ is very very small, then we can ignore it. In that case, (12) will become ¹⁄₆ = ¹⁶⁄₁₀₀
    ♦ But unfortunately, ⁴⁄₆₀₀ is not very small, and we cannot ignore it
• After reaching (12), if we write ¹⁄₆ = 0.16, we are ignoring ⁴⁄₆₀₀
• That is not a good thing to do because, ⁴⁄₆₀₀ is not a small quantity, that can be 'just ignored'
• It may be noted that ⁴⁄₆₀₀ is ten times smaller than ⁴⁄₆₀ , which is causing the problem in (5) 
14. We arrived at (12) by writing ¹⁄₆ as ^(1×100) ⁄_(6×100) in (8).  Now let us write it in a modified form:

15. We know that ¹⁄₆ = ^(1×1000) ⁄_(6×1000). 
16. Let us rearrange the right side: ¹⁄₆ = ¹⁄₁₀₀₀ × ¹⁰⁰⁰⁄₆ 
17. In the above result, we can write ¹⁰⁰⁰⁄₆ as (166 + ⁴⁄₆ )
18. So (16) becomes ¹⁄₆ = ¹⁄₁₀₀₀ × (166 + ⁴⁄₆ ). So we get:
19. ¹⁄₆ = ¹⁶⁶⁄₁₀₀₀ + ⁴⁄₆₀₀₀
20. Look at the above result carefully. We have two fractions on the right side: ¹⁶⁶⁄₁₀₀₀ and ⁴⁄₆₀₀₀
    ♦ Out of these two, ¹⁶⁶⁄₁₀₀₀ is in an 'ideal form'. Because it has a 'power of 10' in the denominator. So it can be readily converted into a decimal form
    ♦ But the other fraction ⁴⁄₆₀₀₀ is causing a problem. It cannot be readily converted into a decimal form
    ♦ If this ⁴⁄₆₀₀₀ is very very small, then we can ignore it. In that case, (19) will become ¹⁄₆ = ¹⁶⁶⁄₁₀₀₀
    ♦ But unfortunately, ⁴⁄₆₀₀₀ is not very small, and we cannot ignore it
• After reaching (19), if we write ¹⁄₆ = 0.166, we are ignoring ⁴⁄₆₀₀₀
• That is not a good thing to do because, ⁴⁄₆₀₀₀ is not a small quantity, that can be 'just ignored'
• It may be noted that ⁴⁄₆₀₀₀ is ten times smaller than ⁴⁄₆₀₀ , which is causing the problem in (12)
• Also it is 100 times smaller than ⁴⁄₆₀ , which is causing the problem in (5)
• So the fractional part is obviously decreasing with each step. It will keep on decreasing with each step, and reach very low values. How low can it reach? 
The lowest value possible is 'zero'. So, with each step, the fractional part gets closer and closer to zero 
21. We arrived at (19) by writing ¹⁄₆ as ^(1×1000) ⁄_(6×1000) in (15)

First we used 10, then 100, and we used 1000 just above. We can proceed using 10000, 100000, etc.,
But we do not have to write the steps. A pattern has already emerged. Based on that pattern, we can write:
■ ¹⁄₆    =  ¹⁄₁₀   +  ⁴⁄₆₀       =   0.1 + ⁴⁄₆₀
■ ¹⁄₆    =  ¹⁶⁄₁₀₀   +  ⁴⁄₆₀₀      =   0.16 +  ⁴⁄₆₀₀
■ ¹⁄₆    =  ¹⁶⁶⁄₁₀₀₀   +  ⁴⁄₆₀₀₀       =   0.166 + ⁴⁄₆₀₀₀
■ ¹⁄₆    =  ¹⁶⁶⁶⁄₁₀₀₀₀   +  ⁴⁄₆₀₀₀₀       =   0.1666 + ⁴⁄₆₀₀₀₀
■ ¹⁄₆     =  ¹⁶⁶⁶⁶⁄₁₀₀₀₀₀   +  ⁴⁄₆₀₀₀₀₀        =   0.16666 + ⁴⁄₆₀₀₀₀₀

All the above results are true. They are exact values of ¹⁄₆ . We can proceed further as long as we wish. But this much is sufficient for us to understand an important property:

When the number of digits on the 'right side of the decimal point' increases, the remaining fractional portion decreases. 
• For example, if we take 4 places on the right side of the decimal point, ¹⁄₆ = 0.1666, the fractional part then is  ⁴⁄₆₀₀₀₀
• If we take 5 places on the right side of the decimal point, ¹⁄₆ = 0.16666, the fractional part then is ⁴⁄₆₀₀₀₀₀, which is smaller than ⁴⁄₆₀₀₀₀₀

As the fractional part becomes smaller and smaller, it can be ignored if :
We take sufficient number of places after the decimal point.

