In the previous section we saw some solved examples which demonstrates the basics of recurring decimals. In this section, we will see a few more problems. Later in this section, we will see an easy method to obtain recurring decimals.
Solved example 6.29
(i) Explain using algebra, that the fractions 1⁄10, 11⁄100, 111⁄1000, ... gets closer and closer to 1⁄9
(ii) Using the general principle above, find the decimal forms of 2⁄9, 3⁄9, 4⁄9, 5⁄9, 6⁄9, 7⁄9, 8⁄9.
(iii) What can we say in general about those decimal forms in which a single digit repeats?
Solution:
(i) Let us first analyse the problem.
• When we find a general principle using algebra, we get a 'method of solution', which will be applicable to any number. Let the number be 'k'.
• We want to find 1⁄9 of 'k'. That is., we want to find k⁄9
• Can we write k⁄9 = k⁄10 ?
• The answer is: k⁄10 is close to k⁄9. But it is not the exact required value
• Can we write k⁄9 = 11k⁄100 ?
• The answer is: 11k⁄100 is close to k⁄9. But it is not the exact required value.
♦ Also, 11k⁄100 is closer to k⁄9 than k⁄10. That is., 11k⁄100 is a better result than k⁄10
• Can we write k⁄9 = 111k⁄1000 ?
• The answer is: 111k⁄1000 is close to k⁄9. But it is not the exact required value.
♦ Also, 111k⁄1000 is closer to k⁄9 than 11k⁄100. That is., 111k⁄1000 is a better result than 11k⁄100
But we have to show proof. It can be done as follows:
1. We know that k⁄9 = (k×10) ⁄(9×10).
2. Let us rearrange the right side: k⁄9 = k⁄10 × 10⁄9
3. In the above result, we can write 10⁄9 as (1 + 1⁄9 )
4. So (2) becomes k⁄9 = k ⁄10 × (1 + 1⁄9). So we get:
5. k⁄9 = k⁄10 + k⁄90
6. Look at the above result carefully. We have two fractions on the right side: k⁄10 and k⁄90
♦ Out of these two, k⁄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 k⁄90 is causing a problem. It cannot be readily converted into a decimal form
♦ If this k⁄90 is very very small, then we can ignore it. In that case, (5) will become k⁄9 = k⁄10
♦ But unfortunately, k⁄90 is not very small, and we cannot ignore it.
• After reaching (5), if we write k⁄9 = 0.3, we are ignoring k⁄90
• That is not a good thing to do because, k⁄90 is not a small quantity, that can be 'just ignored'
7. We arrived at (5) by writing k⁄9 as (k×10) ⁄(9×10) in (1). Now let us write it in a modified form using 100:
8. We know that k⁄9 = (k×100) ⁄(9×100).
9. Let us rearrange the right side: k⁄9 = k⁄100 × 100⁄9
10. In the above result, we can write 100⁄9 as (11 + 1⁄9 )
11. So (9) becomes k⁄9 = k⁄100 × (11 + 1⁄9 ). So we get:
12. k⁄9 = 11k ⁄100 + k⁄900
13. Look at the above result carefully. We have two fractions on the right side: 11k⁄100 and k⁄900
♦ Out of these two, 11k⁄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 k⁄900 is causing a problem. It cannot be readily converted into a decimal form
♦ If this k⁄900 is very very small, then we can ignore it. In that case, (12) will become k⁄9 = 11k⁄100
♦ But unfortunately, k⁄900 is not very small, and we cannot ignore it
• After reaching (12), if we write k⁄9 = 0.11k, we are ignoring k⁄900
• That is not a good thing to do because, k⁄900 is not a small quantity, that can be 'just ignored'
• It may be noted that k⁄900 is ten times smaller than k⁄90, which is causing the problem in (5)
