Contrapositive Math Example 1

Solution

1. 1
   Recall the structure of the conditional: $p \Rightarrow q$ where $p$: '$n^2$ is even' and $q$: '$n$ is even.'
2. 2
   The contrapositive is $\neg q \Rightarrow \neg p$. Form the negations: $\neg q$: '$n$ is not even' (i.e., $n$ is odd), and $\neg p$: '$n^2$ is not even' (i.e., $n^2$ is odd).
3. 3
   Contrapositive: 'If $n$ is odd, then $n^2$ is odd.' By the logical equivalence $p \Rightarrow q \equiv \neg q \Rightarrow \neg p$, this statement has exactly the same truth value as the original — and it is in fact true (odd times odd is odd).