The negation of $\forall \mathrm{x} \in \mathrm{N}, \mathrm{x}^2+\mathrm{x}$ is even number' is
The negation of $\forall \mathrm{x} \in \mathrm{N}, \mathrm{x}^2+\mathrm{x}$ is even number' is
$\forall \mathrm{x} \in \mathrm{N}, \mathrm{x}^2+\mathrm{x}$ is not an even number
$\forall \mathrm{x} \in \mathrm{N}, \mathrm{x}^2+\mathrm{x}$ is not an odd number
$\exists x \in N$ such that $x^2+x$ an even number
$\exists \mathrm{x} \in \mathrm{N}$ such that $\mathrm{x}^2+\mathrm{x}$ is not an even number
Solution
Let $\mathrm{p}: \forall \mathrm{x} \in \mathrm{N}$ and $\mathrm{q}: \mathrm{x}^2+\mathrm{x}$ is even number.
The logical form of given statement is $\mathrm{p} \wedge \mathrm{q}$.
$\sim(\mathrm{p} \wedge \mathrm{q}) \equiv \sim \mathrm{p} \vee \sim$ q i.e. $\exists \mathrm{x} \in \mathrm{N}$ such that $\mathrm{x}^2+\mathrm{x}$ is not an even number.