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I've seen a famous "one-line" proof of irrationality of the square root of two that I don't understand at all.

It says that if $\sqrt{2}=\frac{m}n$ in lowest terms, $\sqrt{2}=\frac{2n-m}{m-n}$ in lower terms.

I have a few questions about this - most fundamentally, how does one get from that first expression to the other? It doesn't seem to be a trivial algebraic manipulation, at least not that I see. Secondly, how do we know that $m-n$ is smaller than $n$?

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$$\sqrt{2} = \frac{m}{n} = \frac{m(\sqrt{2} - 1)}{n(\sqrt{2} - 1)} = \frac{n\sqrt{2}(\sqrt{2} - 1)}{n(\sqrt{2} - 1)} = \frac{2n - n\sqrt{2}}{n\sqrt{2} - n}= \frac{2n-m}{m-n}$$

For the second question:

$m-n$ is smaller than $n$, because we know that $\sqrt{2}$ is somewhere between $1$ and $2$, and therefore $1 < m < 2n$, meaning that $0 < m-n < n$, so $m-n$ is smaller than $n$ (but still positive).

By the way, do you understand that the proof as a whole is a proof by infinite descent?

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  • $\begingroup$ That clears things up! And I understand the proof is saying that, infinitely, you can construct fractions in lower and lower terms if sqrt(2) is rational - is that what you mean by infinite descent? $\endgroup$ – TeaFor2 Mar 18 '18 at 16:13
  • $\begingroup$ @TeaFor2 Yes ... and of course getting lower and lower terms is impossible, since given that $m$ is a natural number, you can only get lower terms a finite number of times. So yes, assuming that square root of $2$ is rational leads to a contradiction. $\endgroup$ – Bram28 Mar 18 '18 at 16:30
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\begin{align}\frac mn=\frac{2n-m}{m-n}&\iff m^2-mn=2n^2-mn\\&\iff\frac{m^2}{n^2}=2,\end{align}which is true.

On the other hand, $m-n<n\iff m<2n\iff m^2<4n^2$, which is true, since $m^2=2n^2$.

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If $m=\sqrt2 n$, then $(\sqrt2-1)m=2n-m$ and $(\sqrt2-1)n=m-n$. Basically you multiply both the numerator and the denominator by $\sqrt2-1$. This multiplier is in the interval $(0,1)$, so both the numerator and the denominator become smaller.

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This way is a little different and longer from what other suggested but I think this is very interesting and worth posting:

Say we have a triangle, with $90^\circ$ deg and $2$ sides with length $n'$ and hypotenuse with length $m'$.

By Pythagorean theorem $2n'^2=m'^2\implies\sqrt2=\frac{m'}{n'}$. Let's assume that exists a number, say $x$, such that exists $m,n\in\Bbb N$ and $xm=m',xn=n'$ so we left with $\sqrt{2}=\frac m{n}$.

In this part, all I did is that I assumed that both $m',n'$ are rational:

if $m'=a/b,\,n'=c/d$ then $x=\frac1{bd}$ to get $m'=xad=xm,\,n'=xcb=xn$

enter image description here

We can clearly see that $n<m<2n$

Now let's take a point on $xm$ that is $xn$ away from the top corner:

enter image description here

We will take a line from that point that is perpendicular to $xm$ to create similar triangles(I added the length of the sides of the similar triangle): enter image description here

Because the $2$ triangles are similar the ratio between the sides should be constant so $\sqrt2=\frac{m}{n}=\frac{2n-m}{m-n}$ by assumption $m,\,n$ are positive integers so $2n-m<m$ and $m-n<n$ also positive integers(they are positive because of the first inequality:$n<m<2n$). I can repeat this process for the smaller triangle, and then to the smaller triangle from the smaller triangle and so on, until $n$ will be equal to $1$ and then I get contradiction.

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  • $\begingroup$ @TeaFor2 go along from the top corner, to the bottom left corner, now go right to the inner corner and then to the inner $90$ degree, we have Kite, so the hypotenuse is $xn-(x(m-n))=x(n-(m-n))=x(2n-m)$ $\endgroup$ – Holo Mar 19 '18 at 23:34

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