Another important point can also be noted from the above discussion:
• We have written ¹⁄₆ as the sum of a decimal value and a fractional value 
• The left side is always a constant, which is equal to ¹⁄₆
• So the right side must also be a constant. That is., the sum of the decimal value and the fractional value must also be a constant equal to ¹⁄₆
• But we saw that, when the number of places after the decimal point increases, the fractional value decreases
• When the fractional value decreases, the decimal portion must increase in value. Then only will the sum remain a constant. Thus we can say: 
■ As the number of decimal places increases, the decimal portion increases, and gets closer and closer to ¹⁄₆ . This is shown below:

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So now we know that we cannot convert ¹⁄₆ into an exact decimal form. There will always be a small fraction remaining. As in the case of ¹⁄₃, here also we will use the special method to represent it.


In the final pattern that we derived above, we saw 0.1, 0.16, 0.166, 0.1666, and so on. The digit '6' will repeat forever. So ¹⁄₆, when converted into decimal form, will give a recurring decimal. We can represent the decimal by any one of the methods that we saw in the case of ¹⁄₃.

• In Method 1, three dots are placed after the decimal. It indicates that it is a recurring decimal
• In Method 2, a dot is placed above the digit which repeats forever. In our case, 6 repeats for ever. So, the dot is placed over it

• In Method 3, a line is placed above the digit which repeats forever. In our case, 6 repeats for ever. So, the line is placed over it

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Now we will see some solved examples
Solved example 6.28
For each of the fractions given below, write fractions (with denominators powers of 10) getting closer and closer to the original value, and then write the decimal form
(i) ⁵⁄₆ ,  (ii) ³⁄₁₁ ,  (iii) ²³⁄₁₁ ,  (iv) ¹⁄₁₃ 
Solution:
(i) ⁵⁄₆ : 1. We know that ⁵⁄₆ = ^(5×10) ⁄_(6×10). 
2. Let us rearrange the right side: ⁵⁄₆ = ⁵⁄₁₀ × ¹⁰⁄₆ 
3. In the above result, we can write ¹⁰⁄₆ as (1 + ⁴⁄₆ )
4. So (2) becomes ⁵⁄₆ = ⁵⁄₁₀ × (1 + ⁴⁄₆ ). So we get:
5. ⁵⁄₆ = ⁵⁄₁₀ + ²⁰⁄₆₀
Second cycle:
1. We know that ⁵⁄₆ = ^(5×100) ⁄_(6×100). 
2. Let us rearrange the right side: ⁵⁄₆ = ⁵⁄₁₀₀ × ¹⁰⁰⁄₆ 
3. In the above result, we can write ¹⁰⁰⁄₆ as (16 + ⁴⁄₆ )
4. So (2) becomes ⁵⁄₆ = ⁵⁄₁₀₀ × (16 + ⁴⁄₆ ). So we get:
5. ⁵⁄₆ = ⁸⁰⁄₁₀₀ + ²⁰⁄₆₀₀
Third cycle:
1. We know that ⁵⁄₆ = ^(5×1000) ⁄_(6×1000). 
2. Let us rearrange the right side: ⁵⁄₆ = ⁵⁄₁₀₀₀ × ¹⁰⁰⁰⁄₆ 
3. In the above result, we can write ¹⁰⁰⁰⁄₆ as (166 + ⁴⁄₆ )
4. So (2) becomes ⁵⁄₆ = ⁵⁄₁₀₀₀ × (166 + ⁴⁄₆ ). So we get:
5. ⁵⁄₆ = ⁸³⁰⁄₁₀₀₀ + ²⁰⁄₆₀₀₀
Fourth cycle:
1. We know that ⁵⁄₆ = ^(5×10000) ⁄_(6×10000). 
2. Let us rearrange the right side: ⁵⁄₆ = ⁵⁄₁₀₀₀₀ × ¹⁰⁰⁰⁰⁄₆ 
3. In the above result, we can write ¹⁰⁰⁰⁰⁄₆ as (1666 + ⁴⁄₆ )
4. So (2) becomes ⁵⁄₆ = ⁵⁄₁₀₀₀₀ × (1666 + ⁴⁄₆ ). So we get:
5. ⁵⁄₆ = ⁸³³⁰⁄₁₀₀₀ + ²⁰⁄₆₀₀₀₀

