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
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.
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
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
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:
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
In the next section we will see another example.
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
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.
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
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
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:
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 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
In the next section we will see another example.
