We do a proof by contradiction. Let $a=n + \frac{p}{q}$ where $0 < p < q$ are positive integers. Then, $$\frac{1}{a}=a-[a]$$$$\iff \frac{q}{p} = n + \frac{p}{q} - n$$$$\iff \frac{q}{p} = \frac{p}{q}$$$$\iff q^2=p^2$$$$\iff q^2-p^2=0$$$$\iff (q-p)(q+p)=0$$This implies $p=-q$ or $p=q$. If $p=-q$, then this means that both of them cannot be positive, a contradiction. If $p=q$, then this contradicts $p<q$. Q.E.D