Based on the above results we can write:
■ ⁵⁄₆    =  ⁵⁄₁₀   +  ²⁰⁄₆₀       =   0.5 + ²⁰⁄₆₀
■ ⁵⁄₆    =  ⁸⁰⁄₁₀₀   +  ²⁰⁄₆₀₀      =   0.8 +  ²⁰⁄₆₀₀
■ ⁵⁄₆    =  ⁸³⁰⁄₁₀₀₀   +  ²⁰⁄₆₀₀₀       =   0.83 + ²⁰⁄₆₀₀₀
■ ⁵⁄₆    =  ⁸³³⁰⁄₁₀₀₀₀   +  ²⁰⁄₆₀₀₀₀       =   0.833 + ²⁰⁄₆₀₀₀₀
The fractions ⁵⁄₁₀, ⁸⁰⁄₁₀₀, ⁸³⁰⁄₁₀₀₀, ⁸³³⁰⁄₁₀₀₀₀, etc., gets closer and closer to ⁵⁄₆ . Based on this, we can write the decimal form of as:

The result obtained when we divide 5 by 6 using a calculator is shown below:

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In the next section we will solve the second problem ³⁄₁₁.

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Chapter 6.9 - Basics of Recurring decimals

In the previous section we saw how money is expressed as decimals. In this section, we will see some more advanced topics related to decimals.
We have seen the basics about decimals here. We know how to convert fractions like ¹⁄₂ and ³⁄₄ into decimal form. 
We know that  ¹⁄₂ = 0.5 and ³⁄₄  = 0.75
To convert fractions like ¹⁄₈, a little more work is involved. We saw such problems here. Let us analyse them again:
• Take ¹⁄₈.  To convert it into decimal for, we must first convert it into an equivalent fraction
• The denominator of this equivalent fraction should be any one of 10¹, 10², 10³ . . . etc., which ever is suitable
• For obtaining such an equivalent fraction, we must multiply both the numerator and denominator of ¹⁄₈ by a 'suitable number'
• There is a clear procedure to obtain this 'suitable number'. Let us see what it is:
1. We have '8' in the denominator. We must factorise it first
2. We have: 8 = 2×2×2. There are 3 'twos' 
3. We must convert each of these 'twos' into a '10'
4. For that, we give each two, a '5' like this: (2×5)×(2×5)×(2×5). Now it becomes 10×10×10 = 1000
5. So 3 external 'fives' are used to get a power of 10
6. 3 external 'fives' give 5 ×5 ×5 = 125.   So the 'suitable number' is 125. If we multiply the denominator by 125, we will get the 'required power of 10'. But the numerator must also be multiplied by 125. So we can write:
7. ¹⁄₈ = ^(1×125) ⁄_(8×125) = ¹²⁵⁄₁₀₀₀ = 0.125

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The above result will find application in another situation also:
If we have a fraction with denominator 125, we can multiply both the numerator and denominator by '8'. We will get the 'required power of 10' in the denominator.

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Another example: Convert ³⁄₁₆₀  into decimal form
1. We have '160' in the denominator. We must factorise it first
2. We have: 160 = 2×2×2×2×2×5. There are 5 'twos', and a five. We will separate 1 two and the five
3. So we can write: 160 = (2×2×2×2)×(2×5) =  (2×2×2×2)×(10). So we have 4 twos remaining
4. We must convert each of these 'twos' into a '10'
5. For that, we give each two, a '5' like this: (2×5)×(2×5)×(2×5)×(2×5)×(10). Now it becomes 10×10×10×10×10 = 100000 =10⁵
6. So 4 external 'fives' are used to get a power of 10. 
7. 4 external 'fives' give 5 ×5 ×5×5 = 625. So the 'suitable number' is 625. If we multiply the denominator by 625, we will get the 'required power of 10'. But the numerator must also be multiplied by 625. So we can write:
8. ³⁄₁₆₀ = ^(3×625) ⁄_(160×625) = ¹⁸⁷⁵⁄₁₀₀₀₀₀ = 0.01875

