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This is false : $(\forall n \in N)[2n|n^2]$ but how can be proved ?

$b|a$ if and only if $ \exists q[a=bq] , b \neq 0 $

This is my proof :

To prove $(\forall n \in N)[2n|n^2]$ is false take $n=3$

So $2(3)|3^2$ = $6/9$ = $2/3$ which is not an integer.

To prove $(\forall n \in N)[2n|n^2]$ by contradiction , is this correct :

To prove a statement is false prove it's negation is true.

Negating $(\forall n \in N)[2n|n^2]$

= $\neg(\forall n \in N)[2n|n^2]$

Negating a $\forall$ becomes $\exists$ , Negating a $\exists$ becomes $\forall$

= $(\exists n \forall N)[2n|n^2]$

So now need to prove $(\exists n \forall N)[2n|n^2]$ is true

Is $2(3)|3^2$ = $6/9$ = $2/3$ now a contradiction ?

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    As you remarked "to prove $(\forall n \in \mathbb{N})[2n|n^2]$ is false take $n=3$", so this proves that $$(\exists n \in \mathbb{N})\overline{[2n|n^2]}$$ is true and that suffices ! – 2017-02-26
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    Your first proof is not right, as you interpret $a|b$ incorrectly. In your second attempt, the negation does not cause another quantifier. You just move the negetation across the universal, change that t9o existential, finally put negation on statement. – 2017-02-26

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The formula $(\forall n \in N)[2n|n^2]$ is false if $\exists n\in N$ such that $[2n\not| n^2]$. In other words, the former formula is false when $\exists n\in N[n\not|n^2]$ is true.