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e is equal to 1+(1/1!)+(1/2!)+(1/3!)+.... (assume this as a definition of e. it works out)
e = p / q assuming the contrary, that e can be expressed as a ratio of two positive integers, a rational number.
multiply the def. of e by q! . q!e = q! + (q!/1!) + (q!/2!) + ... + (q! / q!) + ... (other numbers of the sum)
because e is assumed rational, = p / q , q!e is an integer, as is the sum q! + (q!/1!) + (q!/2!) + ... + q! / q!
thus the "other numbers of the sum" are equal to the difference of two integers. call these numbers P.
P = q! ( (1 / (q! + 1)) + (1 / (q! + 2)) + ...)
multiplying in the q!, P = (1 / (q + 1)) + (1 / (q + 2)) + ... which is less than (1 / (q + 1)) + (1 / (q + 1)2) + (1 / (q + 1)3) + ...
P is less than (1 / (q + 1)) ( 1 + (1 / (q + 1)) + (1 / (q + 1)2) + ...)
P is less than (1 / (q + 1)) * (1 / (1 / (1 - (1 / (q + 1)))) (this with use of the infinite geometric sum formula, S = (the first term of series) / 1 - r
rearranged as
P is less than (1 / (q+1)) * ((q+1) / q)
reduced to be:
P is less than 1 / q
P and q are by our definition positive, valuing P as an integer between 0 and 1 / q
THIS IS A CONTRADICTION. Thus, the assumption that e is rational is false, which characterizes e as irrational.
QED.
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(the html on this thing was a bitch. sorry if it's difficult to read.)

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