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A very brief proof to show that $\sqrt{2}$ is irrational goes like this:

We know that $\sqrt{2}$ is an algebraic integer. If it is rational it should be an integer which is not therefore it is irrational.

Now the question is that can we use a similar proof for showing that $e$ or $\pi$ are irrational?

I mean is there a proof of existence or non-existence of an $n \times n$ matrix $A$ of integers whose eigenvalue is $e$ or $\pi$?

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    $e$ and $\pi$ are transcendental...so no such matrix exists. But this is not an elementary result. – lulu Oct 24 '16 at 11:51
  • Elementary result? So you mean we can write a proof for non existence? –  Oct 24 '16 at 11:52
  • [here](http://math.stackexchange.com/questions/12872/how-hard-is-the-proof-of-pi-or-e-being-transcendental) is a good general discussion of some of the ideas. $e$ is easier to work with and a sketch of the proof can be found [here](https://en.wikipedia.org/wiki/Transcendental_number). – lulu Oct 24 '16 at 11:53
  • there are elementary proofs that $e$ is irrational. [here](https://en.wikipedia.org/wiki/Proof_that_e_is_irrational) for instance. Transcendence is more subtle. – lulu Oct 24 '16 at 11:55
  • Great reference, thanks! –  Oct 24 '16 at 11:59
  • Showing that $e$ is irrational using that $e$ is transcendental? – GEdgar Oct 24 '16 at 12:34
  • @GEdgar That would be trivial, since all transcendental numbers are also irrational. – u8y7541 Dec 16 '17 at 23:12

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