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}<j+1 then assuming that \sqrt{k}=m/n is in lowest terms also implies that \sqrt{k}=(kn-jm)/(m-jn) is in lower terms.


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Filed under Combinatorics, Miscellaneous

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