Proof of square root of two irrationality origin

[TheAbdicatedKing]

What is the earliest known accepted proof of the irrationality of the square root of two or the n root of any prime preferably one that is attributed to someone?

[nathi123]

Let's assume that [tex]\sqrt{2}=\frac{p}{q}[/tex] where p,q[tex]\in Z; (p,q)=1 \Rightarrow 2=\frac{p^{2}}{q^{2}}\Leftrightarrow p^{2}=2q^{2}; (p,q)=1\Rightarrow
p=2m;m\in Z[/tex]
[tex]\Rightarrow p^{2} = 4m^{2} \Rightarrow 4m^{2} = 2q^{2} \Leftrightarrow q^{2}=2m^{2} \Rightarrow q=2n; n\in Z \Rightarrow p=2m ; q=2n[/tex]
contradiction with (p,q)=1 [tex]\Rightarrow \sqrt{2}[/tex] is an irrational number.

[HallsofIvy]

That doesn't really answer the question asked! nathi123, Pythagoras knew that [tex]\sqrt{2}[/tex] and that bothered him to the extent that he wanted to ban [tex]\sqrt{2}[/tex] as a number at all (he wanted a philosophy where everything was based on small integers and ratios of them) but it was his own "Pythagorean Theorem" that showed that the length of the hypotenuse of a right triangle with legs of length 1 was precisely [tex]\sqrt{2}[/tex].

[Raiden Mitchell]

Thank you very much for providing evidence and describing everything in as much detail as possible. Indeed, in this case, the Pythagorean theorem turned out that the length of the hypotenuse of a right-angled triangle with legs of length 1 is exactly equal to the root of 2.