Elementary proof that e is irrational(fermatslibrary.com)
fermatslibrary.com
Elementary proof that e is irrational
http://fermatslibrary.com/s/elementary-proof-that-e-is-irational
46 comments
"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
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
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.
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 site doesn't seem to mention it but the article is from the Monthly [1], which has a focus on exposition. Still, assuming some analysis seems fair.
[1] http://www.jstor.org/stable/2308411
[1] http://www.jstor.org/stable/2308411
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!"
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!"
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.
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.
[deleted]
> 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.
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.
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 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.
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
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 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!
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!
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).
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).
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.
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.
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.
Thank you for confirming this. These kind of errors drive me crazy when I try to work through proofs on my own.
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.
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.
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.)
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?
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].
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].
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).
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).
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).
It's a contradiction because there are no integers strictly between 0 and 1.
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
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/713467/prove-that-e-...
http://math.stackexchange.com/questions/425963/is-there-a-si...
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/713467/prove-that-e-...
http://math.stackexchange.com/questions/425963/is-there-a-si...
Does someone have a link to the notation to decipher this?
You'll also need to know that e^x = sum{0 -> infinity} x^n / n!.
Which particular notation? Maybe Factorial [0] and Summation [1]?
[0] https://en.wikipedia.org/wiki/Factorial [1] https://en.wikipedia.org/wiki/Summation
[0] https://en.wikipedia.org/wiki/Factorial [1] https://en.wikipedia.org/wiki/Summation
Are the big braces in the second equation just supposed to be parentheses?
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.
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.
Yeah
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.
He writes 'but the left side is an integer'....
IMO the left side of the last equation is equal 0....
Any comments?
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.)
[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.)
Could you explain why the left hand side must be zero?
I had something in mind like, assume that the last line is not a contradiction, then we have c( e- a) on the left and something smaller than one on the right. However, assume the last line is not a contradiction is false, so I can't build the argument atop of it.
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.