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Looking for Easy proof of Pi Irrational

  1. Nov 2, 2003 #1


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    Looking for "Easy" proof of Pi Irrational

    Hi, I just got to this forum after searching for an easy proof that Pi is irrational. The thread I found (google) was this one HERE. I wanted to reply, but since it is now “archived” I thought it would be better to post a new thread. Sorry if this topic has already been done to death or anything.

    I’ve been trying to come up with a simpler proof (of Pi irrational) than the ones I’ve been able to find so far on the net. Most of the proofs I’ve found are similar to the following LINK which I can follow but find unappealing (as in I would have trouble repeating as the steps are not intuitive for me).

    I’ve just come up with (what I think is) a proof that I like a lot better. If it is correct then it has almost certainly been done before. Basically I just want to know if this would be considered a valid proof (or at least if it could be massaged into one).

    I was trying to get a proof of Pi^2 irrational out of the series 1 + 1/4 + 1/9 + ... 1/n^2 + ... = Pi^2 / 6. I know that this cant be a self contained proof unless it includes a proof of the above series limit, but I thought that if I used the series as a starting point then it might lead to a fairly easy proof from that point on.

    The idea I was working on was that if I tried to write a (PI^2) / 6 = b then somehow there would be too many prime factors in the Pi^2/6 (series) denominator for integer a to be bounded. At first I couldn’t make it work but then I remembered that Sum{n=1..Infinity : 1/n^2} = Product{k=1..Infinity : (p_k)^2 / ((p_k)^2 - 1)}, where p_k is the kth prime number, and that seemed to give a pretty straight forward proof. So here it is.

    Assume Pi^2 / 6 = a / b for some finite integers a and b

    Then a = b ( 1 + 1/4 + 1/9 + 1/16 + ... )

    or a = b 2^2 / (2^2 - 1) 3^2 / (3^2 - 1) 5^2 / (5^2 - 1) ..... (p_k^2 / (p_k^2 - 1)) ...

    But since p^2 - 1 = (p-1) (p+1), both of which are even factors for primes p > 2, then it follows that none of the denominator product terms, up to and including the (p_k^2 - 1) term, can cancel either of the two p_k prime factors in the numerator (for p_k > 2). It may be the case that the denominators of further product terms do eventually have prime factors which cancel this double p_k numerator prime factor, but these further product terms will themselves introduce yet larger numerator prime factors and so on. So the largest prime factor of a is unbounded and therefore a is unbounded.

    Is this a valid proof? And if not then can it be fixed up?

    Thanks :)
    Last edited: Nov 2, 2003
  2. jcsd
  3. Nov 2, 2003 #2


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    I think your proof is fatally flawed. Fractions and topology don't mix; there's no reason to think that the limit of a sequence must be irrational just because the numerators and denominators of the sequence grow without bound.

    e.g. consider the sequence


    Which converges to 1/1
  4. Nov 3, 2003 #3


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    Thanks for the feedback Hurkyl. The proof may have other flaws but I'm pretty sure that it’s ok to contradict the assumption of rationality by proving that the integers a and b necessarily become unbounded (as in become unbounded even in their simpilist multually prime form). Perhaps I should have stipulated at the very start of the proof that a and b are mutually prime (no common factors). This can be done without any real loss of generality.

    You see if you let a and b increase without bound then you really can have any number, irrational or rational (ie Infinity / Infinity = anything to put it very loosely).
    Last edited: Nov 3, 2003
  5. Nov 3, 2003 #4


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    Did you not understand Hurkyl's point?

    The sequence of fractions 9/10, 99/100, 999/1000, ...

    is precisely a sequence of fractions in which the numerator and denominator are MUTUALLY PRIME, cannot be reduced, and both grow without bound. That seems to be exactly what you are requiring. Yet, that sequence converges to 1, a remarkably rational number!
  6. Nov 3, 2003 #5


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    The most elementary proof I know of the irrationality of pi is

    y= cos(x) is positive on the open interval from 0 to pi and continuous for all x (so, especially on the closed interval from 0 to pi). All "anti-derivatives" of cos(x) can be taken to be cos(x), -cos(x), sin(x), or -sin(x) by appropriate choice of "constant of integration. These all have values of 0, 1, or -1 at both 0 and pi. Therefore pi is irrational!

    Of course, that requires that one know the theorem:

    Let c be a positive real number. If there exist a function, f, positive on the open interval from 0 to c, continuous on the closed interval from 0 to c, and such that all its anti-derivatives can be taken to be integer valued at 0 and c (by appropriate choice of the "constant of integration") then c is irrational.

    The proof of that is elementary but long.

    You can prove the irrationality of pi using a power series for arctan(x) but I don't recall the details right now.
  7. Nov 3, 2003 #6


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    Whoops actually I mis-read it :eek:. I read it too quickly and thought Hurkyl was saying 9/10, 90/100, 900/1000 etc. My mistake, I should have read the reply more carefully.

    Actually the geometric series like the counter-example suggested was one of the things that had me a little unsure when I was trying to figure out if a “proof" along these lines could work at all, especially when I was looking at the infinite sum form of Pi^2/6. The way the geometric series keeps adding new prime factors (actually only new powers of existing prime factors for the denominator) but in the limit they all disappear leaving only 1/1 (in this example).

