Prove the following statement by proving its contrapositive:
“If $n^3 + 2n + 1$ is odd then n is even”
Therefore: $\lnot q \rightarrow \lnot p =$ "if $n^3 + 2n + 1$ is even then $n$ is odd.
So for this I began assuming that: $n=2k+1$
$(2k+1)^3 +2(2k+1)+1 = 8k^3+12k^2 +10k+4 = 2k(4k^2 +6k+5)+4$
The last statement: $2k$ is even, therefore $2k(4k^2 -6k+5)$ is also even and 4 is $2\cdot 2$ which is also even.
Now, my question is, when proving the contrapositive, what's your final conclusion? If it works for the contrapositive, then your theorem holds? Or is there something else?
Best Answer
Let p be the boolean value for the statement "$n^3+2^n+1$ is odd", q be the boolean value for "n is even". $$p \rightarrow q$$ $$1 \rightarrow 1$$ $$0 \rightarrow 1$$ $$0 \rightarrow 0$$ (as you can see, only $1 \rightarrow 0$ will indicate $p \rightarrow q$ is false)
The contrapositive of $p \rightarrow q$ is $\lnot q \rightarrow \lnot p$, which is "if n is not even, $n^3+2^n+1$ is not odd."
The statement "if $n^3+2n+1$ is even then n is odd" is implying $\lnot p \rightarrow \lnot q$ and it is not equivalent to $p → q$. $$¬p → ¬q$$ $$¬0 → ¬0$$ $$¬1 → ¬0$$ $$¬1 → ¬1$$
($\lnot p \rightarrow \lnot q$ will still be valid when $p=1$ and $q=0$, which is not the case in $p \rightarrow q$)
If it works for the contrapositive, your statement definitely holds.
The statements $p \rightarrow q$ and $\lnot q \rightarrow \lnot p$ are logically equivalent.