# Two one sentence proofs

Continuing in the same vein here are two one sentence proofs.

Theorem: For $x,y\in\mathbb{C}, n\in\mathbb{N}$ we have $(x+y)^n=\sum_{k=0}^{n}\tbinom{n}{k}x^ky^{n-k}$.

Proof: The coefficient of $x^ky^{n-k}$ in the left hand side of the given expression, i.e. in $\underbrace{(x+y)(x+y)\cdots (x+y)}_{\text{n terms}}$, is obtained by selecting exactly $k$ parenthesis out of the total $n$ available to yield the $x's$, (the remaining parenthesis yield the $y's$), which can be done in $\tbinom{n}{k}$ ways following which the coefficient is exactly $\tbinom{n}{k}$.$\Box$

Theorem: $\sqrt{2}$ is irrational.

Proof: If $\sqrt{2}=m/n$ is in lowest terms then $\sqrt{2}=(2n-m)/(m-n)$ is in lower terms.$\Box$

Remark: The above proof generalizes to the case when $k$ is a non-square positive integer. Indeed, if $j<\sqrt{k} then assuming that $\sqrt{k}=m/n$ is in lowest terms also implies that $\sqrt{k}=(kn-jm)/(m-jn)$ is in lower terms.