If $\frac{a}{b}$ is a non-unit fraction between 0 and 1, in lowest terms let $\frac{1}{n}$ be the largest unit fraction less than $\frac{a}{b}$, and form the fraction $\frac{a^{'}}{b^{'}}=\frac{a}{b}-\frac{1}{n}$. Assume $\frac{a^{'}}{b^{'}}$ is in lowest terms, show that $0<a^{'}<a$.
Progress so far:
Since $\frac{a^{'}}{b^{'}}=\frac{a}{b}-\frac{1}{n}\Rightarrow\frac{a^{'}}{b^{'}}<\frac{a}{b}$
I managed to show that if $\frac{a^{'}}{b^{'}}$ is a unit fraction, $0<a^{'}<a$, as it is stated that $\frac{1}{n}$ is the largest unit fraction.
I can't find a way to show that $0<a^{'}<a$ when $\frac{a^{'}}{b^{'}}$ is a non-unit fraction.
Hint: $\frac{a}{b}-\frac{1}{n}>0$, but $\frac{a}{b}-\frac{1}{n-1}<0$. What do you see when you clear denominators?
Further hint: Note $\frac{a'}{b'}=\frac{an-b}{bn}$. So $a'\leq an-b$ (why?).