Fractions and decimals

Comparing two fractions directly is awkward because they may be written with different denominators; multiplying through by both denominators clears the fractions and turns the comparison into an ordinary integer equation.

Direction 1 ($\Rightarrow$). Suppose $\dfrac{a}{b}=\dfrac{c}{d}$. Since $b,d\neq0$, we may multiply both sides by the nonzero number $bd$: $\frac{a}{b}\cdot bd = \frac{c}{d}\cdot bd$. On the left, $\frac{a}{b}\cdot bd = a\cdot d$ because the factor $b$ cancels; on the right, $\frac{c}{d}\cdot bd = c\cdot b$ because the factor $d$ cancels. So $ad=bc$.

Direction 2 ($\Leftarrow$). Suppose $ad=bc$. Divide both sides by the nonzero number $bd$: $\dfrac{ad}{bd}=\dfrac{bc}{bd}$. Cancel the common factor $d$ from the left fraction and the common factor $b$ from the right fraction: $\dfrac{a}{b}=\dfrac{c}{d}$.

Both directions hold, so the two statements are equivalent: $\dfrac{a}{b}=\dfrac{c}{d}$ exactly when $ad=bc$.

Decimal place value is built from powers of $10=2\times5$. A fraction can be rewritten with a denominator that is a power of $10$ exactly when its denominator's only prime factors are $2$ and $5$ — any other prime factor can never be "absorbed" into a power of $10$.

($\Leftarrow$) 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 $m,n\ge0$.