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There is a set $A$ with positive integers $x$ such that there exists $y$ s.t.$ x^2=2y^2$. Show that if A is non-empty, it violates the well ordering principle.

I don't even know how to start this.

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    http://math.stackexchange.com/questions/917983/the-proof-of-sqrt2-is-not-rational-number-via-fundamental-theorem-of-arithm2015-11-20

3 Answers 3

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If $x^2=2y^2$, then $(2y-x)^2=2(x-y)^2$.

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    This way of doing it used to be not very widely known. I wonder if that's changing. The way Bill Dubuque does it in his answer posted here is the one that everyone knows.2012-10-24
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Hint $\ $ Show that if $\rm\,A\,$ is nonempty then it has no least element, contra $\rm\,A\subseteq \Bbb N\,$ is well-ordered. Note $\rm\:a\in A\:\Rightarrow\:a^2 = 2\,b^2\:\Rightarrow\:a $ even so cancelling $\,2\,$ yields $\rm\:b^2 = 2\,(a/2)^2$ so $\rm\,b\in A,\,$ and $\rm\,b < a.\:$

Remark $\ $ This contrapositive form of induction is known as infinite descent (Fermat). More conceptually the proof is a descent on the set $\rm\,A\,$ of numerators of fractions $\rm = \sqrt{2},\:$ since $\rm\:(a/b)^2\! = 2\:$ $\Rightarrow$ $\rm\:a/b = 2b/a = b/(a/2),\:$ yielding a fraction with smaller numerator $\rm\,b < a$.

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    @JackF $\ $ It's just $\ \ \dfrac{1}2\, =\, 2\cdot \dfrac{1}4\, =\, 2\cdot \left(\dfrac{1}2\right)^2\ $ scaled by $\rm\:a^2\ \ $2012-10-24
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What you want to do is show that if $\sqrt 2 = m/n$, where $m$ and $n$ are positive integers, then there are positive integers $p$ and $q$ with $q < n$ such that $\sqrt 2 = p/q$.

The key trick is to write $\sqrt 2 = \sqrt 2 \frac{\sqrt 2 - 1}{\sqrt 2 - 1} = \frac{2-\sqrt 2}{\sqrt 2 - 1} = \frac{2-m/n}{m/n-1} = \frac{2n-m}{m-n}$.

Now set $p = 2n-m$ and $q = m-n$. Since $m-n < n$, this denominator is smaller, and well-ordering can be used.

This can be used to prove that $\sqrt k$ is irrational for any non-square positive integer $k$.

I don't know if a variation of this is known for the cube root.

Of course, this is not original with me.

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    Gerry: That's ni$c$e. Somehow it doesn't seem that it sho$u$ld be that easy. I g$u$ess it just requires "straight thinking". Thanks2012-10-27