- Show that it’s possible
a^b=cwhereaandbare irrational, andcis rational.
Sry for the gap I ran out of ideas.
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e^(log 2) = 2 is rational
that is simply genius
(i suppose it didnt come to me when i think of “irrational”)
this one got some table slams from my friends
Hint:
spoiler
Find an example which satisfies the equation.
Solution:
solution
e^(i*π) = -1
Also, anything like a^(log(c) / log(a)), for positive rational c and irrational a, to generalize bean_jamming’s answer
I also assert without proof that in the equation x^x = c, x is irrational for most rational values of c
I did start trying out stuff with sqrt(2), thinking back to the tower power problems, but didn’t end up coming up with your solution while doing so ¯\_(ツ)_/¯
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