Irrationals being closed under exponentiation

Question

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?

Answer 1

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.

Answer 2

> are irrationals closed under exponentiation?

No. E.g. e^(ln 2) = 2. It should be clear that both e and ln 2 are irrational.

Answer 3

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.

Answer 4

Not true, consider e and ln2 for example

Answer 5

Think about this intuitively.

What's an "irrational" number? It just mean it's *not* a rational number. Rational numbers are rare, they have a special form compared to the rest of the number. Irrational numbers are, by definition, the vast majority of numbers, because it's defined as those *not* fitting into a very narrow kind of number.

So don't expect it to be closed under anything, really.

The same argument apply to pretty much any class of numbers that is defined to be *not* having a specific form, such as transcendental number, non-period, non-computable number, and so on.