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So I just managed to prove that $e^{m}$ where $m$ is an integer is irrational. I was wondering If I can prove using this fact that $e$ is Transcendental. The only thing I came up with was that if we set this polynomial $P(x)=c_{0}+c_{1}x+c_{2}x^2+\cdots+c_{n}x^{n}$ where $c$ is an integer and then assume that $P(e)=0$ we're saying that: $c_{1}e+c_{2}e^{2}+c_{3}x^{3}+\cdots+c_{n}e^n=-c_0$. Of course every term on left is an Irrational Number. On the right side we have an Integer. If we could only proof that the left side is an Irrational Number We would have a needed contradiction and thus prove that e is an Transcendental Number.

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    $\begingroup$ But of course sums and products of irrational numbers might be rational. How do you want to proof from $e^m\ \forall m\in\mathbb{N}$ that the given sum is also irrational, for all coefficients and all $n$? $\endgroup$ – Václav Mordvinov Apr 19 '18 at 19:46
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Fix $x=\sqrt2+1$.

Then $x^m$ is irrational for every $m>0$, but $x$ is not transcendental.

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Note that $y=\sqrt 2+\sqrt 3$ has the property that $y^m$ is irrational for every non-zero rational integer. And $y$ is algebraic, not transcendental. So your condition is not strong enough to prove your intended conclusion.

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  • $\begingroup$ Yes i know that. But in example of $e$ sum is irrational and I'm asking is there a proof of that. I am not looking for a proof that a sum of iirationals cant be rational beacause you can disprove it immidietly. $\endgroup$ – Filip Kończyk Apr 19 '18 at 20:02
  • $\begingroup$ @FilipKończyk Then you need to use more properties of $e$ than you have so far displayed. $\endgroup$ – Mark Bennet Apr 19 '18 at 20:28

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