14. We arrived at (12) by writing k⁄9 as (k×100) ⁄(9×100) in (8). Now let us write it in a modified form using 1000:
15. We know that k⁄9 = (k×1000) ⁄(9×1000).
16. Let us rearrange the right side: k⁄9 = k⁄1000 × 1000⁄9
17. In the above result, we can write 1000⁄9 as (111 + 1⁄9 )
18. So (16) becomes k⁄9 = k⁄1000× (111 + 1⁄9 ). So we get:
19. k⁄9 = 111k ⁄1000 + k⁄9000
20. Look at the above result carefully. We have two fractions on the right side: 111k⁄1000 and k⁄9000
♦ Out of these two, 111k⁄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 k⁄9000 is causing a problem. It cannot be readily converted into a decimal form
♦ If this k⁄9000 is very very small, then we can ignore it. In that case, (19) will become k⁄9 = 111k⁄1000
♦ But unfortunately, k⁄9000 is not very small, and we cannot ignore it
• After reaching (19), if we write k⁄9 = 0.111k, we are ignoring k⁄9000
• That is not a good thing to do because, k⁄9000 is not a small quantity, that can be 'just ignored'
• It may be noted that k⁄9000 is ten times smaller than k⁄900, which is causing the problem in (12)
• Also it is 100 times smaller than k⁄90, 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 k⁄9 as (k×1000) ⁄(k×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:
■ k⁄9 = k⁄10 + k⁄90 = 0.1k + k⁄90
■ k⁄9 = 11k⁄100 + k⁄900 = 0.11k + k⁄900
■ k⁄9 = 111k⁄1000 + k⁄9000 = 0.111k + k⁄9000
■ k⁄9 = 1111k⁄10000 + k⁄90000 = 0.1111k + k⁄90000
Each of the above results are true. On the right side of the '=' sign, there are two terms. The first term can be readily written in the decimal form because, it has powers of 10 in the denominator. The second term causes problem. But this second term becomes lesser and lesser and becomes closer to zero, with each step. So the first term increases and become closer and closer to k⁄9 with each step.
■ That is., k⁄10, 11k⁄100, 111k⁄1000, ... gets closer and closer to k⁄9
Hence proved.
(ii) We can use the above proof to find 2⁄9, 4⁄9, 5⁄9, etc.,
(a) 2⁄9: • We proved that k⁄10, 11k⁄100, 111k⁄1000, ... gets closer and closer to k⁄9
• Here, k = 2. So 2⁄10, (11×2) ⁄100, (111×2) ⁄1000 , ... gets closer and closer to 2⁄9
• That is., 2⁄10, 22 ⁄100, 222 ⁄1000 , ... gets closer and closer to 2⁄9
• That is., 0.2, 0.22, 0.222 , ... gets closer and closer to 2⁄9
• The digit '2' repeats for ever. So we can write 2⁄9 = 0.222...
(b) 3⁄9: • Here, k = 3. So 3⁄10, (11×3) ⁄100, (111×3) ⁄1000 , ... gets closer and closer to 3⁄9
• That is., 3⁄10, 33 ⁄100, 333⁄1000 , ... gets closer and closer to 3⁄9
• That is., 0.3, 0.33, 0.333 , ... gets closer and closer to 3⁄9
• The digit '3' repeats for ever. So we can write 4⁄9 = 0.333...
(c) 4⁄9: • Here, k = 4. So 4⁄10, (11×4) ⁄100, (111×4) ⁄1000 , ... gets closer and closer to 4⁄9
• That is., 4⁄10, 44 ⁄100, 444 ⁄1000 , ... gets closer and closer to 4⁄9
• That is., 0.4, 0.44, 0.444 , ... gets closer and closer to 4⁄9
• The digit '4' repeats for ever. So we can write 4⁄9 = 0.444...
By the same method we can write:
(d) 5⁄9 = 0.555... , (e) 6⁄9 = 0.666...(f) 7⁄9 = 0.777... , (g) 8⁄9 = 0.888...
(iii) In general, when a single digit repeats in the decimal form, the denominator of the fraction form will be '9'
We have seen the details about recurring decimals, and also we have seen many solved examples. We will now see an easy method to convert those fractions which give recurring decimals.
Consider the fraction 1⁄3 . We can divide 1 by 3, using the usual 'long division method'. Fig.6.36(a) below shows the process:
In fig. 6.36(b), we can see that, in each step, we are getting '1' as the remainder. It is marked with red circles. So in each step, we will get a '3' in the quotient. This will continue for ever. Thus we write: 1⁄3 = 0.333...
Another example:
Consider the fraction 1⁄6 . We can divide 1 by 6, using the usual 'long division method'. Fig.6.36(c) shows the process. In fig. 6.36(d), we can see that, in each step, we are getting '4' as the remainder. It is marked with red circles. So in each step, we will get a '6' in the quotient. This will continue for ever. Thus we write: 1⁄6 = 0.1666...
Another example:
Consider the fraction 2⁄3 . We can divide 2 by 3, using the usual 'long division method'. Fig.6.37(a) below shows the process:
In fig. 6.37(b), we can see that, in each step, we are getting '2' as the remainder. It is marked with red circles. So in each step, we will get a '6' in the quotient. This will continue for ever. Thus we write: 2⁄3 = 0.666...
Another example:
Consider the fraction 1⁄7 . We can divide 1 by 7, using the usual 'long division method'. Fig.6.37(c) shows the process. We can see that, in each step, the remainder obtained is different. So in each step, we will get a different digit in the quotient. But after a few steps, the pattern repeats again. The point of repetition is marked by the second red arrow. The pattern will continue forever. So, in the quotient, we draw a red line over the repeating pattern.
