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.