    Note that in this "counter-example" that all the denominator prime factors are 10 and in the limit they effectively disappear, more or less because 0.9999' and 1 are the same number.

    I’m still not sure that the counter example applies to the case of the infinite product form of Pi^2/6, where the new primes factors are multiplicative and are not just new powers of existing prime factors as in the geometric series counter-example.

    Ok thanks for that HallsofIvy, it looks very interesting. Now I've got to go search for info on that theorem you mention and see if I can understand it. Can you recall any name associated with it to help me search ?
  8. Nov 3, 2003 #7


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    How about


    Each term has a brand new prime in the denominator, yet still converges to 1!

    Your approach of looking at this product may or may not be a good one, but your "goal" is almost certainly not going to be helpful. Infinite products are immune to the fundamental theorem of arithmetic.
  9. Nov 3, 2003 #8
    actually, pi is rational. pi is the ratio of circumference to diameter in any circle. ;)
  10. Nov 4, 2003 #9


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    I was about to say something scathing about phoenixthoth and then I realized "Oh, he's trying to be funny!" Poor guy.
  11. Nov 5, 2003 #10


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    I wish I remembered the author of the proof I mentioned so I could give him credit. It was in the Mathematics Monthly, published by the Mathematical Association of America, eight to 10 years ago.

    It's a lovely proof- very unexpected but elementary. In addition it is a proof by contradiction- the "worst" kind: deny the conclusion and use that to prove "statement A" that doesn't seem to have any direct connection with either conclusion or hypotheses, then prove "statement B" which also doesn't seem to have any direct connection with either conclusion or hypotheses but contradicts "statement A"!

    (This is my memory. Any errors are my fault, not the original author's!)

    Theorem: Let c be a positive real number. If there exist a function f, continuous on the closed interval [0, c], positive on the open interval (0, c), and such that f and its anti-derivatives can be taken to be integer valued at 0 and c (we can always choose the "constant of integration" so that the value at 0, say, is any given number- here we are assuming we can always choose it to make the value of the anti-derivatives integers at both 0 and c), then c is irrational.

    First define P to be the set of polynomials, p(x), such that p and all of its derivatives are integer valued at 0 and c. Notice we are requiring derivatives to be integer valued here while f has anti-derivatives integer valued. This way we can prove
    Lemma 1: ∫0cf(x)p(x)dx is an integer for f(x) as in the theorem and p any polynomial in P.
    Do the integral by repeated integration by parts, repeatedly integrating the f "part" and differentiating the p "part". Since p is a polynomial that will eventually terminate, giving a sum of derivatives of p times anti-derivatives of f. By hypothesis, those are all integer valued.
    Very similar is
    Lemma 2: The set P is closed under multiplication.
    If p and q are in P they are polynomials and, of course, their product pq is also a polynomial. pq(0)= p(0)q(0), pq(c)= p(c)q(c), products of integers. (pq)'= p'q+ pq' by the product rule so (pq)' evaluated at 0 and c is a sum of products of integers. All derivatives can be done by repeated application of the product rule- every derivative is a sum of products of derivatives of p and derivatives of q- all integer valued at 0 and c.

    We want to prove c is irrational so start by assuming c is rational: c= m/n where m and n are integers.
    Let p0= m- 2nx. p0 is a (first order) polynomial. p0(0)= m, an integer. p0(c)= m- 2n(m/n)= -m, an integer. p0' is the constant -2n, an integer, and all succeeding derivatives are 0. p0 is in P.

    For i any positive integer, define pi= (mx- nx2)i/i!. We claim that pi is in P for all i.
    Proof by induction: p1= mx- nx2= x(m- nx).
    p1(0)= 0 and p1(c)= 0 (since m- n(m/n)= 0).
    p1'= m- 2nx= p0. Since we already know p0 is in P, all of its derivatives are integer valued at 0 and c and, thus, so are all succeeding derivatives of p1. p1 is in P.
    Now assume pN is in P and look at pN+1.
    pN+1= (mx- nx2)N+1)/(N+1)!= xN+1(m- nx)N+1/(N+1)!. pN+1(0)= 0 because of the xN+1 factor which is 0 at x=0. pN+1(c)= 0 because of the (m- nx)N+1 factor which is 0 at x= c= m/n.
    By the chain rule, pN+1'= (N(mx- nx2)N/(N+1)!)(m- 2nx)= pN(x)p0(x). Those are both in P and, since that set is closed under multiplication, so is their product. That means that pN+1' and all of its derivatives are integer valued at 0 and c and, thus, so are all succceeding derivatives of pN+1.
    pN+1 is in P and we are done.

