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 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.