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How to prove $\sqrt3$ is irrational using Fermat's infinite descent method?

Like says in Carl Benjamim Boyer's book.

Isnt the same prove to $\sqrt2$, in Boyer's book says something like this.

$\sqrt3=a1/b1$

$1/(\sqrt3-1)=(\sqrt3+1)/2$

$\sqrt3=(3b1-a1)/(a1-b1)$

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    Exactly the same way you prove it for $\sqrt{2}$: write $\sqrt{3}=\frac{a}{b}$, or $b\sqrt{3}=a$. Square both sides, conclude you can find $a'$, $b'$, $a'\lt a$, $b'\lt b$ with $\frac{a'}{b'} = \frac{a}{b}$.2011-10-09

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Because the greatest power of $3$ that divides $p^2$ must be even, whereas the greatest power of $3$ that divides $3q^2$ must be odd. So $(p/q)^2$ can't equal $3$.

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    @CutieKrait: This is just a spite downvote, because I made somebody look foolish in another thread.2013-03-14