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We will see some solved examples:
Solved example 6.27
Write each of the fractions below in the decimal form
(i) ¹⁄₅₀,  (ii) ³⁄₄₀,  (ii) ⁵⁄₁₆,   (iv) ¹²⁄₆₂₅
Solution:
(i) ¹⁄₅₀ : We know that when 50 is multiplied by 2, we will get 100. But we will do the steps to get more acquainted with the process:
1. We have '50' in the denominator. We must factorise it first
2. We have: 50 = 2×5×5. There are 2 'fives', and a 'two'. We will separate 1 five and the two
3. So we can write: 50 = (5)×(2×5) = (5)×(10) . So we have one 'five' remaining
4. We must convert this 'five' into a '10'
5. For that, we give this 'five', a '2' like this: (5×2)×(10). Now it becomes 10×10 = 100 =10²
6. So 1 external 'two' is used to get a power of 10. 
7. So the 'suitable number' is 2. If we multiply the denominator by 2, we will get the 'required power of 10'. But the numerator must also be multiplied by 2. So we can write:
8. ¹⁄₅₀ = ^(1×2) ⁄_(50×2) = ²⁄₁₀₀ = 0.02

(ii) ³⁄₄₀ : 1. We have '40' in the denominator. We must factorise it first
2. We have: 40 = 2×2×2×5. There are 3 'twos', and a five. We will separate 1 two and the five
3. So we can write: 40 = (2×2)×(2×5) =  (2×2)×(10). So we have 2 twos remaining
4. We must convert each of these 'twos' into a '10'
5. For that, we give each two, a '5' like this: (2×5)×(2×5)×(10). Now it becomes 10×10×10 = 1000 =10³
6. So 2 external 'fives' are used to get a power of 10. 
7. 2 external 'fives' give 5 ×5  = 25. So the 'suitable number' is 25. If we multiply the denominator by 25, we will get the 'required power of 10'. But the numerator must also be multiplied by 25. So we can write:
8. ³⁄₄₀ = ^(3×25) ⁄_(40×25) = ⁷⁵⁄₁₀₀₀ = 0.075

(iii) ⁵⁄₁₆ : 1. We have '16' in the denominator. We must factorise it first
2. We have: 16 = 2×2×2×2. There are 4 'twos'.
3. We must convert each of these 'twos' into a '10'
4. For that, we give each two, a '5' like this: (2×5)×(2×5)×(2×5)×(2×5). Now it becomes 10×10×10×10 = 10000 =10⁴
5. So 4 external 'fives' are used to get a power of 10. 
6. 4 external 'fives' give 5 ×5 ×5×5 = 625. So the 'suitable number' is 625. If we multiply the denominator by 625, we will get the 'required power of 10'. But the numerator must also be multiplied by 625. So we can write:
7. ⁵⁄₁₆ = ^(5×625) ⁄_(16×625) = ³¹²⁵⁄₁₀₀₀₀ = 0.3125

(iv) ¹²⁄₆₂₅ : 1. We have '625' in the denominator. We must factorise it first
2. We have: 625 = 5×5×5×5. There are 4 'fives'
3. We must convert each of these 'fives' into a '10'
4. For that, we give each five, a '2' like this: (5×2)×(5×2)×(5×2)×(5×2). Now it becomes 10×10×10×10 = 10000 =10⁴
5. So 4 external 'twos' are used to get a power of 10. 
6. 4 external 'twos' give 2 ×2 ×2×2 = 16. So the 'suitable number' is 16. If we multiply the denominator by 16, we will get the 'required power of 10'. But the numerator must also be multiplied by 16. So we can write:

