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Let $m$ and $n$ be natural numbers. Prove that if $m/n < \sqrt{2}$, then there is another rational number $m'/n'$ with $m/n < m'/n' < \sqrt{2}$

I interpreted this statement as $$\frac{m}{n} < \sqrt{2} \rightarrow \exists \frac{m'}{n'} \in \mathbb{Q} \left( \frac{m}{n} < \frac{m'}{n'} \wedge \frac{m'}{n'} < \sqrt{2} \right)$$ The contrapositive would then be $$\forall \frac{m'}{n'} \in \mathbb{Q} \left(\frac{m'}{n'} < \sqrt{2} \rightarrow \frac{m}{n} \geq \frac{m'}{n'}\right) \rightarrow m/n \geq \sqrt{2}$$

Partial Proof: Suppose that $\forall \frac{m'}{n'} \in \mathbb{Q} \left(\frac{m'}{n'} < \sqrt{2} \rightarrow \frac{m}{n} \geq \frac{m'}{n'}\right)$. Now suppose that $\frac{m}{n} < \sqrt{2}$. Since $\frac{m}{n} \in \mathbb{Q}$ and $\frac{m}{n} < \sqrt{2}$, we have $\frac{m}{n} \leq \frac{m}{n}$...

I'm just wondering why I was not able to draw a contradiction at the end. Is there something wrong with how I interpreted the statement?

Iyeeke
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  • componendo-dividendo may help to prove. – Hussain-Alqatari Jun 09 '20 at 19:40
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    Have a look at this answer ... you can construct one by applying $\frac{m}{n} \to \frac{m+2n}{m+n}$ twice ... – rtybase Jun 09 '20 at 20:14
  • Has $\sqrt2$ already been constructed as a real number, or is the question part of such a construction (using Dedekind cuts)? If the former, then there is nothing special about the numbers $m/n, \sqrt2$ here, so you might as well prove more generally that if $a, b$ are any real numbers (or any positive real numbers, to make it a little bit easier) with $a < b,$ then there is a rational number $m'/n'$ with $a < m'/n' < b.$ That is, the (positive) rational numbers are dense in the (positive) reals. – Calum Gilhooley Jun 09 '20 at 21:03

2 Answers2

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Here is a somewhat different approach: assume that $\dfrac mn < \sqrt 2$. Then \begin{align*} m^2 &< 2n^2\\ 2m^2 &< 4n^2 \\ 2m^2 + 3mn &< 3mn + 4n^2 \\ m(2m + 3n) &< n (3m + 4n) \\ \frac mn &< \frac{3m + 4n}{2m + 3n} \end{align*} and \begin{align*} m^2 &< 2n^2\\ 9 m^2 - 8m^2 &< 18 n^2 - 16n^2 \\ 9m^2 + 16n^2 &< 8m^2 + 18n^2 \\ 9m^2 + 24 mn + 16 n^2 &< 8m^2 + 24mn + 18n^2 \\ (3m + 4n)^2 &< 2 (2m + 3n)^2 \\ \frac{(3m+4n)^2}{(2m+3n)^2} &< 2 \\ \frac{3m+4n}{2m+3n} &< \sqrt{2} \end{align*} so that $$ \frac mn < \frac{3m+4n}{2m+3n} < \sqrt 2.$$

Umberto P.
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If you are allowed to use limits, note that $\displaystyle \lim_{k \to \infty} \frac{k+1}{k} = 1$ and thus $\displaystyle \lim_{k \to \infty} \frac{(k+1)m}{kn} = \frac mn$.

Since $\dfrac{m}{n} < \sqrt 2$, then for sufficiently large $k$ you have $$\frac mn < \frac{(k+1)m}{kn} < \sqrt 2.$$

Umberto P.
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  • This makes sense to me, but I don't think we are allowed to use limits (they are introduced a few chapters later). – Iyeeke Jun 09 '20 at 21:01