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Without using limits or the definition of irrational numbers, how do you solve this? I was thinking proof by contradiction, but I keep running into problems.

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There are many possible answers, but I believe this one complies with the restrictions imposed (assuming $p,q>0$): $$\frac{m}{n} = \frac{4pq}{p^2+2q^2}.$$

  1. Proof of $\frac{p}{q} < \frac{m}{n}$. By hypothesis $p^2 < 2q^2$ then $$\frac{m}{n} = \frac{4pq}{p^2+2q^2} > \frac{4pq}{2q^2+2q^2} = \frac{4pq}{4q^2} = \frac{p}{q}.$$

  2. Proof of $\frac{m}{n} < \sqrt{2}$, or equivalently, $0 < 2n^2 - m^2$: $$2n^2 - m^2 = 2(p^2+2q^2)^2 - 16 p^4q^4 = 2 (2q^2 - p^2)^2 > 0.$$

Q.E.D.

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  • $\begingroup$ My approach was to use Newton's method to solve the equation $x^2 = 1/2$, enter $q/p$ in the algorithm and interpret its output as $n/m$. Since $q/p > 1/\sqrt{2}$ and $x^2-1/2$ is convex, then $n/m$ had to be between $1/\sqrt{2}$ and $q/p$. $\endgroup$ – mlerma54 Nov 21 '18 at 6:12
  • $\begingroup$ Thanks! This helped a lot! $\endgroup$ – Masie Nov 22 '18 at 2:30
  • $\begingroup$ Just out of curiosity, how did you come up with that value for m/n ? $\endgroup$ – Masie Nov 22 '18 at 20:35
  • $\begingroup$ @Masie I used Newton's method for finding approximate roots of $x^2 - 1/2$ - Wikipedia contains a detailed description of the method. If you start with $q/p$ as a first "guess", then you can take $n/m$ as the next approximation generated by the algorithm. I didn't use $x^2-2$ and $p/q$ because the next approximation would overshoot the root. $\endgroup$ – mlerma54 Nov 23 '18 at 3:47
  • $\begingroup$ Ah! I see. Got it. Thanks $\endgroup$ – Masie Nov 27 '18 at 20:37
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Let $a$ be a rational, close to, but below $\sqrt2$. Then $b=2/a$ is a rational, close to, but above $\sqrt2$. Consider $c=\frac12(a+b)$. That is rational and should be even closer to $\sqrt2$. But it turns out that $c>\sqrt2$. Why not try then $d=2/c$? Can you prove $a<d<\sqrt2$?

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Let $\frac{p}{q}=r$ and

if $r>\sqrt{2} - 1$, consider $(r+h)^2<2$ where $0<h<1$.

Then $r^2+2rh+h^2<2.$ Let $r^2+2hr+h=2 <r^2+2rh+h^2$.

Then $h=\frac{2-r^2}{2r+1}<1$

if $r<\sqrt{2} - 1$, take $h=1$.

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