A proof that $e$ or $\pi$ is irrational via algebraic integers

Asked Oct 24 '16 at 11:48

Active Oct 24 '16 at 11:48

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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 $\pi$?

asked Oct 24 '16 at 11:48

2

$e$ 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 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. – lulu Oct 24 '16 at 11:53
there are elementary proofs that $e$ is irrational. here 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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