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Define e as the sum 1/0! + 1/1! + 1/2! + ...
Suppose that e is equal to some fraction p/q, in lowest terms. Then
e = p/q = 1/0! + 1/1! + 1/2! + 1/3! + ... + 1/q! + 1/(q+1)! + ....
Multiplying by q!,
(q-1)!p = (q!/0!+ q!/1! + ... q!/q!) + 1/(q+1) + 1/[(q+1)(q+2)] + ...
and noting that q!/x! is integer as long as x is less than or equal to q, the left side of the following equation is an integer
(q-1)!p - (q!/0!+ q!/1! + ... q!/q!) = 1/(q+1) + 1/[(q+1)(q+2)] + ...
Therefore, we know that 1/(q+1) + 1/[(q+1)(q+2)] + ... is some integer, and it's obvious that it is greater than 0. But,
1/(q+1) + 1/[(q+1)(q+2)] + ... < 1/(q+1) + 1/(q+1)2 + 1/(q+1)3 + ... = 1/q ≤ 1
Therefore, there is an integer between 0 and 1, which is a contradiction.
Perfect Squares -- proof that sqrt(n) is irrational, as long as n isn't a perfect square.
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