Sincer, cred ca este un fel de axioma, deci nu poti demonstra.

Verified answer

[tex]\it Fie\ k\in\mathbb{N},\ k\ne\ p\breve a trat\ perfect \Rightarrow \exists\ n\in\mathbb{N}\ a.\ \hat\imath.\ n^2 < k < (k+1)^2 \Rightarrow \\ \\ \Rightarrow \sqrt{n^2} < \sqrt k < \sqrt{(n+1)^2} \Rightarrow n < \sqrt k < n+1 \Rightarrow \sqrt k\not\in\mathbb{N}\subset\mathbb{Q}[/tex]