User:Karlhahn/User e-irrational

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e \ne \frac {p} {q} This user can prove that the number, e, is irrational 



Original source:
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PROOF:

If \scriptstyle e were rational, then

e = \frac {p} {q}

where \scriptstyle p and \scriptstyle q are both positive integers. Hence

qe = p

making \scriptstyle qe an integer. Multiplying both sides by \scriptstyle (q-1)!,

q!e = p(q − 1)!

so clearly \scriptstyle q! e is also an integer. By Maclaurin series

e = \sum_{k=0}^\infty \frac {1} {k!}

Multiplying both sides by \scriptstyle q!:

q! e = \sum_{k=0}^\infty \frac {q!} {k!}

The first \scriptstyle q+1 terms of this sum are integers. It follows that the sum of the remaining terms must also be an integer. The sum of those remaining terms is

r = \sum_{k=1}^\infty \frac {q!} {(q+k)!}

making \scriptstyle r an integer. Observe that

\frac {q!} {(q+k)!} < \frac {q!} {q! q^k} = \frac {1} {q^k}


So

r = \sum_{k=1}^\infty \frac {q!} {(q+k)!} < \sum_{k=1}^\infty \frac {1} {q^k}

But

\sum_{k=1}^\infty \frac {1} {q^k} = \frac {1} {q-1}

This means that \scriptstyle 0 < r < \frac {1} {q-1} \le 1, which requires that \scriptstyle r be an integer between zero and one. That is clearly impossible, hence \scriptstyle e is irrational

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