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A Simple Proof That Pi Is Irrational
- exacube 11y agoI feel really bad when I read stuff like this and don't understand, especially when it's titled "Simple" (granted I didn't spend more than 5 minutes trying to following the proof)
- mlitchard 11y agoWhat is it you don't understand?
- jordigh 11y agoUsually people who don't understand stuff like this are unable to even begin to explain what they don't understand. I don't really understand people like this. I have tried to guess many times what the problem is. I think they may be too embarrassed to admit they don't know what an integral or factorial sign means, or perhaps something like why f is a polynomial, or what a polynomial is.
- madprof 11y agoBut once they get all that they get stuck on what a Taylor series is, what the integrals of trigonometric functions are or what the squeeze rule is.
- mauricioc 11y agoThis is not directed at you specifically (I get your point), but here's an attempt at an explanation for those parts: 1) Verifying that f^(j) = 0 is 0 for all j doesn't require Taylor series (though, as 'dnautics pointed out, it does motivate the construction): 1.1) f^(j)(0) is zero for 0 <= j < n because every term of the polynomial f has degree at least n (and therefore you won't get a nonzero constant coefficient if you derive fewer than n times). 1.2) f^(j)(0) is zero for j >= n because once you derive n times you will get a factor of n! in each term, thus cancelling the only source of "non-integerness". 2) Point taken, you have to know how to differentiate a product and what the derivative of the sine and cosine functions is. With this in mind, checking the equation before equation (1) is routine, though. You then apply the fundamental theorem of calculus to get equation (1). 3) He is not applying the squeeze rule here, as it would not produce a contradiction. The squeeze rule would say that the limit of f(x) sin(x) (there's an implicit dependence on n here) is zero, which would say that F(pi) - F(0) is zero, which is not a contradiction. The argument requires less machinery: For large enough n (not in the limit), f(x) sin(x) is strictly between zero and 1 (and is thus not an integer), because pi^n * a^n / n! is smaller than 1 if n is large enough. The simplest way of explaining this is that n! >= (n/2)^(n/2), as there are n/2 terms each larger than n/2 in the definition of n!. Thus the expression is at most ((pi a)^2/(n/2))^(n/2), and thus taking taking any n such that n/2 >= (pi a)^2 works for making the right side less than 1. (If you know the definition of the Euler constant e as a series, you'll see that (pi a)^n / n! appears in the expansion of e^(pi a). Since the series converges, this means that (pi a)^n / n! is less than 1 if n is large enough. But this requires more previous knowledge.)
- hasenj 11y agoI don't understand calculus and have no problem admitting to it. I hated calculus throughout highschool and college.
- jameshart 11y agoI'm not the OP, but your attitude here is unnecessarily dismissive. Not everybody has the calculus skills to understand this, let alone an awareness of mid-twentieth century calculus notation conventions. For my part, I do know what polynomials are, what a Taylor series is, and so on, and in theory I can trace through the steps here and agree that yes, one follows from the other. Yet I find this proof unsatisfying because it doesn't demonstrate clearly, to me, which particular properties of pi it is using that bring about the contradiction. When the proof makes use of pi, it doesn't explain why the statements it is making are specifically true for pi, and not for some other number. Take this same proof, substitute the number 3 for every occurrence of pi. Now pinpoint for me the place in the proof where it is clear and obvious that the proof makes an invalid claim about the number three (but where for pi, it was clearly and obviously correct). If you can't find it, then this proof structure equally serves as a convincing argument that three is irrational. That's quite unsatisfying - though of course, a proof doesn't have to be convincing, it just has to be right. Nevertheless, to qualify as a 'simple' proof, I think it does have to appeal to intuitions and concepts in a way that simply convinces you of its truth. This proof is short, but it is not simple. It doesn't satisfy because it doesn't show me how the ratio of a circumference to a diameter has to be irrational - only how a number called pi which has particular relationships (not specified clearly in the proof) to the sin and cos functions, has to be irrational.
- jordigh 11y agoI didn't mean my attitude to be dismissive. It's just that truly a lot of people respond this way. "I don't understand it at all." And when you ask what don't they understand, they're unable to say it. Witness for example how exacube seems to have vanished and will probably never tell us what she or he did not understand. I think what happens is that people are so overwhelmed with unfamiliar ideas when they encounter a proof like this that they just grind to a halt, curl up into a ball, and scream how much they hate it all and don't understand a bit of it. We have at least a couple of other people in this thread who have expressed their hatred of calculus. Starting from that it seems pretty hopeless to try to explain to them this proof. Yet I find this proof unsatisfying because it doesn't demonstrate clearly, to me, which particular properties of pi it is using that bring about the contradiction. Only one: that it's a root of sin(x). The proof actually works for any nonzero root of sin(x). In fact, that's a great definition of pi: the least positive root of sin. It's a much easier definition to work with than ratio of circumference to diameter (how do you define cirumference? What is length? What is a curve?)