This completes the discussion on Recurring decimals. In the next chapter we will learn about 'Ratios'.
Solved example 6.29
(i) Explain using algebra, that the fractions 1⁄10, 11⁄100, 111⁄1000, ... gets closer and closer to 1⁄9
(ii) Using the general principle above, find the decimal forms of 2⁄9, 3⁄9, 4⁄9, 5⁄9, 6⁄9, 7⁄9, 8⁄9.
(iii) What can we say in general about those decimal forms in which a single digit repeats?
Solution:
(i) Let us first analyse the problem.
• When we find a general principle using algebra, we get a 'method of solution', which will be applicable to any number. Let the number be 'k'.
• We want to find 1⁄9 of 'k'. That is., we want to find k⁄9
• Can we write k⁄9 = k⁄10 ?
• The answer is: k⁄10 is close to k⁄9. But it is not the exact required value
• Can we write k⁄9 = 11k⁄100 ?
• The answer is: 11k⁄100 is close to k⁄9. But it is not the exact required value.
♦ Also, 11k⁄100 is closer to k⁄9 than k⁄10. That is., 11k⁄100 is a better result than k⁄10
• Can we write k⁄9 = 111k⁄1000 ?
• The answer is: 111k⁄1000 is close to k⁄9. But it is not the exact required value.
♦ Also, 111k⁄1000 is closer to k⁄9 than 11k⁄100. That is., 111k⁄1000 is a better result than 11k⁄100
But we have to show proof. It can be done as follows:
1. We know that k⁄9 = (k×10) ⁄(9×10).
2. Let us rearrange the right side: k⁄9 = k⁄10 × 10⁄9
3. In the above result, we can write 10⁄9 as (1 + 1⁄9 )
4. So (2) becomes k⁄9 = k ⁄10 × (1 + 1⁄9). So we get:
5. k⁄9 = k⁄10 + k⁄90
6. Look at the above result carefully. We have two fractions on the right side: k⁄10 and k⁄90
♦ Out of these two, k⁄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 k⁄90 is causing a problem. It cannot be readily converted into a decimal form
♦ If this k⁄90 is very very small, then we can ignore it. In that case, (5) will become k⁄9 = k⁄10
♦ But unfortunately, k⁄90 is not very small, and we cannot ignore it.
• After reaching (5), if we write k⁄9 = 0.3, we are ignoring k⁄90
• That is not a good thing to do because, k⁄90 is not a small quantity, that can be 'just ignored'
7. We arrived at (5) by writing k⁄9 as (k×10) ⁄(9×10) in (1). Now let us write it in a modified form using 100:
8. We know that k⁄9 = (k×100) ⁄(9×100).
9. Let us rearrange the right side: k⁄9 = k⁄100 × 100⁄9
10. In the above result, we can write 100⁄9 as (11 + 1⁄9 )
11. So (9) becomes k⁄9 = k⁄100 × (11 + 1⁄9 ). So we get:
12. k⁄9 = 11k ⁄100 + k⁄900
13. Look at the above result carefully. We have two fractions on the right side: 11k⁄100 and k⁄900
♦ Out of these two, 11k⁄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 k⁄900 is causing a problem. It cannot be readily converted into a decimal form
♦ If this k⁄900 is very very small, then we can ignore it. In that case, (12) will become k⁄9 = 11k⁄100
♦ But unfortunately, k⁄900 is not very small, and we cannot ignore it
• After reaching (12), if we write k⁄9 = 0.11k, we are ignoring k⁄900
• That is not a good thing to do because, k⁄900 is not a small quantity, that can be 'just ignored'
• It may be noted that k⁄900 is ten times smaller than k⁄90, which is causing the problem in (5)
