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# the square root of 2

The number that equals 2 when multiplied by itself.

It's irrational, yes. Lets prove it with a rather number-theoretical approach. Assuming it can be written as an irreducible fraction $${a\over b}$$, we can safely say that all prime factors in $$a$$ are not in $$b$$ and vice versa (otherwise the fraction is not irreducible). When squaring the fraction, we should get 2. But, the prime factors in $$a$$ and in $$b$$ didn't change, so the fraction is still irreducible, but also equal to 2. This would mean $$b=1$$, but obviously, the square root of 2 is not an integer.

The continued fraction of the square root of 2 is also interesting. We can easily prove with recurrence that$\sqrt{2}=1+{1\over 2+{1\over 2+{1\over 2+{1\over \ddots}}}}$which is sometimes written as $$[1;\bar{2}]$$. I also, for the sake of this page, created the infinite sums$\sqrt{2}=\sum_{k=0}^{\infty}{(2k)!\over 8^kk!^2}$and (this one converges really slow)${1\over\sqrt{2}}=\sum_{k=0}^{\infty}{(2k)!\over (-4)^kk!^2}$Or we connect it to $$\pi$$ by $\sqrt{2}=2\sum_{k=0}^{\infty}{(-\pi^2)^k\over 16^k(2k)!}$

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