- wehadfun 11y agoLets start with those polynomials, then those integral coefficients.
- dnautics 11y agoIt would have been nice for the author to explain the motivation for this construction. Note that the polynomials are the construction for the Taylor series of exp(ix) around the point (a/b), and remember Euler's magic formula.
- umanwizard 11y agoWell, you have to spend as long as it takes to understand, and you have to understand calculus, and probably have at least a non-zero amount of practice following proofs from a university level or strong high school level math course. It's "simple" relative to the rest of mathematics, not relative to daily life. The point is -- don't feel bad; math doesn't come easily to anyone :)
- joe_the_user 11y agoThe one other thing about this proof compared to the exposition of upper division college math is that it doesn't follow an easy linear path but is instead using the theories and tools in an ad-hoc manner. Getting an idea of theories and proof methods as a toolbox rather than just a linear progression of ideas is one of the milestones of "mathematical sophistication" - especially because it's important to not lose sight of the linear development of the field as well.
- davvolun 11y agoNote: Not a mathematician, by any means I'm not a fan of how the proof is explained, specifically why are we doing x or y. I would prefer this -- Let pi be a rational number thus ( * by the definition of rational numbers) pi = a/b, the quotient of positive integers. We will show no such a and b can exist, therefore pi cannot be rational. ( * is not completely necessary, since the definition of rationals is so elementary) <Next Paragraph, and so on> I suspect a lot of mathematicians prefer the format given because it is more obtuse...
- jordigh 11y ago> I suspect a lot of mathematicians prefer the format given because it is more obtuse... No, because as you say, defining what a rational number is seems pretty pointless here, as why would you be reading the proof of something whose definition you don't even know? And you also want an explanation of what a proof by contradiction is, which also seems to be way too elementary. Proof by contradiction is one of the most basic techniques. Spivak's version of this proof explains it a bit more, but still requires work from the reader. Any proof does. Mathematics cannot be a spectator sport.
- Natsu 11y agoA lot of people struggle at math because they're missing things like that. With today's technology, there's no reason things can't be broken down to arbitrarily small steps, so that people can be filled in on the cracks in the foundations of their understanding of math. The usual reason for not including everything is that it becomes ridiculously laborious. With mathematics being abstraction piled atop abstraction, this is quite reasonable. Just look at how much work it is to prove 2+2=4 -- http://us.metamath.org/mpegif/mmset.html#trivia http://us.metamath.org/mpegif/mmset.html#trivia But, as you can see from the above, there's technology to simplify things. I wonder if someday we'll be able to break things down in a friendly way so that for any piece a student doesn't understand, they can get a proof in terms of things they do understand?
- chx 11y agoErm, the linked metamath page asks what's the longest path one can take. Well, geez, it's long. Proving that 2+2=4 from the usual Peano axioms isn't that hard or long. We define addition by a+0=a,a+S(b)=S(a+b). Now what we mean by a positive integer number is just a shorthand for that number of successive S(). So 2+2=4 is just S(S(0)) + S(S(0)) = S(S(S(S(0)))). To prove this, let's apply the second part of the addition to S(S(0)) + S(S(0)) we get S(S(S(0)) + S(0)). Let's repeat this "move the S() from the second to the outside" and you will get S(S(S(S(0)) + 0)). Now we can use our first half of definition where +0 can be left out: S(S(S(S(0)))). Now, that wasn't that long, was it?
- hcarvalhoalves 11y ago"Simple" in the sense it's minimal, not as in "easy to grasp".
- jackmaney 11y agoI started reading it, and then there was a nagging interstitial, begging for my email address so that I could be sent spam. Closing that garbage, various buttons and unnecessary margin bits ("Click here to see more!" [bounce] [bounce] [bounce]...) kept jittering for my attention. So, I closed the tab. I have no idea who is responsible for the design of fermatslibrary.com, but they should feel ashamed of what they've done. This is one of the most infuriatingly infantile designs I've ever seen. The old "punch the monkey!" ads have nothing on this.
- selestify 11y agoWeird, I didn't experience any of that when I visited it.
- jackmaney 11y agoI don't know what to tell you, except that when I viewed the page, there was a reasonably well-formatted PDF in the center of the screen surrounded by unnecessary, fiddly, bouncing bullshit "click on me!" buttons.
- mgdo 11y agoI think these are simply the paper annotations. It's not that obtrusive.
- jackmaney 11y agoThere are sites that allow for annotations and manage to do so without bouncing, distracting "click on me!" buttons. Medium does a great job of this. I have no idea what the site's creator was thinking when they decided not to emulate the established, non-distracting examples instead of doing what they did.
- deleted 11y ago[deleted]
- GnarfGnarf 11y ago
- selestify 11y agoFinally, no more "The proof is left as an exercise for the reader."
- itistoday2 11y agoCommenting requires signing up for Facebook or G+? That's a shame, would have liked to use this site.
