# certain proofs of the irrationality of $\sqrt{2}$

I had the impression that there might be proofs of the irrationality of $\sqrt{2}$ that showed that $$\left|\frac a b - \sqrt{2} \right| \ge (\text{something possibly depending on a or b}) >0$$ where $a,b\in\mathbb{Z}$. But the one I saw in Wikipedia's article on the square root of $2$ spoke of whether the multiplicity of $2$ as a factor of $a$ or $b$ is even or odd, and that makes it seem not all that different from the old-fashioned proof that we all learned in childhood (you know, in 500 BC when we were children). Is there a short and simple proof of that kind to which this last criticism will not apply?

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Hint: $|\sqrt2 - \frac{a}{b}| > \frac{1}{3b^2}$ for all rational $a/b$

Opp, sorry, I do not think there is a proof of rationality of $\sqrt2$ that uses the inequality. I saw 5,6 proofs of the rationality of $\sqrt2$ on the first day of my class but most proofs are in the article that Nbubis posted, and this inequality above on the next day, is to study how well rational, irrational and transcendental numbers are approximated.

PS: One cute proof for the rationality of $\sqrt{2}$ is to use the fundamental theorem of arithmetic, ie: every integer can be factored into product of prime numbers. For a square, every prime will appear an even number of times. $a^2 = 2 b^2$ The prime $2$ appears even times for the right and odd times for left sides, contradiction.

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care to explain why? – nbubis Apr 28 '12 at 18:44
@nbubis I side with you. – Peter Tamaroff Apr 28 '12 at 18:45
you can prove it by square both sides, do some algebra, and you will get: 18*b^4 + 9*b^2*a^2 >1 + 18*sqrt(2)*a*b^3, and you can use: 18*b^4 + 9*b^2*a^2 >= 2* sqrt(18*a^4* 9*b^2*a^2) = sqrt(2)* 18*a*b^3. You know that the equality is not happening, since sqrt(2) is different from a/b – Long Mai Apr 28 '12 at 18:56
@LongMai Isn't that last statement in your comment what you're trying to prove? – Mike Apr 28 '12 at 19:02
@LongMai But the inequality is true if and only if there is no $a/b$ such that $\sqrt 2=a/b$! – Peter Tamaroff Apr 28 '12 at 19:56
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You can use that if $a^2=2b^2$ then $(2b-a)^2=2(a-b)^2$ to show that if $\frac ab$ is a square root of 2 then so is $\frac {2b-a}{a-b}$. So there is not a fraction with smallest denominator.

Also if $\frac ab$ is an approximation to $\sqrt 2$, then $\frac {a+2b}{a+b}$ is in general a better one, which may be what you are remembering.

If $$2-\frac{a^2}{b^2}=\epsilon$$ then

$$\frac{(a+2b)^2}{(a+b)^2}-2 = \frac{2b^2-a^2}{(a+b)^2}=\epsilon \left(\frac b{a+b}\right)^2$$

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You might like the geometric proof posted here: http://blog.plover.com/math/sqrt-2-new.html

I think it can be converted into the form you presented.

Edit: 20 proofs of the irrationality of $\sqrt{2}$ can be found here, but none with the exact form you mentioned.

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I saw that one, and presented it in classrooms, before Tom Apostol published it. I learned it from a book by Otto Toeplitz, published maybe around 1960 or so. – Michael Hardy Apr 28 '12 at 18:32