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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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