1. Write the following in decimal form and say what kind of decimal expansion each has :
(i) ` \36/100\ `
(ii) ` \1/11\ `
(iii) ` \4\frac{1}{8}\ `
(iv) ` \3/13\ `
(v) ` \2/11\ `
(vi) ` \329/400\ `
solution :
(i) ` \36/100\ `
= 0.36 terminating.
(ii) ` \1/11\ `
= 0.0909 non-terminating repeating.
(iii) ` \4\frac{1}{8}\ `
= 4.125 Terminating.
(iv) ` \3/13\ `
= 0.230769 non-terminating repeating.
(v) ` \2/11\ `
= 0.18 non-terminating repeating.
(vi) ` \329/400\ `
= 0.8225 Terminating.
2. You know that ` \1/7= 0.\overline{142857}\ ` you
predict what the decimal expansions of ` \2/7, 2/7, 3/7, 4/7, 5/7, 1/7,
6/7\ ` are, without actually doing the long division? If so, how?
[Hint
: Study the remainders while finding the value of ` \1/7\ ` carefully.]
solution :
` \frac{1}{7} = 0.\overline{142857}\ `
∴ ` \2 × \frac{1}{7}=2 × 0.\overline{142857}=0.\overline{285714}\ `
∴ ` \3 × \frac{1}{7}=3 × 0.\overline{142857}=0.\overline{428571}\ `
∴ ` \4 × \frac{1}{7}=4 × 0.\overline{142857}=0.\overline{571428}\ `
∴ ` \5 × \frac{1}{7}=5 × 0.\overline{142857}=0.\overline{714285}\ `
∴ ` \6 × \frac{1}{7}=6 × 0.\overline{142857}=0.\overline{857142}\ `
3. Express the following in the form ` \p/q\ `, where p and q are integers and q ≠ 0.
(i) 0.6
(ii) 0.47
(iii) 0.001
solution :
(i) 0.6
0.6 = 0.666....
let ` \x = 0.66...\ `.
we will multiply both side by 10.
So, ` \10x = 6.66...\ `.
Or, ` \10x = 6 + 0.66...\ `.
Or, ` \10x = 6 + x\ `
Or, ` \10x - x = 6\ `
Or, ` \9x = 6\ `
Or, ` \x = \frac{6}{9}\ `
So, ` \x = \frac{2}{3}\ `
(ii) 0.47
0.47 = 0.4777....
let ` \x = 0.4777...\ `.
we will multiply both side by 10.
So, ` \10x = 4.777...\ `.
Or, ` \10x = 4.3 + 0.477...\ `.
Or, ` \10x = 4.3 + x\ `
Or, ` \10x - x = 4.3\ `
Or, ` \9x = 4.3\ `
Or, ` \x = \frac{4.3}{9}\ `
So, ` \x = \frac{43}{90}\ `
(iii) 0.001
0.001 = 0.001001....
let ` \x = 0.001001...\ `.
we will multiply both side by 1000.
So, ` \1000x = 1.001...\ `.
Or, ` \1000x = 1 + 0.001...\ `.
Or, ` \1000x = 1 + x\ `
Or, ` \1000x - x = 1\ `
Or, ` \999x = 1\ `
Or, ` \x = \frac{1}{999}\ `
So, ` \x =\frac{1}{999}\ `
4. Express 0.99999.... in the form ` \p/q\ `. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
solution :
0.99999....
let ` \x = 0.99999...\ `.
we will multiply both side by 10.
So, ` \10x = 9.9999...\ `.
Or, ` \10x = 9 + 0.9999...\ `.
Or, ` \10x = 9 + x\ `
Or, ` \10x - x = 9\ `
Or, ` \9x = 9\ `
Or, ` \x = \frac{9}{9}\ `
Or, ` \x = \frac{1}{1}\ `
So, ` \x = 1\ `
The difference between 1 and 0.99999 is 0.00001 which is negligible.
Hence, we can say that, 0.999 is too much near 1, therefore, 1 as the answer can be justified.
5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of` \1/17\ `? Perform the division to check your answer.
solution :
` \1/17\ `
Dividing 1 by 17
` \1/17\ ` = 0.0588235294117647
There are 16 digits in the repeating block of the decimal expansion of ` \1/17\ `.
6. Look at several examples of rational numbers in the form ` \p/q\ ` (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
solution :
We observe that when q is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10....... Then the decimal expansion is terminating.
For example :
` \frac{1}{1}=1\ `, denominator q = 1 × 1
` \frac{1}{2}=0.5\ `, denominator q = 2 × 1
` \frac{1}{3}=0.\overline{3}\ `, denominator q = 3 × 1
` \frac{1}{4}=0.25\ `, denominator q = 2 × 2
` \frac{1}{5}=0.2\ `, denominator q = 5 × 1
` \frac{1}{6}=0.\overline{16}\ `, denominator q = 2 × 3
` \frac{1}{7}=0.\overline{142857}\ `, denominator q = 7 × 1
` \frac{1}{8}=0.125\ `, denominator q = 2 × 2 × 2
` \frac{1}{9}=0.\overline{1}\ `, denominator q = 3 × 3
` \frac{1}{10}=0.1\ `, denominator q = 2 × 5
We can say that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.
7. Write three numbers whose decimal expansions are non-terminating non-recurring.
solution :
We know that all irrational numbers are non-terminating non-recurring.
Three numbers are,
(i) ` \sqrt{5}=\ `2.2360679774997896964091736687313..................
(ii) ` \sqrt{3}=\ `1.7320508075688772935274463415059..................
(iii) ` \sqrt{7}=\ `2.6457513110645905905016157536393..................
8. Find three different irrational numbers between the rational numbers ` \frac{5}{7}\ ` and ` \frac{9}{11}\ `.
solution :
` \frac{5}{7}=0.\overline{714285}\ `
` \frac{9}{11}=0.\overline{81}\ `
Three different irrational numbers are,
(i) 0.7507500750007500007500000...........
(ii) 0.7707700770007700007700000...........
(iii) 0.7907900790007900007900000...........
9. Classify the following numbers as rational or irrational :
(i) ` \sqrt{23}\ `
(ii) ` \sqrt{255}\ `
(iii) 0.3796
(iv) 7.478478
(v) 1.101001000100001.....
solution :
(i) ` \sqrt{23}\ `
` \sqrt{23}=4.79583152331......\ `.
Since the number is non-terminating non-recurring.
Therefore, it is an irrational number.
(ii) ` \sqrt{225}\ `
` \sqrt{225}=15=\frac{15}{1}\ `
Since the number can be represented in ` \frac{p}{q}\ ` form.
Therefore, it is a rational number.
(iii) 0.3796
0.3796 = 0.3796
Since the number is terminating.
Therefore, it is a rational number.
(iv) 7.478478
7.478478 = 7.478478
Since the number is non-terminating but recurring.
Therefore, it is a rational number.
(v) 1.101001000100001.....
1.101001000100001..... = 1.101001000100001.....
Since the number is non-terminating non-recurring.
Therefore, it is an irrational number.
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