    By hypothesis, f is continuous on [0,c] and so attains a maximum on that set: let M be the maximum value of f on [0,c]. That, is f<= M on [0,c]
    For all i, pi is a polynomial and not only is continuous but is differentiable for all x (and so on [0,c]). pi not only has a maximum on [0,c] but must achieve that maximum either at an endpoint or in the interior where pi'= 0. We already know that pi(0)= pi(c)= 0.
    The derivative of pi is pi-1p0. pi-1 is 0 only at 0 and c so in order for pi' to be 0 we must have p0(x)= m- 2nx= 0 or x= m/(2n)= c/2.
    At x= m/(2n), pi(m/2n)= (m(m/2n)- n(m/2n)2)i/i!= (m/4n)i/i!. Replacing both f and pi by their maxima, &int;f(x)pi(x)dx<= M(m/4n)i/i! &int;0cdx= (Mc)(m/4n)i/i!. That is, of course, independent of x but depends on i. It is a constant to the i power over i! and, as i goes to infinity, goes to 0. Thus we arrive, finally, at

    Statement A: For some (sufficiently large) i, &int;0cf(x)pidx< 1.
    (Since it goes to 0, we can make it less than any positive number by choosing sufficiently large i. 1 is convenient.)

    You may be able to see the rest, now.

    By hypothesis, f is positive on (0,c). For all i, pi is positive at c/2 and 0 only at 0 and c. pi is positive on (0,c). Therefore, for all i, &int;0cf(x)pidx is a positive integer (remember lemma 1?) or

    Statement B: For all i, &int;0cf(x)pidx>= 1.

    Since Statement A and Statement B are contradictory, we are done.
    Last edited: Nov 5, 2003
  12. Nov 5, 2003 #11


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    [I edited this: obviously cos(x) is NOT positive on (0,&pi;)!]

    Corollary to Theorem: &pi; is a positive real number. sin(x) is continuous for all x and so on [0,&pi;]. It is positive on (0,&pi;).
    All anti-derivatives of sin(x) can be taken, by choosing the constant of integration to always be 0, to be cos(x), -cos(x), sin(x), or -sin(x), all of which have values of 0, 1, or -1 at x=0 and x= &pi;. Therefore &pi; is irrational.

    Wow, that was easy!
    Last edited: Nov 7, 2003
  13. Nov 5, 2003 #12


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    Yet another corollary:

    Let p be a positive real number, NOT equal to 1. If ln(p) is rational, then p itself is irrational.

    First, since 1/p is rational if and only if p is, and ln(1/p)= -ln(p) is rational if and only if ln(p) is, it is sufficient to prove this for p>1 (if p<1, apply the theorem to 1/p).

    Again, a proof by contradiction: assume p is rational- p= m/n.
    Since p>1, ln(p)> 0. Let c= ln(p) and apply the theorem with
    f(x)= nex.
    Of course, f is continuous and positive for all x and so on [0,c] and (0,c). We can take all anti-derivatives of f to be nex by taking the constant of integration always to be 0.
    f(0)= ne0= n, an integer. f(c)= f(ln(m/n))= n eln(m/n)= n(m/n)= m, an integer. Thus, by the theorem, ln(p) is irrational which contradicts the hypothesis.

    Corollary to the corollary:

    e is a positive real number, not equal to 1. ln(e)= 1, a rational number. Therefore, e is irrational.
  14. Nov 13, 2003 #13


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    Thanks very much for taking the time to post that proof HallsofIvy, it was very interesting - I'll have to read it a few times to take it all in, but it looks convincing.
  15. Nov 15, 2003 #14


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    uart: If you are still checking on this thread, here's a method I just saw on another forum. The poster was asking about it- I suspect it was "analysis class" homework.

    Define In to be q^(2n)/n! &int;-pi/2pi/2(((pi^2)/4-x^2)^n)*cos(x) dx.

    (a) Show that I0= 2 and I1=2q

    (b) Show that I_n = (4n -2)*q^2*I_n-1 - p^2*q^2*I_n-2 for n>= 2 where we are assuming pi = p/q (that is, that pi is rational)

    You can do this by induction on n, using integration by parts twice in the induction step.

    (c) Deduce that I_n is an integer
    Once you have the formula in (a) use induction on n.

    (2) show that 0 < I_n < (p/q)*(p/2)^(2n) /n! n = 0, 1, 2, ....
    Obviously (pi^2/4- x^2)<= pi^2/4 and cos(x)<= 1. Put those upper bounds into the integral.

    (e) If (p/q)*(p/2)^(2n) /n! < 1, deduce that pi is irrational.

    Since that can't be an integer of course.
  16. Nov 26, 2003 #15


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    Ok, that's an interesting proof. I still have to convince myself of step b (I made a stuff up the first time and will have to look at it again) but the rest seems very straight foward.

    BTW, I think [tex]I_1 = 2q[/tex] is a typo in step a, it should of course be [tex]I_1 = 4 q^2[/tex]

    Now the thing that really interests me is step e, the final step. Here the proof is essentually saying that if we can find an interger, n, for which [tex]0 < \frac{p}{q}*(\frac{p}{2})^{(2n)} /n! < 1 [/tex] then the said term cannot be an integer and hence we have contradiction.

    But, doesn't that require p to be bounded, exactly the same as in my attempted proof at the start of this thread ?. For if p is not assummed to be bounded then we cannot claim that [tex](p/q)*(p/2)^{(2n)} /n! < 1 [/tex] for some n.
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