If $p: \forall \in N, n^2+n$ is an even number $q: \forall n \in N, n^2-n$ is an odd number, then the truth values of $p \wedge q, p \vee q$ and $p \rightarrow q$ are respectively

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Aicharya · Accepted Answer

$\because$ product of two natural numbers is an even natural number Hence, $n^2+n=n(n+1)$ and $n^2-n=n(n-1)$ are even numbers So, $p$ is true and $q$ is false $\Rightarrow p \wedge q$ is false and $p \vee q$ is true and $p \rightarrow q$ is false

If $p: \forall \in N, n^2+n$ is an even number $q: \forall n \in N, n^2-n$ is an odd number, then the truth…

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