### Chapter 1: Real Numbers

Q
##### Real Numbers Solutions

Question:

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i)   $\frac{13}{3125}$           (ii)   $\frac{17}{8}$               (iii)  $\frac{64}{455}$          (iv)   $\frac{15}{1600}$        (v)  $\frac{29}{343}$

(vi)   $\frac{23}{2^35^2}$        (vii)   $\frac{129}{2^25^77^5}$       (viii)  $\frac{6}{15}$          (ix)   $\frac{35}{50}$             (x)  $\frac{77}{210}$

Solutions: We know that   x = $\frac{p}{q}$  be a rational number, such that the prime factorization of q is of the form $2^{n}5^{m}$ , where n, m are non-negative integers. Then, x  has a decimal expansion which terminates.

And   x= $\frac{p}{q}$   be a rational number, such that the prime factorization of q is not of the form   $2^{n}5^{m}$ , where n, m are non-negative integers. Then, x  has a decimal expansion which non-terminating repeating.

(i)   $\frac{13}{3125}$

If we factorize the denominator, we get
3125 = 5 × 5 × 5 × 5 × 5 = $5^{5}$

So, the denominator is in the form of  $5^{5}$ so, โ $\frac{13}{3125}$   is terminating decimal expansion.

(ii)   $\frac{17}{8}$

If we factorize the denominator, we get
8 = 2 × 2 × 2 = $2^{3}$
So, the denominator is in the form of   $2^{3}$  so, โ $\frac{17}{8}$   is terminating decimal expansion.

(iii)  $\frac{64}{455}$

If we factorize the denominator, we get

455 = 5 × 7 × 13

So, the denominator is not in the form of  $2^{n}5^{m}$ so,โ  $\frac{64}{455}$

is non-terminating repeating decimal expansion.

(iv)   $\frac{15}{1600}$

If we factorize the denominator, we get

1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 =  $2^{6}5^{2}$

So, the denominator is in the form of  $2^{n}5^{m}$ . So, $\frac{15}{1600}$   is terminating decimal expansion.

(v)  $\frac{29}{343}$

If we factorize the denominator, we get

343 = 7 × 7 × 7 =  $7^{3}$

So, the denominator is not in the form of  $2^{n}5^{m}$ . So,   is non-terminating repeating decimal expansion.

(vi)   $\frac{23}{2^35^2}$

Here, the denominator is in the form of    $2^{n}5^{m}$ . So, $\frac{23}{2^35^2}$   is terminating decimal expansion.

(vii)   $\frac{129}{2^25^77^5}$

Here, the denominator is not in the form of   $2^{n}5^{m}$ only. So,  $\frac{129}{2^25^77^5}$ is non-terminating repeating decimal expansion.

(viii)  $\frac{6}{15}$

If we divide nominator and denominator both by 3 we get $\frac{3}{5}$
So, the denominator is in the form of $5^{m}$.   So, $\frac{6}{15}$  is terminating decimal expansion.

(ix)    $\frac{35}{50}$

If we divide nominator and denominator both by 5, we get $\frac{7}{10}$
If we factorize the denominator, we get

10 = 2 × 5

So, the denominator is in the form of  $2^{n}5^{m}$ .  So,    $\frac{35}{50}$     is terminating decimal expansion.

(x)    $\frac{77}{210}$

If we divide nominator and denominator both by 7, we get
If we factorize the denominator, we get

30 = 2 × 3 × 5

So, the denominator is not in the form of $2^{n}5^{m}$ .  So,  $\frac{77}{210}$   is non-terminating repeating decimal expansion.

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