Elementary proof that e is irrational

This proof's use of "clear" is sorta reasonable, since they are probably assuming the reader is familiar with alternating series with decreasing terms.

I would agree that it is sometimes fairly annoying. "Proof by intimidation"

A: "it is clear that X"

B: "Sorry, why is it clear?"

A: "No, it is CLEAR that X!"

It's a general fact that any series satisfying the list of conditions in http://calculus.subwiki.org/wiki/Alternating_series_theorem is bounded by any two consecutive partial sums; the author probably expects everyone to have seen the proof of this before

when I read it it wasn't clear to me either, so I looked at what was written and saw that the fractions on the right side were clearly getting smaller and smaller fast because each successive denominator consisted of the multiplication of a clearly increasing number of clearly increasingly large integers so within a matter of seconds it was clear to me that within that small number of terms that each additional term would have a vanishingly small effect on the (alternating sign) summation so what they said was had to be true.

So, I do sympathize with your overall point, but this isn't really a good example of it. It's less clear to me what they said about the left side of the (in)equality because that expression to me is much more eye-glazing...but it doesn't take long to convince yourself of that one too. I think it's fair for them to say it's "clear", but you have to not let yourself be put off by the large number of algebraic symbols strung together.

"According to the Nobel Prize-winning physicist Richard Feynman (Feynman 1997), mathematicians designate any theorem as "trivial" once a proof has been obtained--no matter how difficult the theorem was to prove in the first place. There are therefore exactly two types of true mathematical propositions: trivial ones, and those which have not yet been proven." http://mathworld.wolfram.com/Trivial.html

Anecdote has it that a professor at my alma mater was presenting a proof to students, stating that X was obvious, before halting and wondering aloud: "Is this really obvious?" He examined the blackboard and started pacing for a good 15 minutes mumbling and thinking, before exclaiming with a triumphant expression on his face: "Indeed it IS obvious!"

That said, obviously no proof can function without assuming some mathematical background knowledge. This is a paper from a research journal, not a textbook and the readership of the journal where it was published would presumably find this clear.

The problem is, if you actually do ever single 'atomic' step even very simple proofs become huge (see formal methods). To avoid this, we use "clearly" to mean something like "obvious t anyone with sufficient training, ie the target audience.

There are two problems with this - if you are trying to obtain that training it may no yet be obvious. Worse, sometimes subtle problems slip through while you are nodding your head!

I take "clearly" to be a flag to come back to that point later. Especially if it is not clear to me. It is often a flag for points that the author feels contributes to the intuition of the proof. Which is not that it is easy. Just that it is an educational part of the process.

This is why the Fermat's Library website is awesome: the notes on the margin explain this to you.

To be fair, the author says that the value is between 0 and 1 since _the alternating series clearly converges_ to a value between the 1st and 1st+2nd.

So in reality it explains pretty well why it's between 0 and 1, and leaves the "clearly" to something which in his opinion is common knowledge: https://en.wikipedia.org/wiki/Alternating_series_test

The proof is quite terse but did you find the 'clearly' bit harder to digest and verify than the series expansion itself? You know a is an integer >= 1 and every time you subtract a term bigger than the one your subsequently add so there's no way to get past the sum of the first two terms.

The bit of algebra and the series expansion is shown in the margin note and not touched upon at all in the text - it seems not nearly as simple to check by simply staring at the thing.

> the author says "clearly" the right hand side is between 0-1

No, the author says 'the alternating series clearly converges to a value between its first term and the sum of its first two terms', which is a late-first-year-calculus sort of thing.

Really nice proof! The "clearly" thing isn't too hard to see even if you aren't aware of the "alternating series theorem":

Consider the sequence a1, a2, a3... with the sum being a1 - a2 + a3..., note we can write the sum as: a1 + (-a2 + a3) + (-a4 + a5) hence the sum is less than a1.

We can also write it as: (a1-a2) + (a3-a4) + (a5-a6) .... hence the sum is greater than (a1-a2).

Looks to me like there are errors in the transformation. The sum term corresponding to the resulting equation of the proof has a wrong sign in the second summand of the left term (+ instead of -). Also in the original manuscript the right side always starts with a positive alternating series term, which is correct. In the version on the margin the sign of the first term of the right side depends on the evenness of a.

If 0 < a < 1, then a can't be equal to 0 or 1. There's no "arbitrarily close" for numbers, just sequences of numbers (like the partial sums in the series).

It's a contradiction because there are no integers strictly between 0 and 1.

If the proof were just that the sum converges to some number 0 <= x <= 1, then you'd be right.

But it's showing specifically that the sum lies between the first term s_0 (which is strictly less than 1) and the sum of the first two terms s_0 - s_1 (which is strictly greater than 0). So there's no possibility for it to converge to a value outside of the interval [s_0-s_1,s_0], which is itself strictly contained inside [0,1].

Which particular notation? Maybe Factorial [0] and Summation [1]?

[0] https://en.wikipedia.org/wiki/Factorial [1] https://en.wikipedia.org/wiki/Summation

You'll also need to know that e^x = sum{0 -> infinity} x^n / n!.

Zero is an integer. The point is the right side is not. Since it's strictly greater than zero and less than 1.

Erm, nope, it would be if the sum went to inf but it doesn't.

It is clear that the left hand side has to be zero, however I do not see a straightforward way to prove that without using the contradiction itself and he does not need it to be zero to choose that there is a contradiction, so the claim the left side is an integer sidesteps an entirely optional technical difficulty.

[Edit:] The argument I had in mind is bad, my point is that the value of the left hand side does not matter for the proof, so it is not calculated. (I leave the comment up for context of the answers below.)