- phkahler 11y agoThis doesn't work. The term (a-bx) is zero since bx = a. That turns the whole equation into a big fat zero.
- omaranto 11y ago(a-bx) is not zero, x is a variable. when x=pi, a-bx is zero, otherwise not. a-bx=0 is an equation whose solution is x=pi. The function f(x) is not identically zero, it is however zero when evaluated at x=pi: f(pi)=0.
- phkahler 11y agoAnd he does talk about f(pi) quite a bit. It's not like there are limits being discussed where we should consider something in the neighborhood of a 0/0 situation. He evaluates it at pi, and in doing so could just put a zero in there.
- gizmo686 11y agoa-bx is only 0 when x=π.
- pierrebai 11y agoBut nowhere does he proves that all derivatives of f(x) are not zero for x=0 and x=pi. f(x) clearly is zero there. So saying that F(pi) + F(0) is an integer proves nothing if both are equal to zero...?
- gizmo686 11y agoYou don't actually need F(pi) + F(0) to be non-zero. The proof shows that the integral is positive, but arbitrarily small. The means that F(pi)+F(0) must be positive, but arbitrarily small. If F(pi)+F(0) is a positive integer, it fails the arbitrarily small test. If F(pi)+F(0) is a non-positive integer (including 0), it fails the positive test.
- asgard1024 11y ago
- jordigh 11y agoOh, I get it. "Fermat's library," because we're writing on the margin, which, for once, is large enough to contain our marginalia. Cute. Btw, this is the same proof that is in Spivak's Calculus, but Niven explains it a little less than Spivak does.
- semi-extrinsic 11y agoSince the linked proof was published in a mathematical journal [Bull. Amer. Math. Soc. Volume 53, Number 6 (1947), 509], I think it's safe to assume that Niven was first.
- tzs 11y agoIf the annotations are not working right when you expand them, particularly the second one, and you are using Safari, try reading in Chrome or Firefox. Those appear to work fine. In Safari, it looks like when the height of the expanded annotation display region is calculated for dealing with scrolling, it is including the area under the sign up doohickey. It thinks you can see more than you can. This means the scrolling limits are set too small, and so you cannot scroll the bottom of the annotation up into view. If your window size and text size make it so that the second annotation fits in the visible region, this won't matter.
- thams 11y agoLove this, want more papers though, there are a few computer science papers that I'm going to suggest
- hebdo 11y agoNeat! I remember that presenting (from memory) a slightly longer proof that pi is irrational was the final question on my Analysis oral exam. It is surprising how many people know that pi is not rational, yet how non obvious is the proof. Like the first one ever presented. This one is also a nice piece of cake, pulls this polynomial f(x) from a hat.
- hasenj 11y agoSomehow, the "original proof" seems simpler. > The first rigorous proof that π is irrational is from Johann Heinrich Lambert in 1761. He proved that if x≠0 is rational, then tan(x) must be irrational. Since tan(π/4)=1 is rational, then π must be irrational. Of course, the proof for (tan(x) is irrational for rational x != 0) might be complicated, but at least it's easy to see why this proves that PI is irrational. This one, on the other hand, I can't even follow in principle what it's trying to say. That's of course because I don't understand integrals and differentials.
- jordigh 11y agoThat just wraps up a mystery inside another. Why would tan(x) be irrational if x is rational? That seems like an even bigger mystery than why π is irrational.
- dnautics 11y agoThats easy if you remember arctan(y/x) is the angle from x axis to (x,y)
- hebdo 11y agoI fail to see how this helps. Actual proof is more complicated, it is based on continued fractions.
- hasenj 11y agoIt's all mysteries. The point of "simple" is to create a series of simple steps, each step should be straight forward and easy to understand. Terse != simple
- hebdo 11y ago> Of course, the proof for (tan(x) is irrational for rational x != 0) might be complicated It is: http://www.pi314.net/eng/lambert.php http://www.pi314.net/eng/lambert.php
- pdpi 11y agoIt's worth noting that there's a number of annotations added that make the proof much easier to follow for someone whose calculus is not quite as... hum... fresh as the proof expects.
- izzydata 11y agoEvery time I tried studying from a math textbook back in school I got the impression that examples and explanations were purposefully obscure to make it look more sophisticated. Then a friend comes along and explains it in 10 seconds and it makes perfect sense.
- amelius 11y agoIt would be interesting to see this as a formal proof, and to compare the complexity and lengths of both approaches.
- throwaway1967 11y agoSocrates (in Meno) helped a slave kid prove to himself that the square root of 2 is irrational, using nothing but a stick and the sand on the ground. I don't think that kid would understand this "simple proof".
- Houshalter 11y agoThis isn't simple at all. Is there an explanation anywhere of this proof?
- wolfgke 11y agoNow I wish a simple proof that pi is transcendental (i.e. not algebraic).