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) $$\dfrac{13}{3125}$$ (ii) $$\dfrac{17}{8}$$ (iii) $$\dfrac{64}{455}$$ (iv) $$\dfrac{15}{1600}$$ (v) $$\dfrac{29}{343}$$ (vi) $$\dfrac{23}{2^{3}5^{2}}$$ (vii) $$\dfrac{129}{2^{2}5^{7}7^{5}}$$ (viii) $$\dfrac{6}{15}$$ (ix) $$\dfrac{35}{50}$$ (x) $$\dfrac{77}{210}$$

Solution

## Theorem: Let $$x = \dfrac{p}{q}$$ be a rational number, such that the prime factorisation 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.(i) $$\dfrac{13}{3125}$$Factorise the denominator, we get$$3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5$$So, denominator is in form of $$5^m$$ so, $$\dfrac{13}{3125}$$ is terminating.(ii) $$\dfrac{17}{8}$$Factorise the denominator, we get$$8 = 2 \times 2 \times 2 = 2^3$$So, denominator is in form of $$2^n$$ so, $$\dfrac{17}{8}$$ is terminating.(iii) $$\dfrac{64}{455}$$Factorise the denominator, we get$$455 = 5 \times 7 \times 13$$So, denominator is not in form of $$2^{n} 5^{m}$$ so, $$\dfrac{64}{455}$$ is not terminating.(iv) $$\dfrac{15}{1600}$$Factorise the denominator, we get$$1600 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 = 2^{6} 5^{2}$$So, denominator is in form of $$2^{n}5^{m}$$ so, $$\dfrac{15}{1600}$$ is terminating.(v) $$\dfrac{29}{343}$$Factorise the denominator, we get$$343 = 7 \times 7 \times 7 = 7^{3}$$So, denominator is not in form of $$2^{n}5^{m}$$ so, $$\dfrac{29}{343}$$ is not terminating.(vi) $$\dfrac{23}{2^{3}5^{2}}$$Here, the denominator is in form of $$2^{n}5^{m}$$ so, $$\dfrac{23}{2^{3}5^{2}}$$ is terminating.(vii) $$\dfrac{129}{2^{2}5^{7}7^{5}}$$Here, the denominator is not in form of $$2^{n}5^{m}$$ so, $$\dfrac{129}{2^{2}5^{7}7^{5}}$$ is not terminating.(viii) $$\dfrac{6}{15}$$Divide nominator and denominator both by 3 we get $$\dfrac{3}{15}$$So, denominator is in form of $$5^{m}$$ so, $$\dfrac{6}{15}$$ is terminating.(ix) $$\dfrac{35}{50}$$ Divide nominator and denominator both by 5 we get $$\dfrac{7}{10}$$Factorise the denominator, we get$$10 = 2 \times 5$$So, denominator is in form of $$2^{n}5^{m}$$ so, $$\dfrac{35}{50}$$ is terminating.(x) $$\dfrac{77}{210}$$Divide nominator and denominator both by 7 we get $$\dfrac{11}{30}$$Factorise the denominator, we get$$30 = 2 \times 3 \times 5$$So, denominator is not in the form of $$2^{n}5^{m}$$ so $$\dfrac{6}{15}$$ is not terminating.MathematicsRS AgarwalStandard X

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