proof that $\sqrt{2}$ is irrationalSquareRootOf2IsIrrationalProof2002-05-20 03:05:392007-08-23 19:34:01Proofsquare root of 20% this is the default PlanetMath preamble. as your knowledge
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\def\N{{\mathbb N}}\PMlinkescapeword{argument}
\PMlinkescapeword{hypothesis}
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Assume that the \PMlinkname{square root of $2$}{SquareRootOf2} is rational. Then we can write
\begin{equation*}
\sqrt{2} = \frac{a}{b},
\end{equation*}
where $a,b\in\N$ 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 $c\in\N$. 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 $k,m\in\N$ such that $n=p^km$, where $p\nmid m$ and $k$ is odd. Assume that $\sqrt{n}=a/b$, where $a,b\in\N$ 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.