7. ¹²⁄₆₂₅ = ^(12×16) ⁄_(625×16) = ¹⁹²⁄₁₀₀₀₀ = 0.0192

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Now let us try to write ¹⁄₃ in decimal form. The above method will not work because, there is no natural number, which when multiplied with 3, will give any 'power of 10'. So we will use another method:
1. We know that ¹⁄₃ = ^(1×10) ⁄_(3×10). 
2. Let us rearrange the right side: ¹⁄₃ = ¹⁄₁₀ × ¹⁰⁄₃ 
3. In the above result, we can write ¹⁰⁄₃ as (3 + ¹⁄₃ )
4. So (2) becomes ¹⁄₃ = ¹⁄₁₀ × (3 + ¹⁄₃ ). So we get:
5. ¹⁄₃ = ³⁄₁₀ + ¹⁄₃₀
6. Look at the above result carefully. We have two fractions on the right side: ³⁄₁₀ and ¹⁄₃₀ 
    ♦ Out of these two, ³⁄₁₀ is in an 'ideal form'. Because it has a 'power of 10' in the denominator. So it can be readily converted into a decimal form
    ♦ But the other fraction ¹⁄₃₀ is causing a problem. It cannot be readily converted into a decimal form
    ♦ If this ¹⁄₃₀ is very very small, then we can ignore it. In that case, (5) will become ¹⁄₃ = ³⁄₁₀
    ♦ But unfortunately, ¹⁄₃₀ is not very small, and we cannot ignore it. 
• After reaching (5), if we write ¹⁄₃ = 0.3, we are ignoring ¹⁄₃₀
• That is not a good thing to do because, ¹⁄₃₀ is not a small quantity, that can be 'just ignored'
7. We arrived at (5) by writing ¹⁄₃ as ^(1×10) ⁄_(3×10) in (1).  Now let us write it in a modified form: 

8. We know that ¹⁄₃ = ^(1×100) ⁄_(3×100). 
9. Let us rearrange the right side: ¹⁄₃ = ¹⁄₁₀₀ × ¹⁰⁰⁄₃ 
10. In the above result, we can write ¹⁰⁰⁄₃ as (33 + ¹⁄₃ )
11. So (9) becomes ¹⁄₃ = ¹⁄₁₀₀ × (33 + ¹⁄₃ ). So we get:
12. ¹⁄₃ = ³³⁄₁₀₀ + ¹⁄₃₀₀
13. Look at the above result carefully. We have two fractions on the right side: ³³⁄₁₀₀ and ¹⁄₃₀₀
    ♦ Out of these two, ³³⁄₁₀₀ is in an 'ideal form'. Because it has a 'power of 10' in the denominator. So it can be readily converted into a decimal form
    ♦ But the other fraction ¹⁄₃₀₀ is causing a problem. It cannot be readily converted into a decimal form
    ♦ If this ¹⁄₃₀₀ is very very small, then we can ignore it. In that case, (12) will become ¹⁄₃ = ³³⁄₁₀₀
    ♦ But unfortunately, ¹⁄₃₀₀ is not very small, and we cannot ignore it
• After reaching (12), if we write ¹⁄₃ = 0.33, we are ignoring ¹⁄₃₀₀
• That is not a good thing to do because, ¹⁄₃₀₀ is not a small quantity, that can be 'just ignored'
• It may be noted that ¹⁄₃₀₀ is ten times smaller than ¹⁄₃₀ , which is causing the problem in (5) 
14. We arrived at (12) by writing ¹⁄₃ as ^(1×100) ⁄_(3×100) in (8).  Now let us write it in a modified form:

15. We know that ¹⁄₃ = ^(1×1000) ⁄_(3×1000). 
16. Let us rearrange the right side: ¹⁄₃ = ¹⁄₁₀₀₀ × ¹⁰⁰⁰⁄₃ 
17. In the above result, we can write ¹⁰⁰⁰⁄₃ as (333 + ¹⁄₃ )
18. So (16) becomes ¹⁄₃ = ¹⁄₁₀₀₀ × (333 + ¹⁄₃ ). So we get:
19. ¹⁄₃ = ³³³⁄₁₀₀₀ + ¹⁄₃₀₀₀
20. Look at the above result carefully. We have two fractions on the right side: ³³³⁄₁₀₀₀ and ¹⁄₃₀₀₀
    ♦ Out of these two, ³³³⁄₁₀₀₀ is in an 'ideal form'. Because it has a 'power of 10' in the denominator. So it can be readily converted into a decimal form
    ♦ But the other fraction ¹⁄₃₀₀₀ is causing a problem. It cannot be readily converted into a decimal form
    ♦ If this ¹⁄₃₀₀₀ is very very small, then we can ignore it. In that case, (19) will become ¹⁄₃ = ³³³⁄₁₀₀₀
    ♦ But unfortunately, ¹⁄₃₀₀₀ is not very small, and we cannot ignore it
• After reaching (19), if we write ¹⁄₃ = 0.333, we are ignoring ¹⁄₃₀₀₀
• That is not a good thing to do because, ¹⁄₃₀₀₀ is not a small quantity, that can be 'just ignored'
• It may be noted that ¹⁄₃₀₀₀ is ten times smaller than ¹⁄₃₀₀ , which is causing the problem in (12)
• Also it is 100 times smaller than ¹⁄₃₀ , which is causing the problem in (5)
• So the fractional part is obviously decreasing with each step. It will keep on decreasing with each step and reach very low values. How low can it reach? 
The lowest value possible is 'zero'. So, with each step, the fractional part gets closer and closer to zero
21. We arrived at (19) by writing ¹⁄₃ as ^(1×1000) ⁄_(3×1000) in (15)

