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