The proof of $\sqrt{2}$ is not rational number via fundamental theorem of arithmetic.

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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)).