Question

8/343,State whether given rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it.

Answer 1

Hey there!

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Q. $\frac{8}{343}$

⇒ Given number is $\frac{8}{343}$ and HCF (8, 343) = 1

∴ $\frac{8}{343} = \frac{8 ÷ 1}{343 ÷ 1} = \frac{8}{343}$

⇒ Now, 343 = $(7^{3})$ and 7 is not a factor of 8.

∴ $\frac{8}{343}$ is in its simplest form.

⇒ Also, 343 = $(7^{3}) ≠ (2^{m} × 5^{n})$

∴ $\frac{8}{343}$ is a non-terminating repeating decimal.

⇒ $\frac{8}{343}$ does not have terminating decimal because it has one more prime factor 7 other than 2 and 5.

Answer 2

Hi ,

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Let x = p/q be a rational number ,

such that the prime factorisation of

q is not of the form 2^n× 5^m , where

n and m are non - negative integers .

Then x has a decimal expansion which

is non - terminating and repeating.

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Now ,

8/343

= 8/( 7 × 7 × 7 )

= 8/7³

= p/q

denominator q is not of the form

2^n5^m.

Therefore ,

8/343 is a non - terminating and

repeating decimal .

I hope this helps you.

: )