Gujarat Board GSEB Textbook Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.3 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.3
Question 1.
Write the following in decimal form and say what kind of decimal expansion each has
Solution:
(i) \(\frac { 36 }{ 100 }\)
(ii) \(\frac { 1 }{ 11 }\)
(iii) 4\(\frac { 1 }{ 8 }\)
(iv) \(\frac { 3 }{ 13 }\)
(v) \(\frac { 2 }{ 11 }\)
(vi) \(\frac { 329 }{ 400 }\)
Solution:
(i) We have
\(\frac { 36 }{ 100 }\) = 0.36
∴ The decimal expansion is terminating.
(ii) We have \(\frac { 1 }{ 11 }\)
The decimal expansion is non-terminating repeating.
(iii) We have 4\(\frac { 1 }{ 8 }\) = \(\frac { 33 }{ 8 }\)
The decimal expansion is terminating.
(iv) We have \(\frac { 3 }{ 13 }\)
The decimal expansion is non-terminating repeating.
(v) We have \(\frac { 2 }{ 11 }\)
The decimal expansion is non-terminating repeating.
(vi) We have \(\frac { 329 }{ 400 }\)
The decimal expansion is terminating.
Question 2.
You know that \(\frac { 1 }{ 7 }\) = 0.\(\overline { 142857 }\). Can you predict what the decimal expansions of \(\frac { 2 }{ 7 }\), \(\frac { 3 }{ 7 }\), \(\frac { 4 }{ 7 }\), \(\frac { 5 }{ 7 }\), \(\frac { 6 }{ 7 }\) are, without actually doing the long division? If so, how?
(Hint. Study the remainders while finding the value of \(\frac { 1 }{ 7 }\) carefully.)
Solution:
Yes, we can predict the decimal expansions of \(\frac { 2 }{ 7 }\), \(\frac { 3 }{ 7 }\), \(\frac { 4 }{ 7 }\), \(\frac { 5 }{ 7 }\), \(\frac { 6 }{ 7 }\) without actually doing the long division.
We know \(\frac { 1 }{ 7 }\)
\(\frac { 1 }{ 7 }\) = 0.142857142857…
∴ \(\frac { 1 }{ 7 }\) = 0.\(\overline { 142857 }\)
Now
\(\frac { 2 }{ 7 }\) = 2 x \(\frac { 1 }{ 7 }\) = 2 x 0.\(\overline { 142857 }\)
∴ \(\frac { 2 }{ 7 }\) = 0.\(\overline { 285714 }\)
Similarly,\(\frac { 3 }{ 7 }\) = 3 x \(\frac { 1 }{ 7 }\) = 3 x 0.\(\overline { 142857 }\)
⇒ \(\frac { 3 }{ 7 }\) = 0.\(\overline { 428571 }\)
\(\frac { 4 }{ 7 }\) = 4 x \(\frac { 1 }{ 7 }\) = 4 x 0.\(\overline { 142857 }\)
⇒ \(\frac { 4 }{ 7 }\) = 0.\(\overline { 571428 }\)
\(\frac { 5 }{ 7 }\) = 5 x \(\frac { 1 }{ 7 }\) = 5 x 0.\(\overline { 142857 }\)
⇒ \(\frac { 5}{ 7 }\) = 0.\(\overline { 714285 }\)
\(\frac { 6 }{ 7 }\) = 6 x \(\frac { 1 }{ 7 }\) = 6 x 0.\(\overline { 142857 }\)
⇒ \(\frac { 6}{ 7 }\) = 0.\(\overline { 857142 }\)
Question 3.
Express the following in the form \(\frac { p }{ q }\), where p and q are integers and q ≠ 0.
(i) 0.\(\overline { 6 }\)
(ii) 0.4\(\overline { 7 }\)
(iii) 0.\(\overline { 0.001 }\)
Solution:
(i) 0.\(\overline { 6 }\)
(ii) We have 0.4\(\overline { 7 }\)
(iii) We have 0.\(\overline { 0.001 }\)
Question 4.
Express 0.99999… in the form of \(\frac { p }{ q }\). Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Solution:
Let x = 0.99999… …(1)
Multiplying by 10 on both sides, we get
10x = 9.99999… …(2)
Subtracting equation (1) from eqn. (2),
Thus 0.99999… = 1 = \(\frac { 1 }{ 1 }\)
Here we get p = 1 and q = 1
Since 0.99999… goes on forever. Hence there is no gap between 1 and 0.99999… and hence both are equal.
