I am currently working alone through the book The Foundations of Mathematics (Stewart&Tall,2015) and I am completely stuck. The question is as such.
Let p/q be a fraction in lowest terms such that $$\frac{1}{n+1} \lt \frac{p}{q} \lt \frac{1}{n}$$ for a natural number n. Show that $\frac{p}{q}-\frac{1}{n+1}$ is a fraction, which, in its lowest terms, has numerator less than p. Hence, by induction, prove that every proper fraction p/q, where p < q, can be written as a finite sum of distinct reciprocals $$\frac{p}{q} = \frac{1}{n_1} + \ldots + \frac{1}{n_k}$$ where $n_1,\ldots,n_k$ are natural numbers.
I've given it a go by simply trying to create an inequality from the final numerator and p, i.e. $ p \gt p(n+1)-q $ but that's not helping at all. I tried substituting q for n+k for some positive k as well ($p<q$ for obvious reasons) but this approach is leading nowhere. Any advice?
Thanks!