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.
.
.
.
(the html on this thing was a
bitch. sorry if it's difficult to read.)