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 2^{m} x 5^{n} 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.