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Proof

Suppose $\sqrt{2} = \frac{p_o}{q_o}$, where $q_o$ is as small as possible.

We have $q_o < p_o < 2q_o$ (from $1 < \sqrt{2} < 2$, which follows from $1 < 2 < 4$)

let $q_1 = p_o - q_o$, and $p_1 = 2q_o - p_o$ (where both $p_1$ and $q_1$ are natural numbers)

Then, $\frac{p_1}{q_1} =\frac{2q_o-p_o}{p_o-q_o}$ $=\frac{(2q_o-\sqrt{2}qo)}{\sqrt{2}q_o - q_o}$ $=\frac{\sqrt{2}q_o(\sqrt{2} - 1)}{q_o(\sqrt{2} - 1)} $ $= \sqrt{2}$

but $q_1 < q_o$, so contradiction.

Questions:

  1. How does $q_o < p_o <2q_o$ follow from $1 < \sqrt{2} < 2$? I know it can also be written as $\frac{p_o}{\sqrt{2}} < p_o < \sqrt{2}p_o$, but I don't know how this is related to $1 < \sqrt{2} < 2$.

  2. How does $q_1 < q_o$ prove $\sqrt{2} is irrational? Which premise does it contradict?

IamKnull
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David Durham
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    Here's a MathJax tutorial :) – Shaun Sep 07 '19 at 18:31
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    @MostafaAyaz Not a duplicate since David is trying to understand this proof, which is different from that in the other question. – Jamie Radcliffe Sep 07 '19 at 18:33
  • You should also require that $q_0>0$ (and say that both $q_0$ and $p_0$ are integers, you could assume both are positive integers). Otherwise there is ambiguity $\frac{p_0}{q_0}=\frac{-p_0}{-q_0}$, and if $0<q_0<p_0$ then indeed $q_1=p_0-q_0<2q_0-q_0=q_0$, but $-q_0<-q_1$. Re the question you erased: $A = { x \in\Bbb R : \forall y \in [5, 7], x < y }$, rewrite $A = { x \in\Bbb R : x < 5 }$ – Mirko Sep 10 '19 at 02:39

1 Answers1

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For your question 1, we take the known fact $1<\sqrt{2}<2$ and multiply everything by $q_0$, giving us $q_0 < q_0 \sqrt{2} < 2q_0$. Then we note that by hypothesis $q_0 \sqrt{2} = p_0$.

For your question 2, we supposed (in your very first line) that $q_0$ was the smallest (positive) denominator of any fraction equalling $\sqrt{2}$. You then proceeded to produce another representation $\sqrt{2}=p_1/q_1$ with a smaller denominator. That's the contradiction.

Jamie Radcliffe
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