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Homework Help: Is this proof valid? (e is irrational)

  1. Jan 9, 2014 #1
    Is this proof that e is an irrational number valid?

    e = ∑[itex]^{∞}_{n=0}[/itex] 1/n! = 1 + 1/1! + 1/2! + 1/3! + ... + 1/n! +...

    Let e = a + b where

    a = Sn = 1 + 1/1! + 1/2! + 1/3! + ... + 1/n!

    b= 1/(n+1)! + 1/(n+2)! + 1/(n+3)! +...

    Multiply both sides by (n!) giving e(n!) = a(n!) + b(n!)

    For proof by contradiction, take e to be an integer, such that e = m/n where m and n are both positive integers and in their lowest terms.

    We can take n to be a positive integer and as such, n! is a positive integer.

    a(n!) = (1 + 1/1! + 1/2! + 1/3! + ... + 1/n!)(n!) = (n! + n!/1! + n!/2! + n!/3! + ... + n!/n!)

    We can see a(n!) is positive integer as n > 1,2,3 ... ,n-1,n and as such each fraction in the parentheses will cancel down to produce a positive integer. The sum of positive integers is yet another positive integer.

    We can note that according to our equation, e = a + b, b is simpy the difference of two positive integers, which in turn is a positive integer.

    However, it can be shown that b(n!) is not a positive integer:

    b(n!) = [1/(n+1)! + 1/(n+2)! + 1/(n+3)! +...](n!) = [n!/(n+1)! + n!/(n+2)! + n!/(n+3)! +...]
    = 1/(n+1) + 1/(n+1)(n+2) + 1/(n+1)(n+2)(n+3) + ...

    We know n is a positive integer ∴ n ≥ 1, n+1 ≥ 2, n+2 > 2, n+3 >2.

    So we have:

    b(n!) < 1/2 + 1/22 + 1/23 +...

    b(n!) < a/1-r
    b(n!) < 0.5/0.5
    b(n!) < 1

    We know b(n!) > 0 as e - a ≠ 0
    ∴ 0 < b(n!) < 1 which means b(n!) cannot be an integer and contradicts the statement above, that b(n!) is the difference of two integers.

    Therefore e must be irrational.

    This was how it was taught to me, but I don't feel confident on the proof. I haven't proven that b is irrational, which would tell me that e is irrational, as e would become the summation of a rational and an irrational number multiplied by n!, where n! is an integer and a rational number. Can someone tell me if the proof is valid?
  2. jcsd
  3. Jan 9, 2014 #2


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    The order and some steps are a bit messy, but the general idea is valid.

    That follows from the proof, but you don't need it for the contradiction.
    Last edited: Jan 9, 2014
  4. Jan 9, 2014 #3
    Ahh, okay, I see. Thank you very much for the confirmation. Anything I can do to tidy it up a bit?
  5. Jan 9, 2014 #4


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    - don't use n for two different things (as index in the sum and as denominator for e afterwards)
    - start with the assumption e=m/n, otherwise your split e=a+b is not well-defined (where are we supposed to do it?)

    You cannot tell "positive" from the difference, the difference of two positive integers can be negative. "Integer" is sufficient here, you can show b>0 afterwards.

    "a" was used for something else before, "a" and "r" are not introduced and the denominator needs brackets (or a proper fraction: ##\frac{a}{1-r}##).
    The first part does not follow from the second part.
  6. Jan 9, 2014 #5
    If I start with the assumption e=##\frac{p}{q}## can I state that p,q must either both be positive or both be negative?

    I'll start with my assumption in the future.
    This was meant to be ≥, apologies.

    That's the same thought I had.

    How do I create a proper fraction? I couldn't find it in the list of symbols.

    Would b(n!) > 0 as e(n!) - a(n!) ≠ 0 be correct?

    Dividing by (n!):

    b > 0 as e - a ≠ 0

    Thanks for the help.
  7. Jan 9, 2014 #6


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    You can even assume p and q to be positive, as you know e is positive (or if you don't know this, it is easy to see based on the sum), but it does not matter as long as q is positive and that is always possible.

    This forum supports LaTeX.

    Is -4.3(n!) > 0 because 2.7(n!) - 7(n!) ≠ 0?
  8. Jan 9, 2014 #7
    No. In what way do we know b > 0?

    Also I found a much more concise explanation here: http://cazelais.disted.camosun.bc.ca/250/e-irrational.pdf. However some of the notation is confusing me. I believe Rn is the value I have labelled b in my original post, but I'm unsure as to why it takes a value such that 0 < Rn < ##\frac{3}{(n+1)!}##

    And I'm also unsure what "Choose n > max(b,3)" means.
    Last edited: Jan 9, 2014
  9. Jan 10, 2014 #8


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    In the same we can know a>0: it is a sum of positive numbers.

    You can get this in the same way you found 1/n! as upper limit - it is just a bit more general as it covers the case n=0 (which is not needed afterwards).

    n is chosen to be larger than b, and at least 4.
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