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[parent] proof that $\sqrt{2}$ is irrational (Proof)

Assume that the square root of $ 2$ is rational. Then we can write

$\displaystyle \sqrt{2} = \frac{a}{b},$    

where $ a,b\in{\mathbb{N}}$ and $ a$ and $ b$ are relatively prime. Then $ \displaystyle 2=(\sqrt{2})^2=\left(\frac{a}{b}\right)^2=\frac{a^2}{b^2}$. Thus, $ a^2=2b^2$. Therefore, $ 2\mid a^2$. Since $ 2$ is prime, it must divide $ a$. Then $ a=2c$ for some $ c\in{\mathbb{N}}$. Thus, $ 2b^2=a^2=(2c)^2=4c^2$, yielding that $ b^2=2c^2$. Therefore, $ 2\mid b^2$. Since $ 2$ is prime, it must divide $ b$.

Since $ 2\mid a$ and $ 2\mid b$, we have that $ a$ and $ b$ are not relatively prime, which contradicts the hypothesis. Hence, the initial assumption is false. It follows $ \sqrt{2}$ is irrational.

With a little bit of work, this argument can be generalized to any positive integer that is not a square. Let $ n$ be such an integer. Then there must exist a prime $ p$ and $ k,m\in{\mathbb{N}}$ such that $ n=p^km$, where $ p\nmid m$ and $ k$ is odd. Assume that $ \sqrt{n}=a/b$, where $ a,b\in{\mathbb{N}}$ and are relatively prime. Then $ \displaystyle p^km=n=(\sqrt{n})^2=\left(\frac{a}{b}\right)^2=\frac{a^2}{b^2}$. Thus, $ a^2=p^kmb^2$. From the fundamental theorem of arithmetic, it is clear that the maximum powers of $ p$ that divides $ a^2$ and $ b^2$ are even. Since $ k$ is odd and $ p$ does not divide $ m$, the maximum power of $ p$ that divides $ p^kmb^2$ is also odd. Thus, the same should be true for $ a^2$. Hence, we have reached a contradiction and $ \sqrt{n}$ must be irrational.

The same argument can be generalized even more, for example to the case of nonsquare irreducible fractions and to higher order roots.



"proof that $\sqrt{2}$ is irrational" is owned by Wkbj79. [ full author list (2) | owner history (1) ]
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See Also: irrational, surd


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Cross-references: roots, fractions, irreducible, contradiction, even, clear, fundamental theorem of arithmetic, odd, square, integer, positive, irrational, divide, prime, relatively prime, rational
There are 3 references to this entry.

This is version 8 of proof that $\sqrt{2}$ is irrational, born on 2002-05-20, modified 2007-08-23.
Object id is 2920, canonical name is SquareRootOf2IsIrrationalProof.
Accessed 7932 times total.

Classification:
AMS MSC: 11J72 (Number theory :: Diophantine approximation, transcendental number theory :: Irrationality; linear independence over a field)
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