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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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marked as duplicate by Arturo Magidin, Asaf Karagila, Rasmus, Srivatsan, Chris Eagle Oct 9 '11 at 20:44

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Quite similarly to this answer: math.stackexchange.com/questions/5/… –  Asaf Karagila Oct 9 '11 at 20:38
    
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 $\frac{a'}{b'} = \frac{a}{b}$. –  Arturo Magidin Oct 9 '11 at 20:41

1 Answer 1

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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I don't understand the downvote. This is a valid proof! –  user59671 Mar 13 '13 at 22:04
    
@CutieKrait: This is just a spite downvote, because I made somebody look foolish in another thread. –  TonyK Mar 14 '13 at 15:46

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