Hey guys. I understand that the set of irrational numbers aren't closed under addition, subtraction, multiplication, and division. However, I was wondering if it is closed under exponentiation. When I Google it, I see unanimously that it's not closed because you can simply square an irrational number such as sqrt(2) to get a rational number. However, I was under the impression that both the base and exponent had to be irrational numbers to test if it was closed. This is how I remember showing whether or not other number sets are closed under a particular operation.

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So, do both the base and exponent need to be irrational numbers to test for closedness?

If not, I understand how irrationals aren't closed, since raising numbers to positive integer powers is just repeated multiplication, which is already known not to be closed.

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If they do have to be, are irrationals closed under exponentiation?
There's a famous proof of this

Consider the number x = sqrt(2)^sqrt(2)

Consider two cases:

Case 1: x is rational

case 2: x is irrational

If case 1 is true, then we're done. An irrational (sqrt(2)) to the power of an irrational (sqrt(2)) is rational

In case 2, we have x^sqrt(2) = 2 (basic rules of exponents, also true in case 1 of course). Again, an irrational (x) to the power of an irrational (sqrt(2)) is rational.
> are irrationals closed under exponentiation?

No. E.g. e^(ln 2) = 2. It should be clear that both e and ln 2 are irrational.
Take any integer n > 1, then log n is irrational. The set of numbers log n / x, where x > 0 is irrational is uncountable, hence it must contains uncountably many irrational numbers w.

So there must uncountably many irrational numbers in the form y=exp(w), where w=log n/x is irrational and x irrational. Hence n = y^(x) for some (uncountably many) irrational numbers x and y.
Not true, consider e and ln2 for example