What is a proof that there are two irrational numbers where raising one to the power of the other produces a rational number?
Here is a proof by cases that there exist two irrational numbers a and b such that a^b is rational. First, consider a = b = sqrt(2). Recall that sqrt(2) is irrational, so If a^b = sqrt(2) ^ sqrt(2) is rational, we are done because we have found irrational a and b with a^b being a rational number. Otherwise, sqrt(2) ^ sqrt(2) is irrational. Let a = sqrt(2) ^ sqrt(2) and b = sqrt(2). Then a ^ b = (sqrt(2) ^ sqrt(2)) ^ sqrt(2) = sqrt(2) ^ (sqrt(2) * sqrt(2)) = sqrt(2) ^ 2 = 2, which is rational. So either way, we can find irrational a and b with a^b being rational.
Instruction: What is a proof that there are two irrational numbers where raising one to the power of the other produces a rational number?
Context: 
Response: Here is a proof by cases that there exist two irrational numbers a and b such that a^b is rational. First, consider a = b = sqrt(2). Recall that sqrt(2) is irrational, so If a^b = sqrt(2) ^ sqrt(2) is rational, we are done because we have found irrational a and b with a^b being a rational number. Otherwise, sqrt(2) ^ sqrt(2) is irrational. Let a = sqrt(2) ^ sqrt(2) and b = sqrt(2). Then a ^ b = (sqrt(2) ^ sqrt(2)) ^ sqrt(2) = sqrt(2) ^ (sqrt(2) * sqrt(2)) = sqrt(2) ^ 2 = 2, which is rational. So either way, we can find irrational a and b with a^b being rational.