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I assume that $\sqrt{2}$ is positive number satisfies $(\sqrt{2})^2=2$.

proof) Let $m$, $n$ as natural number,$\ $ $M$ is the number of prime factor of $m$,$\ $ $N$ is also the number of prime factor of $n$. For example, $m=12=2^2\cdot3$, $M$ is $3$.

Then, if $\sqrt{2}$ were rational number, it could be expressed as a fraction $\frac{m}{n}$ in lowest terms.

If $\sqrt{2}=\frac{m}{n}$, $m^2=2\cdot n^2$. Then, $m^2$ has $2M$ prime factor, $n^2$ has $2N$ prime factor. LHS has even prime factor, RHS has odd prime factor. This is a contradiction to fundamental theorem of arithmetic.

Therefore the initial assumption—that $\sqrt{2}$ can be expressed as a fraction—must be false.

Is there any problems? If not, I think this proof is simple and easier than ordinary proof -contradiction to the property of lowest terms.

p.s. This proof may apply to any root of $n$-th power $\sqrt[n]{a}$, $n\in\mathbb{N}$ iff $a$ is prime.

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  • $\begingroup$ Looks nice to me! $\endgroup$ – Matthias Sep 3 '14 at 9:23
  • $\begingroup$ Typo: N^2 should be n^2? $\endgroup$ – user117644 Sep 3 '14 at 9:32
  • $\begingroup$ absolutely, thank you! $\endgroup$ – Rupert Sep 3 '14 at 9:34
  • $\begingroup$ This looks good to me too, and it generalises to any $\sqrt a$ if $a$ has an odd number of prime factors. To generalise to non-square numbers $a$ with an even number of prime factors, you have to look at the number of repetitions of a particular prime in the factorisations of $m$ and $n$ $-$ specifically, a prime that occurs an odd number of times in the factorisation of $a$. Such a prime must exist if $a$ is non-square. $\endgroup$ – TonyK Sep 3 '14 at 9:34
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    $\begingroup$ Note that the FTA is required to deduce that $m^2$ has twice as many prime factors as does $m,\,$ and this needs to be explicitly mentioned for the proof to be correct/complete. In rings with nonunique prime factorizations, $m^2$ may have a prime factorization with an odd number of primes, so the proof fails. $\endgroup$ – Bill Dubuque Sep 3 '14 at 13:19
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It looks OK, apart from one little typo:

$\sqrt{2}=\frac{n}{m}\implies 2m^2=n^2$, you got it the other way around.

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  • $\begingroup$ Thank you for your check! I'll edit. $\endgroup$ – Rupert Sep 3 '14 at 9:28
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As to why it is easier than the usual proof: The usual proof doesn't use the fundamental theorem of arithmetic (or at least, can be avoided pretty easily), which actually takes some work to prove.

Using the fundamental theorem of arithmetic for $\mathbb{Q}$ (i.e: Any rational number can be uniquely written as a product of INTEGER powers of primes, this is an easy corollary of the statement for integers) you can prove a much more general claim:

If $r$ is a positive rational number and $n \in \mathbb{Z}$, then $r^{1/n}$ is rational if and only if the prime factorization of $r$ is of the form $$\prod_{p} p^{n e_p}$$ for some $e_p \in \mathbb{Z}$, with only finitely many $e_p \neq 0$.

In more fancy words: The multiplicative group of positive rationals is a free abelian group, where the primes form a basis.

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  • $\begingroup$ Thank you for your answer. The reason why I thought this is easier than usual, this proof needn't proposition "$m^2$ is even, $m$ is even". Certainly fundamental theorem of arithmetic is difficult. But we and students can "feel" the fundamental theorem of arithmetic is clear. So I said "easier". $\endgroup$ – Rupert Sep 3 '14 at 10:18
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The step Bill Dubuque refers to can be made more explicit with this argument:

If P(x) is the prime factorization of x then by definition P(x)P(x) is equal to xx. The combination of the sets P(x) and P(x) [allowing duplication of elements] is when multiplied out, by FTA, the prime factorization of only one number, which because it is P(x)P(x) can only be xx. Consequently if M(P(x)) is the size of P(x), M(P(xx)) = M(P(x)) + M(P(x)).

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