若a,b都是無理數,則a的b次方都是無理數?

No. We would like to show that one can raise an irrational number to an irrational power and get a rational number out.
Consider (√2)^(√2) and [(√2)^(√2)]^(√2).
Suppose the first of these is irrational. Show that the second gives an example of an irrational number raised to an irrational power yielding a rational number.

Calculating, we have
[(√2)^(√2)]^(√2).=2.
Thus it is rational. If we suppose that
[(√2)^(√2)]
is irrational, then we would have an example of an irrational number raised to an irrational power giving a rational number.
Now, using the above, we argue that there exist irrational numbers a and b such that a^b is rational.
We will prove that there exists a pair (a; b) of irrational numbers such that a^b is rational. We will not construct the pair, however. First, suppose We would like to show that one can raise an irrational number to an irrational power and get a rational number out.
Consider
(√2)^(√2) and [(√2)^(√2)]^(√2).
suppose [(√2)^(√2)]
is rational. Then we have an example taking a = b =√2
since we know by class that √2 is irrational.
On the other hand, if
[(√2)^(√2)]
is irrational, the previous paragraph gives us that choosing a =[(√2)^(√2)]
and b =√2 then a^b is rational with each irrational.
Q.E.D.
The question is (√2)^(√2) rational, algebraic, or transcendental? is a very important question in 20th century mathematics. David Hilbert announced it as the 10th of his 23 problems at the International Mathematics Congress of 1900. It wasn't solved for about 30 years until Gelfand and Schneider proved that (√2)^(√2) is transcendental. They showed something considerably harder.

考慮(√2)^(√2)

如果此數為有理數，則有「無理數的無理數次方為有理數」

如果此數為無理數，則考慮 [(√2)^(√2)]^(√2)

容易計算
[(√2)^(√2)]^(√2)=(√2)^2=2 為有理數

於是也有「無理數的無理數次方為有理數」。

此法最厲害的地方，就是不論(√2)^(√2)為有理數或是無理數，
也能推出同樣結果：存在無理數a, b使得a^b為有理數。