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We know e (exponential) is a irrational number

  1. Apr 2, 2003 #1
    we know e (exponential) is a irrational number....
    how can we prove it??
     
    Last edited by a moderator: Feb 4, 2013
  2. jcsd
  3. Apr 2, 2003 #2

    emu

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    The way we calculate e can be similiar to Pi.
    One way yields an infinite series of non-repeating rational numbers. The sum is therefore irrational.

    Try proving the sqrt(5) is irrational.
     
  4. Apr 2, 2003 #3

    HallsofIvy

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    Emu, do you know of any way of proving directly that the digits in the decimal expansion of e are NOT repeating? I'm not saying it can't be done, only that I think it's easier to prove e cannot be written as a fraction.

    One standard method is to use the Taylor's series (which may be what emu meant): e= 1+ 1/2 + 1/6+ ...+ 1/n!+ ...
    If j is any positve integer then e*j!= integer+ 1/(q+1)+ 1/(q+1)(q+2)+ ... which is not an integer so e cannot be written as a fraction with denominator j for any j.

    A theorem I saw years ago was this: If c> 0, and there exist a function f(x), continuous on [0,c], positive on (0,c) and such that f(x) and its iterated anti-derivatives can be taken to be integer valued at both 0 and c, the c is irrational!

    Taking f(x)= sin(x) in this theorem shows that pi is irrational.

    It can also be used to prove: If c is a positive number other than 1 and ln(c) is rational, then c is irrational.

    Since e is a positive number, not equal to 1, and ln(e)= 1 is rational, it follows that e is irrational.
     
  5. Apr 2, 2003 #4
  6. Apr 2, 2003 #5
    While e cannot be written as a fraction, e to its first 2 million decimal places can. I'm just not going to.
     
  7. Apr 2, 2003 #6

    Hurkyl

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    That's not obvious... I don't see why the infinite sequence there cannot add up to an integral value.

    Hurkyl
     
  8. Apr 3, 2003 #7
    Let An=1+1/2!+...+1/n!;
    It's quite simple to prove that 1/(n+1)!<e-An<1/(n!*n);
    Let's suppose e is rational, so it's equal to p/q, where p and q are integers.
    1/(n+1)!<e-(1+1/2!+...+1/n!)<1/(n!*n); | *n!;
    1/(n+1)<n!*p/q-n!*(1+1/2!+...+1/n!)<1/n;
    But between 1/(n+1) and 1/n is no integer...
    n!*p/q must be an integer because for n big enough n! is a multiple of q;
    So e is not rational...
    QED
     
  9. Apr 3, 2003 #8
    that's kind of a neat prove, thanks for posting it. I don't think I would've caught that last part about for n big enough...
     
  10. Apr 6, 2003 #9
    thank you....:)
     
  11. Apr 6, 2003 #10

    Hurkyl

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    e-An<1/(n!*n)

    That bit isn't obvious either.... but the ordinary taylor remainder formula gives e/(n+1)! for that term which is sufficient for the proof. Don't tell me how to get that end of the inequality, it'd be a good exercise to figure it out myself!

    Hurkyl
     
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