First we used 10, then 100, and we used 1000 just above. We can proceed using 10000, 100000, etc.,
But we do not have to write the steps. A pattern has already emerged. Based on that pattern, we can write:
■ ¹⁄₃    =  ³⁄₁₀   +  ¹⁄₃₀       =   0.3 + ¹⁄₃₀
■ ¹⁄₃    =  ³³⁄₁₀₀   +  ¹⁄₃₀₀      =   0.33 + ¹⁄₃₀₀
■ ¹⁄₃    =  ³³³⁄₁₀₀₀   +  ¹⁄₃₀₀₀       =   0.333 + ¹⁄₃₀₀₀
■ ¹⁄₃    =  ³³³³⁄₁₀₀₀₀   +  ¹⁄₃₀₀₀₀       =   0.3333 + ¹⁄₃₀₀₀₀
■ ¹⁄₃     =  ³³³³³⁄₁₀₀₀₀₀   +  ¹⁄₃₀₀₀₀₀        =   0.33333 + ¹⁄₃₀₀₀₀₀

All the above results are true. They are exact values of ¹⁄₃ . We can proceed further as long as we wish. But this much is sufficient for us to understand an important property:

When the number of digits on the 'right side of the decimal point' increases, the remaining fractional portion decreases. 
• For example, if we take 4 places on the right side of the decimal point, ¹⁄₃ = 0.3333, the fractional part then is  ¹⁄₃₀₀₀₀
• If we take 5 places on the right side of the decimal point, ¹⁄₃ = 0.33333, the fractional part then is ¹⁄₃₀₀₀₀₀, which is smaller than ¹⁄₃₀₀₀₀

As the fractional part becomes smaller and smaller, it can be ignored if we take sufficient number of places after the decimal point. In various fields of science and engineering, there are strict rules that tell us the 'number of places' that we have to take after the decimal point.

Another important point can also be noted from the above discussion:
• We have written ¹⁄₃ as the sum of a decimal value and a fractional value 
• The left side is always a constant, which is equal to ¹⁄₃
• So the right side must also be a constant. That is., the sum of the decimal value and the fractional value must also be a constant equal to ¹⁄₃
• But we saw that, when the number of places after the decimal point increases, the fractional value decreases
• When the fractional value decreases, the decimal portion must increase in value. Then only will the sum remain a constant. Thus we can say: 
■ As the number of decimal places increases, the value of the decimal portion gets closer and closer to ¹⁄₃ . This is shown below:

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So now we know that we cannot convert ¹⁄₃ into an exact decimal form. There will always be a small fraction remaining. We use a special method to represent such decimals.

In the final pattern that we derived above, we saw 0.3, 0.33, 0.333, and so on. The digit '3' will repeat for ever. Such decimals are called recurring decimals. There are three different ways to represent recurring decimals. We will see the details of those methods by taking ¹⁄₃ as an example:

• In Method 1, three dots are placed after the decimal. It indicates that it is a recurring decimal
• In Method 2, a dot is placed above the digit which repeats forever. In our case, 3 repeats for ever. So, the dot is placed over it
• In Method 3, a line is placed above the digit which repeats forever. In our case, 3 repeats for ever. So, the line is placed over it

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In the next section we will see another example.

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