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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.

Rupert
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  • Looks nice to me! – Matthias Sep 03 '14 at 09:23
  • Typo: N^2 should be n^2? –  Sep 03 '14 at 09:32
  • absolutely, thank you! – Rupert Sep 03 '14 at 09:34
  • 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. – TonyK Sep 03 '14 at 09:34
  • Just to confirm - the number of prime factors of a square is always twice the number of the root? Does this generalize in any way to higher powers? –  Sep 03 '14 at 09:39
  • To use modern notation, I think Rupert means $\Omega(x)$ when he says "the number of prime factors of $x$". – Jose Arnaldo Bebita Dris Sep 03 '14 at 09:41
  • I was wrong about last notation, because if $\sqrt{6}=\frac{m}{n}$, $2\cdot3 n^2=m^2$. Hence, iff $a$ is prime ($\sqrt[n]{a}$) , this proof make a sense. – Rupert Sep 03 '14 at 10:02
  • 2 is prime; the purpose is to show that irrationals exist. Is the number of prime factors of a square always twice the number of the root? –  Sep 03 '14 at 10:17
  • Your proof could be changed as follows: The prime power factorization of $m^2$ has an even number (possibly $0$) of occurrences of the prime $2$, while the factorization of $2n^2$ has an odd number of occurrences of the prime $2$. Phrased that way, it applies word for word to the case $a=6$. – André Nicolas Sep 03 '14 at 10:18
  • @mistermarko It is always. Because let $a^p$ has $A$ prime factors, $b^q$ has $B$ prime factors, then $(a^p\cdot b^q\cdots )^2=a^{2p}\cdot b^{2q}\cdots$. Here,$a^{2p}$ has $2A$ prime factor,$b^{2q}$ has$2B$ prime factor. First, it has $A+B$ prime factor, after squared, $2(A+B)$ prime factor. – Rupert Sep 03 '14 at 10:35
  • @AndréNicolas What I wanted to say is if $a=6$, equation is $2\cdot 3\cdot n^2=m^2$, LHS has $2M+2=2(M+1)$ prime factor. In this situation, we can't apply fundamental theorem of arithmetic. In above proof, essential is "the number of prime factor is even or add ". So I should think when $a$ is composite number. – Rupert Sep 03 '14 at 10:47
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    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. – Bill Dubuque Sep 03 '14 at 13:19

3 Answers3

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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.

BlackAdder
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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.

Mathemagician
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  • 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". – Rupert Sep 03 '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)).