Seen it. But I did not know it was a classic.
The "classic" solution goes like this:
Theorem: . can be rational.
Proof
There are only two possibilities:
. . (1) is rational.
. . (2) is irrational.
If (1) is rational, the theroem is verified.
If (2) is irrational, consider: .
. . an irrational number raised to an irrational power.
Then we have: . . . . a rational number.
Either way, an irrational raised to an irrational power can be rational.
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I've always found this proof amusing.
Note that we still don't know if is rational or irrational
. . but it doesn't matter . . .