When does a fraction terminate as a decimal?

($\Leftarrow$) Suppose $b=2^m5^n$. Let $k=\max(m,n)$. Multiply numerator and denominator by $2^{k-n}5^{k-m}$ (both exponents are $\ge0$ since $k$ is the max): $\dfrac{a}{b}=\dfrac{a\cdot2^{k-n}5^{k-m}}{2^m5^n\cdot2^{k-n}5^{k-m}}=\dfrac{a\cdot2^{k-n}5^{k-m}}{2^k5^k}=\dfrac{N}{10^k}$, where $N=a\cdot2^{k-n}5^{k-m}$ is an integer. Writing $N$ in ordinary digits and placing the decimal point $k$ digits from the right (padding with leading zeros if $N$ has fewer than $k$ digits) gives exactly the decimal expansion of $\frac{a}{b}$, and it stops after at most $k$ digits: it terminates.

($\Rightarrow$) Suppose $\frac{a}{b}$ terminates after $k$ decimal digits, i.e. $\dfrac{a}{b}=\dfrac{N}{10^k}$ for some integer $N$. Cross-multiplying (Theorem 1 above): $a\cdot10^k = b\cdot N$, so $b$ divides $a\cdot10^k$. Because $\frac{a}{b}$ is in lowest terms, $\gcd(a,b)=1$, meaning $b$ shares no prime factor with $a$; hence any prime factor $p$ of $b$ cannot divide $a$, and since $p$ divides $a\cdot10^k$ it must divide $10^k$ instead. But $10^k=2^k5^k$, whose only prime factors are $2$ and $5$. So every prime factor of $b$ is $2$ or $5$, which means $b=2^m5^n$ for some $m,n\ge0$.