![Theorem. If $n$ is an integer and $n^2$ is even, then $n$ is itself even.
Proof. Contrapositives are for cowards, so assume $n$ is an integer and $n^2$ is even. Then $n^2=2k$ for some integer $k$, and thus $n^2-2k=0$. Behold:
\[ n = n + (n^2-2k) = n(n+1)-2k. \]
Both $n(n+1)$ and $2k$ are even, so $n$ is even.
QED.](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:4wrxafbsjycxbpjo4e42ht6q/bafkreihucnue4bdyzulcrzddytan5ttdx56rg5ci6cn4vhxw5lebnbdzma@jpeg)
Theorem. If $n$ is an integer and $n^2$ is even, then $n$ is itself even. Proof. Contrapositives are for cowards, so assume $n$ is an integer and $n^2$ is even. Then $n^2=2k$ for some integer $k$, and thus $n^2-2k=0$. Behold: \[ n = n + (n^2-2k) = n(n+1)-2k. \] Both $n(n+1)$ and $2k$ are even, so $n$ is even. QED.
Contrapositives are for cowards. Behold.
22.01.2025 17:58 — 👍 135 🔁 38 💬 6 📌 9
