$x$ has terminating decimal represantation

Question

Hey to all mathematicians out there. I'm new here so please be gentle if i have made error or something.The problem is this : $x\in (0,1)$ has terminating decimals if there exist $m,n\in\mathbb N \bigcup \{0\} $, so that $2^m5^nx$ is integer. I can't find any solution to the problem. My only thought was that for this to happen m must be equal to n.

Answer 1

Observe that if $2^m5^n x=p$ with $p\in\Bbb N$ and $2^m 5^n\in\Bbb N$ for $n,m\in\Bbb N$, then we have that $x=p/(2^m 5^n)$.

If you want that $x\in (0,1)$ then:

• First choose any $n,m\in\Bbb N$

• After choose some $p\in\Bbb N$ distinct of zero such that $p<2^m 5^n$

Under these conditions you have the desired $x\in(0,1)$ with terminating decimals.