If p: forall n in N, n^2+n is an even number | vedclass

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(C) The expression $n^2+n = n(n+1)$ is the product of two consecutive natural numbers,which is always even. Thus,$p$ is true.The expression $n^2-n = n

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$F, T, F$

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(C) The expression $n^2+n = n(n+1)$ is the product of two consecutive natural numbers,which is always even. Thus,$p$ is true.The expression $n^2-n = n(n-1)$ is also the product of two consecutive natural numbers,which is always even. Thus,$q$ is false.Now,evaluating the truth values:$p \wedge q = T \wedge F = F$$p \vee q = T \vee F = T$$p$ $\rightarrow q = T$ $\rightarrow F = F$Therefore,the truth values are $F, T, F$.

If $p: \forall n \in N, n^2+n$ is an even number and $q: \forall n \in N, n^2-n$ 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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