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 Skip Navigation LinksMath Help > First Principles > Logic and Proof > Irrationality Proofs > Irrationality Proof - e

Proof that e is irrational

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.

Internet references

http://www.meikleriggs1.free-online.co.uk/pi/index.htm 

Related Pages in this website

Perfect Squares -- proof that sqrt(n) is irrational, as long as n isn't a perfect square.

 


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