Question 5.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of \(\frac { 1 }{ 17 }\)? Perform the division to check your answer.
Solution:
Long division method
Thus \(\frac { 1 }{ 17 }\) = 0.\(\overline { 0.0588235294117647 }\)
We observe that by long division method maximum number of digits in repeating block in the decimal expansion of \(\frac { 1 }{ 17 }\) is 16, thus answer is verified.
Question 6.
Look at several examples of rational numbers in the form \(\frac { p }{ q }\) (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:
(i) \(\frac { 1 }{ 2 }\) = \(\frac { 1 × 5 }{ 2 × 5 }\) = \(\frac { 5 }{ 10 }\) = 0.5
(ii) \(\frac { 3 }{ 4 }\) = \(\frac { 3×5 × 5 }{ 2 × 2 × 5 × 5 }\) = \(\frac { 75 }{ 100 }\) = 0.75
(iii) \(\frac { 7 }{ 8 }\) = \(\frac { 7 × 5 × 5 × 5 }{ 2 × 2 × 2 × 5 × 5 × 5 }\) = \(\frac { 875 }{ 1000 }\) = 0.875
(iv) \(\frac { 13 }{ 25 }\) = \(\frac { 13 × 2 × 2 }{ 5 × 5 × 2 × 2 }\) = \(\frac{52}{5^{2} \times 2^{2}}\)
= \(\frac{52}{(10)^{2}}\) = \(\frac { 52 }{ 100 }\) = 0.52
(v) \(\frac { 3 }{ 125 }\) = \(\frac { 3 }{ 5 × 5 × 5 }\) = = \(\frac{3}{5^{3}}\) = \(\frac{3 \times 2^{3}}{5^{3} \times 2^{3}}\)
= \(\frac{3 \times 8}{(5 \times 2)^{3}}\) = \(\frac { 24 }{ 1000 }\) = 0.024
(vi) \(\frac { 27 }{ 16 }\) = \(\frac{27 \times 5^{4}}{2^{4} \times 5^{4}}\) = \(\frac{27 \times 5^{4}}{(2 \times 5)^{4}}\)
= \(\frac{27×625}{(10)^{4}}\) = \(\frac{16875}{(10)^{4}}\) = 1.6875
We observe that the denominator of all the above rational numbers are of the form 2m x 5n i.e., the prime factorization of denominators has only powers of 2 or powers of 5 or both.
Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution:
(i) 0.012012001200012…
(ii) 0.21021002100021000021…
(iii) 0.32032003200032000032…
Question 8.
Find three different irrational numbers between the rational numbers \(\frac { 5 }{ 7 }\) and \(\frac { 9 }{ 11 }\).
Solution:
\(\frac { 5 }{ 7 }\)
Thus \(\frac { 5 }{ 7 }\) = 0.714285…
\(\frac { 5 }{ 7 }\) = 0.\(\overline { 714285 }\)….
Now \(\frac { 9 }{ 11 }\)
Thus \(\frac { 9 }{ 11 }\) = 0.8181… = 0.\(\overline { 81 }\)
Thus three irrational numbers between the rational numbers \(\frac { 5 }{ 7 }\) and \(\frac { 9 }{ 11 }\) can be taken as
0. 73073007300073000073…
0. 757075700757000757…
and 0.808008000800008…
Question 9.
Classify the following numbers as rational or irrational.
(i) \(\sqrt{23}\)
(ii) \(\sqrt{225}\)
(iii) 0.3796
(iv) 7.478478…
(v) 1.101001000100001…
Solution:
(i) \(\sqrt{23}\), 23 is not a perfect square so \(\sqrt{23}\) will not give an integral value.
Hence it is not a rational number.
(ii) \(\sqrt{225}\)
∴ \(\sqrt{225}\)
Here p = 15
and q = 1 (q ≠ 0)
(iii) 0.3796
The decimal expression is terminating.
Hence 0.3796 is a rational number.
(iv) 7.478478…
∴ 7.478478… = 7.\(\overline { 748 }\)
The decimal expansion is non-terminating recurring.
∴ 7.478478… is a rational number.
(v) 1.101001000100001…
∵ The decimal expansion is non – terminating non-recurring.
∴ 1.101001000100001… is an irrational number.