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Let $(x_n)_{n> 0}$ a sequence of $\{ 0,1 \}$ and

$$ x=\sum_{n=1}^{\infty}\frac{x_n}{10^n}. $$

Prove that if $x$ is irrational then $x$ is transcendental.
I tried to first start by proving that $x$ can't be the root of a second degree polynomial with integer coefficients, but I couldn't do it.

Ernie060
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  • The basic tool for proving that numbers are transcendental is Liouville's result. I don't know if it will work here but it's certainly the first thing I would try. – JonathanZ Nov 17 '18 at 16:16
  • Well this is a very courageous conjecture given that I don't know if the standard approach applies to non Liouville numbers all I can do is think but I will always fail with transcendental numbers even algebraic irrationals – Adam Ledger Jan 05 '19 at 19:21
  • Perhaps I guess we could start by looking at the kinds of $x_n$ necessary to ensure rationality of $x$, for example $x_n=\frac{1}{2}(1+(-1)^n)$ will give us $x=\frac{1}{99}$ – Adam Ledger Jan 05 '19 at 19:26
  • So we must partition the set of all arithmetic functions into two subsets to begin with, those that will give us rational $x$ if we set them as our $x_n$,and those that will not. perhaps you ought to provide these sequences in mod 2 that you are certain will provide us with irrational $x$ when they are selected as our $x_n$, lets worry about rationality before we stress about algebraicity, it is not so obvious as to just state the sequence as you have when both outcomes of rational and irrational can occur – Adam Ledger Jan 05 '19 at 19:31

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