I know the proof for this one with square root of 2, but is they any other examples where you don't have to do manuipulation like actually a and b are irrational.
a nbe b irrational numbers hnce a^b is rational
-1
$\begingroup$
alternative-proof
irrational-numbers
-
4I don't understand the question. Could you please clarify? Are you asking if *whenever* $a$ and $b$ are irrational, the number $a^b$ must be rational? – 2017-02-18
-
0For an example of an irrational number to the power of another irrational number being irrational, see [this post](http://mathoverflow.net/questions/40145/irrationality-of-pi-e-pi-pi-and-e-pi2) which shows at least one of $e^e$ and $e^{e^2}$ must be irrational. – 2017-02-18
1 Answers
1
Quick example for $(irrational)^{(irrational)}=rational$ -
$e^{\ln 5} = 5$
where $e=2.71828182846...$ and $ln 5=1.60943791243...$