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I'm not looking for a proof, but rather an explanation, because I know there's something wrong with my thinking.

So, I know that $$ e = \lim_{n\to\infty} (1+\dfrac{1}{n})^n $$ And also $$ e = \sum_{k=0}^{\infty} \dfrac{1}{k!} $$

And I'm confused as to why $e$ can be irrational, since both of those definitions are a rational number(?).

I know that addition under rationals is closed, so I'm confused as both of these can be rearranged to something with rationals (e.g. sum of rationals, or $(\dfrac{n+1}{n})^n$), so I guess my question is, is why these rationals converge to an irrational?

Thanks.

José Carlos Santos
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talbi
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    For every irrational number $x$ you have a sequence of rational numbers converging to $x$. – Martin R Nov 05 '20 at 10:56
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    $e$ is limit of a sequence of rational numbers. This does not imply that $e$ is rational. (Actually, every real number is limit of a sequence of rational numbers !) – TheSilverDoe Nov 05 '20 at 10:57
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    Every number is the limit of rational numbers, and so it should not be shocking that $e$ can be written in such a way even though it is irrational. If you want an actual proof, wikipedia has several. – Aaron Nov 05 '20 at 10:57
  • Here (https://math.stackexchange.com/questions/2030524/a-very-short-proof-of-e-is-irrational) you can find a short ptoof of irrationality of $e$ using, precisely that $e=\sum\frac{1}{n!}$. – Tito Eliatron Nov 05 '20 at 10:58
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    The set of rationals is closed under addition. However it is not closed under limits and, in particular, it is not closed under infinite sums. – halrankard2 Nov 05 '20 at 10:59
  • @halrankard2 I did think of that, but is it not possible to add two rationals and increase the lower bound on $k$? I mean, add $\dfrac{1}{1!} + \dfrac{1}{2!}$ and then continue repeating that? – talbi Nov 05 '20 at 11:01
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    @TheSilverDoe Thanks! It makes more sense now. – talbi Nov 05 '20 at 11:05
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    @talbi, addition is really only defined as addition of finitely many terms. We mistakenly sometimes think of adding infinitely many terms... But what we are really doing is taking partial sums of finitely many numbers, and taking the limit of that sequence. This 'limit taking' is a legitimately new operation outside of addition. So we can't simply extend our intuition that finitely many rational terms add to a rational... to infinitely many terms. We have this operation of limit taking which is something new. – Ameet Sharma Nov 05 '20 at 11:09
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    @AmeetSharma This explains why my intuition from before wouldn't work. Thanks! – talbi Nov 05 '20 at 11:13

3 Answers3

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The error lies in the assertion that “both of those definitions are a rational number”. Not true. Both of them express $e$ as a limit of a sequence of rational numbers. Your error lies in the hidden (and false) assumption that the limit of a convergent sequence of rational numbers must be rational too.

José Carlos Santos
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That the rationals are closed under addition means that the sum of any two rationals is a rational again. By induction you can prove that this works for any finite number of rationals. But the rationals are not closed under infinite sums.

Tchaikovski
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Just because every partial (finite) sum is a rational number, that does not mean the limit is.

Take the well known constant as an example: $\pi = 3+0.1+0.04+...$

Every partial sum is rational, but the limit is a well known irrational number.

Similarly, for every finite natural number $n$, $1/n > 0$ but the limit $\lim_{n\to\infty} 1/n \not > 0$.

Eff
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