Proof of Transcendental number

[eemilwallin]

The proof from ProofWiki is quite elegant:

Let q = 10^(n!), so that we can write L = p/q + sum for appropriate choice of p. The sum term then goes from k=n+1 instead of k=1. Now we can take away p/q from both sides and since the sum is increasing, we can take the absolute value only on the left: |L-p/q| = sum. This is in the form that we want by Liouville's theorem for the reciprocals 1/x^n, which were shown in the video. Because evaluating the sum is hard, we can take a sandwich inequality of 1/q^(n+1) < |L-p/q| < 2/q^(n+1), the righthandside inequality is evident from the sum because 0.2 > L and so on. Now we're almost there.

But now we can also notice that 2/q^(n+1) < 1/q^n (trivial but a high-schooler can prove this at home) so we get 1/q^(n+1) < |L-p/q| < 1/q^n, where the righthandside is in the form of contradiction to Lioville's theorem. So for any degree polynomial n, there exists a counter example to L being a root of a irreducible polynomial of deg n. So L is not an algebraic number, so it is transcendental.

[Shreyan-sdtd]

thats very easy number theory problem let me give you a moderate one
prove that root(2)+root(3) is also transcendental number
and when you are done with that i have another problem that asks to prove that if a number is rational then the decimal expansion is either terminating or repeating

[nirvanaxoxo]

greatest gen-z kid ever seen not sure but the only bad thing is his english is not so clear and he stops and does some mistakes and he cant explain much well