10 ms·
Elementary proof that e is irrational
- cjjuice 10y agoDoes someone have a link to the notation to decipher this?
- monochromatic 10y agohttps://en.wikipedia.org/wiki/Summation#Capital-sigma_notation https://en.wikipedia.org/wiki/Summation#Capital-sigma_notati...
- widdma 10y agoWhich 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 10y agoAre the big braces in the second equation just supposed to be parentheses?
- zodiac 10y agoYeah
- Someone 10y agoAs 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.
- jeffcoat 10y agoYou'll also need to know that e^x = sum{0 -> infinity} x^n / n!.
- davekeck 10y agoBut 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 10y agoIf 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 10y agoBut 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).
- siegelzero 10y agoIt's a contradiction because there are no integers strictly between 0 and 1.
- nilkn 10y agoIf 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].
- vadiml 10y agoHe writes 'but the left side is an integer'.... IMO the left side of the last equation is equal 0.... Any comments?
- widdma 10y agoZero is an integer. The point is the right side is not. Since it's strictly greater than zero and less than 1.
- SidiousL 10y agoNo, it's not zero. The sum is up to `a`, not up to infinity.
- chx 10y agoErm, nope, it would be if the sum went to inf but it doesn't.
- yk 10y agoIt 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.)
- andars 10y agoCould you explain why the left hand side must be zero?
- yk 10y agoI 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.
- digler999 10y agovery 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.
- andars 10y agoThis 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!"
- zodiac 10y agoIt's a general fact that any series satisfying the list of conditions in http://calculus.subwiki.org/wiki/Alternating_series_theorem 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
- sweetjesus 10y agowhen 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.
- airesQ 10y agoFor 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 10y agoLooks 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 10y agoI 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.
- mvf0 10y agoThank you for confirming this. These kind of errors drive me crazy when I try to work through proofs on my own.
- mrcactu5 10y agothe "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-irrationality-of-en http://math.stackexchange.com/questions/476939/proving-the-i... http://math.stackexchange.com/questions/713467/prove-that-e-is-irrational http://math.stackexchange.com/questions/713467/prove-that-e-... http://math.stackexchange.com/questions/425963/is-there-a-simple-proof-that-e2-is-irrational-using-a-positional-numeral-syst http://math.stackexchange.com/questions/425963/is-there-a-si...
- leephillips 10y agoThis 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.)
- posterboy 10y agoAm 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.