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The negation of $\forall \mathrm{x} \in \mathrm{N}, \mathrm{x}^2+\mathrm{x}$ is even number' is
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$\exists \mathrm{x} \in \mathrm{N}$ such that $\mathrm{x}^2+\mathrm{x}$ is not an even number
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 $\mathrm{x}+\mathrm{x}$ is not an 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 $\mathrm{x}+\mathrm{x}$ is not an even number.
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