14. We arrived at (12) by writing k⁄9 as (k×100) ⁄(9×100) in (8). Now let us write it in a modified form using 1000:
15. We know that k⁄9 = (k×1000) ⁄(9×1000).
16. Let us rearrange the right side: k⁄9 = k⁄1000 × 1000⁄9
17. In the above result, we can write 1000⁄9 as (111 + 1⁄9 )
18. So (16) becomes k⁄9 = k⁄1000× (111 + 1⁄9 ). So we get:
19. k⁄9 = 111k ⁄1000 + k⁄9000
20. Look at the above result carefully. We have two fractions on the right side: 111k⁄1000 and k⁄9000
♦ Out of these two, 111k⁄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 k⁄9000 is causing a problem. It cannot be readily converted into a decimal form
♦ If this k⁄9000 is very very small, then we can ignore it. In that case, (19) will become k⁄9 = 111k⁄1000
♦ But unfortunately, k⁄9000 is not very small, and we cannot ignore it
• After reaching (19), if we write k⁄9 = 0.111k, we are ignoring k⁄9000
• That is not a good thing to do because, k⁄9000 is not a small quantity, that can be 'just ignored'
• It may be noted that k⁄9000 is ten times smaller than k⁄900, which is causing the problem in (12)
• Also it is 100 times smaller than k⁄90, 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 k⁄9 as (k×1000) ⁄(k×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:
■ k⁄9 = k⁄10 + k⁄90 = 0.1k + k⁄90
■ k⁄9 = 11k⁄100 + k⁄900 = 0.11k + k⁄900
■ k⁄9 = 111k⁄1000 + k⁄9000 = 0.111k + k⁄9000
■ k⁄9 = 1111k⁄10000 + k⁄90000 = 0.1111k + k⁄90000
Each of the above results are true. On the right side of the '=' sign, there are two terms. The first term can be readily written in the decimal form because, it has powers of 10 in the denominator. The second term causes problem. But this second term becomes lesser and lesser and becomes closer to zero, with each step. So the first term increases and become closer and closer to k⁄9 with each step.
■ That is., k⁄10, 11k⁄100, 111k⁄1000, ... gets closer and closer to k⁄9
Hence proved.
(ii) We can use the above proof to find 2⁄9, 4⁄9, 5⁄9, etc.,
(a) 2⁄9: • We proved that k⁄10, 11k⁄100, 111k⁄1000, ... gets closer and closer to k⁄9
• Here, k = 2. So 2⁄10, (11×2) ⁄100, (111×2) ⁄1000 , ... gets closer and closer to 2⁄9
• That is., 2⁄10, 22 ⁄100, 222 ⁄1000 , ... gets closer and closer to 2⁄9
• That is., 0.2, 0.22, 0.222 , ... gets closer and closer to 2⁄9
• The digit '2' repeats for ever. So we can write 2⁄9 = 0.222...
(b) 3⁄9: • Here, k = 3. So 3⁄10, (11×3) ⁄100, (111×3) ⁄1000 , ... gets closer and closer to 3⁄9
• That is., 3⁄10, 33 ⁄100, 333⁄1000 , ... gets closer and closer to 3⁄9
• That is., 0.3, 0.33, 0.333 , ... gets closer and closer to 3⁄9
• The digit '3' repeats for ever. So we can write 4⁄9 = 0.333...
(c) 4⁄9: • Here, k = 4. So 4⁄10, (11×4) ⁄100, (111×4) ⁄1000 , ... gets closer and closer to 4⁄9
• That is., 4⁄10, 44 ⁄100, 444 ⁄1000 , ... gets closer and closer to 4⁄9
• That is., 0.4, 0.44, 0.444 , ... gets closer and closer to 4⁄9
• The digit '4' repeats for ever. So we can write 4⁄9 = 0.444...
By the same method we can write:
(d) 5⁄9 = 0.555... , (e) 6⁄9 = 0.666...(f) 7⁄9 = 0.777... , (g) 8⁄9 = 0.888...
(iii) In general, when a single digit repeats in the decimal form, the denominator of the fraction form will be '9'
We have seen the details about recurring decimals, and also we have seen many solved examples. We will now see an easy method to convert those fractions which give recurring decimals.
Consider the fraction 1⁄3 . We can divide 1 by 3, using the usual 'long division method'. Fig.6.36(a) below shows the process:
 |
| Fig.6.36 |
Another example:
Consider the fraction 1⁄6 . We can divide 1 by 6, using the usual 'long division method'. Fig.6.36(c) shows the process. In fig. 6.36(d), we can see that, in each step, we are getting '4' as the remainder. It is marked with red circles. So in each step, we will get a '6' in the quotient. This will continue for ever. Thus we write: 1⁄6 = 0.1666...
Another example:
Consider the fraction 2⁄3 . We can divide 2 by 3, using the usual 'long division method'. Fig.6.37(a) below shows the process:
 |
| Fig.6.37 |
Another example:
Consider the fraction 1⁄7 . We can divide 1 by 7, using the usual 'long division method'. Fig.6.37(c) shows the process. We can see that, in each step, the remainder obtained is different. So in each step, we will get a different digit in the quotient. But after a few steps, the pattern repeats again. The point of repetition is marked by the second red arrow. The pattern will continue forever. So, in the quotient, we draw a red line over the repeating pattern.
This completes the discussion on Recurring decimals. In the next chapter we will learn about 'Ratios'.
No comments:
Post a Comment