Question

10. If p/q is a rational number such that the prime
facterization of p/q is of the form 2^a 5^b is where
a and b are non- Negative integer than the
decimal expansion of the rational number​

Answer 1

Case 1:- Let a ≥ b. Then,

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=2^a\times5^b}$

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=2^{a-b+b}\times5^b}$

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=2^{a-b}\times2^b\times5^b}$

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=2^{a-b}\times(2\times5)^b}$

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=2^{a-b}\times10^b}$

Hence the decimal expansion looks like,

$\displaystyle\longrightarrow\sf{\underline{\underline{\dfrac{p}{q}=\left(2^{a-b}\right)\underbrace{000\dots000}_b}}}$

E.g.: If $\displaystyle\sf{\dfrac{p}{q}=2^4\times5^2,}$ then,

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=\left(2^{4-2}\right)\underbrace{00}_2}$

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=400}$

Case 2:- Let a ≤ b. Then,

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=2^a\times5^b}$

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=2^a\times5^{a+b-a}}$

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=2^a\times5^a\times5^{b-a}}$

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=(2\times5)^a\times5^{b-a}}$

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=5^{b-a}\times10^a}$

Hence the decimal expansion looks like,

$\displaystyle\longrightarrow\sf{\underline{\underline{\dfrac{p}{q}=\left(5^{b-a}\right)\underbrace{000\dots000}_a}}}$

E.g.: If $\displaystyle\sf{\dfrac{p}{q}=2^2\times5^4,}$ then,

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=\left(5^{4-2}\right)\underbrace{00}_2}$

$\displaystyle\longrightarrow\sf{\dfrac{p}{q}=2500}$