I am still new to proofs and just getting some practice in so please be kind. I must use the direct method to prove this claim, as stated in the instructions. Any feedback would be much appreciated. Thanks in advance.
Claim: Prove that for all natural numbers $n, n^2-n$ is even.
Proof attempt: If $n^2-n$ is even, then $n^2-n=2k \Rightarrow n^2=2k+n$, for some $k \in \mathbb{Z}$. Multiplying both sides by $2$ we obtain $2n^2=4k+2n \Rightarrow 2n^2=2(2k+n)$. Since $k$ is some integer, $2k+n$ is also some integer. Let $2k+n=m$. Therefore, $2n^2=2m$, which is even.
I know there is something wrong with my proof, but I just can't pinpoint it. Please help!
Should I approach it this way?
If $n^2-n$ is even, then $n(n-1)=2k$, for some $k \in \mathbb{Z}$.
Best Answer
Inductive Proof
Let $n=1$. Then $n^2-n=0$,which is even. This shows the statement holds for $n=1.$ Assume that the statement holds for $n=k$, namely, $k^2-k$ is even. Then $$(k+1)^2-(k+1)=(k^2-k)+2k,$$which is a sum of two even numbers. Hence, this is also even, which shows that the statement holds for $n=k+1$ as well. Consequently, by inductive principle, we may claim the statement holds for all $n=1,2,\cdots.$