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If $m,n\in\mathbb N, m<n, z=\frac{m}{n}, q\ge 2$, the inequality $0<(za-1)a^{q-1}<1$ has solution $a\in\mathbb N$.
Then the solution to the inequality above is unique.

The question is simple: how to prove the statement above?

I am currently studying

Daniel D. Bonar & Michael Khoury Jr. Real Infinite Series. Classroom Resource Materials. The Mathematical Association of America, $2006$. isbn: $0883857456$.

The problem above is here: P.1, P.2.

The book did give a proof to the statement at beginning, as a lemma to that problem in the book.

The book said (I use $q$ instead of $e$ in the book) "$\frac{n}{b}<m\le m(b-a)<\frac{n}{a^{q-1}}$ implies a contradiction", but how? And, how to prove $m(b-a)<\frac{n}{a^{q-1}}$?

What I have thought:

If $a^{q-1}\in\mathbb Z$ again and $q-1>1$, then there may be a contradiction since there are just finite many integer between $a$ and $b$, (similar to proof by infinite descent). But well... I think I thought too much. It just doesn't work.

And I back to the problem itself and try, if $a<b$ are natural numbers satisfying the inequality, \begin{align*} 0<(za-1)a^{q-1}&<(zb-1)b^{q-1}<1\\ 0<(ma-n)a^{q-1}&<(mb-n)b^{q-1}<n\\ 0<ma^q-na^{q-1}&<mb^q-nb^{q-1}<n\\ n(b^{q-1}-a^{q-1})&<m(b^q-a^q)\\ z&>\frac{b^{q-1}-a^{q-1}}{b^q-a^q} \end{align*} But it doesn't seem to be constructive.

I would like a proof to the statement at beginning. New proofs are good, in case there are, but I would like, if possible, someone telling me how the arguments in the book flow.

Tony Ma
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1 Answers1

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If the inequality $$0 < (za-1)a^{q-1}$$ holds, then, since $a^{q-1} > 0$, we must have $za = \frac{m}{n}a > 1$, or equivalently $$m > \frac{n}{a}\,.$$ If $a < b$ are two integers, then $b-a \geqslant 1$, whence $m \leqslant m(b-a)$. If $0 < a < b$ are integer solutions of $$0 < (zx-1)x^{q-1} < 1\,, \tag{$\ast$}$$ then, since the middle expression in $(\ast)$ is monotonic in $x$ for $x > z^{-1}$, we have $$0 < (zb-1)b^{q-1} - (za-1)a^{q-1} < 1\,.$$ Since $b^{q-1} > a^{q-1}$ and $zb > za$, it follows that $$0 < \bigl((zb-1) - (za-1)\bigr)a^{q-1} = \frac{m}{n}(b-a)a^{q-1} \leqslant (zb-1)b^{q-1} - (za-1)a^{q-1} < 1\,,$$ and from that we obtain $$m(b-a) < \frac{n}{a^{q-1}}\,.$$ But from $$m < \frac{n}{a^{q-1}}$$ it follows that $$za = \frac{m}{n}a < \frac{1}{a^{q-2}} \leqslant 1\,,$$ which shows that $a$ can't be a solution of $(\ast)$ by the remark at the top.

Daniel Fischer
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