Question

Show that $\frac{139}{2^3{5}^3}$ will have terninating decimal expression.

Answer 1

A number in form of $\frac{p}{q}$ is terminating,

if (i) p and q are co -prime, and

(ii) q is of the form $2^n{5}^m$

where n and m are non- negative integers.

In questionq is then form of $2^3{5}^3$ and n=3, and m= 3

Check for co- prime

since, 139 is prime number, so there is no common factor between 139 and $2^n{5}^m$

Hence, 139 and $2^n{5}^m$ is co- prime

Thus, $\frac{139}{2^3{5}^3}$ are a terminating decimals.