that $$e$$ is irrational:
From Taylor's theorem, $$\exists\xi\in(0,1)$$ such that $$e=\sum_{n=1}^{N}\frac{1}{n!} + R_{N}$$ where $$0<R_{N}=\frac{e^{\xi}}{(N+1)!}<\frac{3}{(N+1)!}$$. Assume $$e$$ is rational: $$e=\frac{a}{b}, a,b\in\mathbb{Z}, b\neq 0$$ & let $$N>\max(b,3)$$: $$\frac{N!a}{b}-\sum_{n=1}^{N}\frac{N!}{n!} = N!R_{N}$$. Since $$N>b$$, $$N!$$ must contain $$b$$ and $$N!R_{N}\in\mathbb{Z}$$ . Since $$N>3$$: $$0<N!R_{N}<\frac{3}{N+1}<\frac{3}{4}$$ which is a contradiction.

that there are infinitely-many Pythagorean triples (that is, integers $$x,y,z$$ such that $$x^{2}+y^{2}=z^{2}$$):

Let $$a,b$$ be any integers, then:

$$(a^{2}+b^{2})^{2}=(a^{2}-b^{2})^{2} + (2ab)^{2}$$ $$\blacksquare$$

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