`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.)