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.