Please check if I did correctly for my proof by contrapositive question!

Question

Suppose m ∈ ℤ. Prove by contrapositive that if m^2-2m-3 is even, then m is odd.

My working:

Suppose m is any even integer.
Hence,   m = 2k
Thus,    m^2-2m-3 = (2k)^2-2(2k)-3
                  = 4k^2-6k-2k-3
                  = 4k^2-4k-3
                  = 2(2k^2-2k)-3

I am not sure if I am doing it correctly here. How do I continue to show that m^2-2m-3 is odd? Thankyou.

Answer 1

How do I continue to show that m^2-2m-3 is odd?

By showing that there exists some integer, $h$, where $m^2-2m-3 = 2h+1$.

As the comments have mentioned, you are almost done.   You merely need to expand $-3$ into $-4+1$ to get the required form.

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We shall prove by contraposition that any integer $m$ is odd if $m^2-2m-3$ is even.

Suppose that an arbitrary integer $m$ is not odd, aiming to show that $m^2-2m-3$ must not be even if this is so.

By the supposition, there exists some integer $k$ where $m=2k$.

By substitution and algebra we derive: $$\begin{align}m^2-2m-3 &= (2k)^2-2(2k)-3\\&= 4k^2-4k-3\\&=4k^2-4k-4+1\\&=2(2k^2-2k-2)+1\end{align}$$

Let $h=(2k^2-2k-2)$.   Because it is a series of products of integers, this $h$ is an integer.

Therefore the supposition does entail that there exists some integer $h$, where $(m^2-2m-3)=2h+1$.

So demonstrating that: $\bf m^2-2m-3$ cannot be even if $\bf m$ is not odd as we aimed.

Finally, by contraposition we have that: $\bf m$ is odd if $\bf m^2-2m-3$ is even, as was to be proven.

$\blacksquare$