e的平方是无理数的证明 |
送交者: gmann 2007年10月17日07:20:05 于 [灵机一动] 发送悄悄话 |
Suppose e^2 is rational, then (e^2 - 1)/(e^2 + 1) = (e - 1/e)/(e + 1/e) is also rational. Let it be p/q , where p and q are positive integers and q>p and q>1. Expanding (e - 1/e)/(e + 1/e), we have: (1/1! + 1/3! + 1/5! +...)/(1 + 1/2! + 1/4! +...) = p/q multiply q! up and down on the left : q!*(1/1! + 1/3! + 1/5! +...)/q!*(1 + 1/2! + 1/4! +...) = p/q or: (q!*a + q!*b)/(q!*x + q!*y) = p/q (1) where a=1/1! + 1/3! + 1/5! +...+1/r!, (r <= q)
x=1 + 1/2! + 1/4! +...+1/s!, (s <= q-1)
rearrange (1): q!*a - p*(q-1)!*x = p*(q-1)!*y - q!*b (2) The left can be proven to be non-zero integer;
Therefore e^2 is iirational.
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