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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 $p \wedge q, p \wedge q$ and $p \rightarrow q$ are respectively
Options:
$q: \forall n \in N, n^2-n$ is an odd number,
then the truth values of $p \wedge q, p \wedge q$ and $p \rightarrow q$ are respectively
Solution:
2416 Upvotes
Verified Answer
The correct answer is:
F, T, F
$\because$ 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
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
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