
Elementary proof that e is irrational - luisb
http://fermatslibrary.com/s/elementary-proof-that-e-is-irational
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digler999
very cool and concise proof. but minor soapbox rant: practically every other
proof I've read while casually studying number theory and group theory (I
didn't make it very far, especially for GT) seems to use the word "clear".
While the proof is supposed to be the nuts and bolts holding together this
important, indisputable logic, they gloss over how "clearly" some supporting
element is (without proving it). It was irritating to me because it defies the
spirit of proving something.

Even in this proof, the author says "clearly" the right hand side is between
0-1. That's not automatically clear to me. It seems to be true, but at a
glance, I can't automatically be sure that the statement is always true for
all values.

It got really bad in the group theory book, because the "clear" sub-proofs
were very primitive concepts, important things (sorry I can't recall an
example) but that we normally take for granted. So when I saw them gloss over
it with "clearly" I always felt suspect, like they were hand-waving.

~~~
Certhas
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.

~~~
chx
This professor is a widespread urban legend and there's no source to prove it
actually ever happened.

~~~
wfo
It (or something like it) has happened to most working mathematicians.

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airesQ
For those that did not notice: if you click the second comment by the margin,
you get a comment with a step-by-step decomposition of the proof.

~~~
nuclx
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.

~~~
airesQ
I see your point. There are indeed a couple of errors in there. But the point
still holds.

The sign is wrong on the sum of the left hand side. And a (-1)^(a+1) is
missing on the RHS. But these do not change the "this is an integer" claim.

Further down, it does assume that 'a' is even, without saying so . But if 'a'
was odd, things would still be the same but we would have a ]-1,0[ interval.

Anyway, I'm not a mathematician so I'm probably missing stuff.

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leephillips
This is from a volume of The American Mathematical Monthly from 1953. I looked
around this site for some kind of reference but couldn't find it. I find it
hard to believe (even these days) that this site just puts up papers without
saying where they came from, so I assume I'm just finding the UI obscure.
Where's the reference? (I found it thanks to Google Scholar.)

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davekeck
But a number between 0 and 1 could be arbitrarily close to 0 and 1, which I
thought could be proved to be equal to 0 or 1, which are integers. So isn't
this not necessarily a contradiction?

~~~
panic
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).

~~~
irishsultan
But in this case the right side is exactly that, the sum of an infinite
sequence of numbers (which equals the limit of the partial sum for that
sequence if it converges (which this sequence of partial sums does)).

The reason it still works is because it's actually 0 < s_0 < a < s_1 < 1, so
even if a approaches s_1 arbitrarily close it still won't be equal to an
integer (same is true if it goes arbitrarily close to s_0 of course).

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mrcactu5
the "usual" proof can be found on Wikipedia using the argument of Joseph
Fourier (of Fourier series fame)
[https://en.wikipedia.org/wiki/Proof_that_e_is_irrational](https://en.wikipedia.org/wiki/Proof_that_e_is_irrational)

The idea is the same, that all the truncated series 2 < 1 + 1/2! + 1/3! + ...
< 3 lie between two integers can cannot be integer.

This is a different outcome than the geometric series 1 + 1/2 + 1/4 + 1/8 +
... = 1 which converges to a whole number.

[http://math.stackexchange.com/questions/476939/proving-
the-i...](http://math.stackexchange.com/questions/476939/proving-the-
irrationality-of-en)

[http://math.stackexchange.com/questions/713467/prove-
that-e-...](http://math.stackexchange.com/questions/713467/prove-that-e-is-
irrational)

[http://math.stackexchange.com/questions/425963/is-there-a-
si...](http://math.stackexchange.com/questions/425963/is-there-a-simple-proof-
that-e2-is-irrational-using-a-positional-numeral-syst)

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cjjuice
Does someone have a link to the notation to decipher this?

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

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

~~~
dfabulich
Are the big braces in the second equation just supposed to be parentheses?

~~~
Someone
As zodiac said: yes.

I think the printer knew he needed large parentheses, but didn't have them
available (or he had used all of them elsewhere in this edition of this
journal), so he went for something similar that he had instead.

Knuth's rationale for writing TeX is full of this kind of observations on
mathematical typesetting: letters that become larger in some places, italics
that disappear, tiny font changes, etc.

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posterboy
Am I blind or are the papers not sourced? The proper placement for the source,
the date of publication specifically, would be somewhere at the top.

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vadiml
He writes 'but the left side is an integer'.... IMO the left side of the last
equation is equal 0.... Any comments?

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yk
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.)

~~~
widdma
The left side does not need to be zero, and it does not follow from the
assumptions.

The proof only uses the fact that the left side is an integer, since all terms
are integers (a!/n! ∈ Z for a > n)

~~~
yk
You are right, see my edit above.

