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Please clear my doubt

  1. Jan 5, 2006 #1
    Hey guys, how do you prove that 'pi' is irrational? I think that it is related to infinite series? Is there any geometrical method?
     
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  3. Jan 5, 2006 #2

    Tx

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    The series for Pi is Gregory's Series to proove the irrationality is quite difficult, just google it.
     
  4. Jan 5, 2006 #3

    HallsofIvy

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    Here's my favorite proof:

    Lemma 1: Let c be a positive real number. If there exist a function, f, continuous on [0,c] and positive on (0,c) and such that f and its iterated anti-derivatives can be taken to be integer valued at both 0 and c, the c is irrational.
    ("can be taken to be"- we can always choose the constant of integration such that we have any given value at either 0 or c. Here we require that it the function be integer valued at BOTH 0 and c. I will post the proof separately. I remember seeing it in "Mathematics Magazine" many years ago but don't remember the author.)

    Theorem [itex]\pi[/itex] is irrational:
    f(x)= sin(x) is continuous for all x and positive on (0, [itex]\pi[/itex]). All anti-derivatives can be taken to be [itex]\pm sin(x)[/itex] or [itex]\pm cos(x)[/itex], all of which are integer valued (0 or [itex]\pm 1[/itex]) at [itex]x= \pi[/itex]. Therefore, by lemma 1, [itex]\pi[/itex] is irrational.

    One can also use lemma 1 to prove that e is irrational.

    Lemma 2: If a is a positive real number, not equal to 1, such that ln(a) is rational, then a itself is irrational.
    Proof: First note that ln(1/a)= -ln(a) is rational if and only if ln(a) is and 1/a is rational if and only if a is rational so it is sufficient to prove this for a> 1. (If a<1, apply the lemma to 1/a.)
    If a>1 then ln(a)> 0. Suppose that ln(a) is rational and, contradicting the hypothesis, that a is rational: a= m/n reduced to lowest terms. Apply lemma 1 with c= ln(a)= ln(m/n) and f(x)= nex. Then f(x) is positive and continuous for all x and so for the required intervals. We can take ALL anti-derivatives to be f(x)=nex by taking the constant of integration to be 0. f(0)= n, an integer, and f(c)= f(ln(m/n)= neln(m/n)= m, an integer. Therefore, by lemma 1, a is rational, a contradiction.

    Now: Theorem: e is irrational.
    e is a positive real number, not equal to 1 (since 1= e0 and ex is 1 to 1). ln(e)= 1, a rational number. Therefore, by lemma 2, e is irrational.
     
  5. Jan 5, 2006 #4

    HallsofIvy

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    Here is the proof of "lemma 1" above. As I said there, I remember reading it in "Mathematics Magazine" many years ago but don't remember the author. It is certainly not original with me! I find it intriguing for two reasons- first, irrationality is a numeric, not function, property, yet this depends upon calculus methods. Second, it is the "worst" kind of proof by contradiction! Contradicting the conclusion (that c is irrational) leads to two conclusions (I call them "statement A" and "statement B" below) neither of which seems to have much to do with irrationality but which contradict one another.
    Lemma i: let c be a positive real number. If there exist a function f, continuous on [0,c] and positive on (0,c), such that f and all of 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.
    Proof by contradiction:
    First: define the set P of all polynomials, p(x), such that p and all of its derivatives are integer valued at 0 and c.
    Notice "derivatives" rather than "anti-derivatives". That allows us to prove:
    Lemma i: if f(x) is the function above and p(x) is any polynomial in P, then [itex]\int_0^c f(x)p(x)dx[/itex] is an integer.
    To prove this, use repeated integeration by parts, repeatedly integrating the f "part" and differentiating the p "part". Since p is a polynomial and differentiating reduces the degree, that will eventually terminate giving the integral as a sum of anti-derivatives of f times derivatives of p, all of which are integer valued at 0 and c.
    A similar proof gives
    Lemma ii: the set P is closed under multiplication.
    Suppose p and q are both in P. The pq is a polynomial and pq(0)= p(0)q(0) and pq(c)= p(c)q(c) are products of integers. All derivatives of pq can be done by repeated application of the product rule: every derivative is a sum of products of various derivatives of p times derivatives of q- all integer valued at 0 and c.

    Now, suppose c is rational: c= m/n reduced to lowest terms. Let p0(x)= m- 2nx. Clearly p is a polynomial. p(0)= m, an integer and p(c)= p(m/n)= m- 2n(m/n)= -m, an integer. p'(x)= -2n, an integer for all x, and all subsequent derivatives are 0. Therefore, p0 is in P.

    For i any positive integer, let [itex]p_i(x)= \frac{(mx- nx^2)^i}{i!}[/itex]. We will prove, by induction, that pi is in P for all i.,
    If i= 1, p1(x)= mx- nx^2= x(m- nx). p1(0)= 0 because of that 'x' factor and p1(c)= p1(m/n)= 0 because of the 'm-nx' factor. p'1= m- 2nx= p0(x). Since that is in P we have immediately that all derivatives of p1 are integer valued at p and so p1 is in P.

    Assume that pi is in P for some i. [itex]p_{i+1}= \frac{(mx-nx^2)^{i+1}}{(i+1)!}= \frac{x^{i+1}(m-nx)^{i+1}}{(i+1)!}[/itex] is 0 at both x=0 and x= c= m/n because of the factors. [itex]p'(x)= \frac{i(mx-nx^2)^i(m-2nx)}{(i+1)!}= \frac{(mx-nx^2)^i}{i!}(m- 2nx)= p_i(x)p_0(x)[/itex]. Since both pi and p0 are in P and P is closed under multiplication, so is p'i+1- all further derivatives are integer valued at 0 and c and so pi+1 is in P.
    Since f(x) is continuous on [0,c], it takes on a maximum value there: let M= max f(x) on [0,c]. Further, since f is positive on (0, c), M> 0. Since, for all i, pi is differentiable on [0, c] it not only takes on a maximum value but that maximum value occurs either at an endpoint (0 or c) or in the interior where p'i(x)= 0. We have already seen that pi is 0 at 0 and c, for all i, (and only at x= 0 or c) and that p'i(x)= pi-1(x)p0(x). p'i(x)= 0 only where p0(x)= m- 2nx= 0 or x= m/2n= c/2. pi(m/2n)= \frac{\left(\frac{m^2}{4n}\right)^i}{i!}. Since that is positive, that is the maximum value of pi on [0, c].

    Now we can prove:
    [tex]\int_0^c f(x)p_i(x)dx\le \int_0^cM\frac{\left(\frac{m^2}{4n}\right)^i}{i!}dx= Mc\frac{\left(\frac{m^2}{4n}\right)^i}{i!}[/tex]
    That is a constant times a constant to the i power, divided by i!. As i goes to infinity, the factorial dominates and the limit is 0- we can make that as small as we please:
    Therefore, Statement A:
    For some i, [itex]\int_0^c f(x)p_i(x)dx< \frac{1}{2}[/itex].
    But since f(x) and all pi(x) are positive on (0, c) (every pi is 0 only at 0 and c and is positive at c/2),
    [itex]\int_0^cf(x)p_i(x)dx[/itex] is a positive integer and so (statement B) must be larger than or equal to 1 for all i. That contradicts statement A and so the theorem is true.
     
    Last edited: Jan 10, 2006
  6. Jan 5, 2006 #5

    Tx

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    Thanks for posting that, I've never seen a proof for 'lemma 1'.
     
  7. Jan 10, 2006 #6
    Hey thanks For that!
     
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