1. ## Quickie #13

This is a classic (very old).

An irrational number raised to an irrational power is always irrational.

Prove it or provide a counterexample.

2. Seen it. But I did not know it was a classic.

3. Originally Posted by Soroban
This is a classic (very old).

An irrational number raised to an irrational power is always irrational.

Prove it or provide a counterexample.

I'll give several counter examples;

For the most part,

[sqrt(x)^(sqrt(x))]^(sqrt(x)), where x is even, with the exception of when x is a square.

EX:

[sqrt(2)^(sqrt(2))]^(sqrt(2))

[sqrt(2)^(sqrt(2))] = 2^(sqrt(2)/2)

[2^(sqrt(2)/2)]^(sqrt(2)) = 2

4. Originally Posted by AfterShock
I'll give several counter examples;

For the most part,

[sqrt(x)^(sqrt(x))]^(sqrt(x)), where x is even, with the exception of when x is a square.

EX:

[sqrt(2)^(sqrt(2))]^(sqrt(2))

[sqrt(2)^(sqrt(2))] = 2^(sqrt(2)/2)

[2^(sqrt(2)/2)]^(sqrt(2)) = 2
Interesting, but you are assuming that $\sqrt{2}^{\sqrt{2}}$ is irrational, else this isn't a counterexample. I have little doubt that it is, but this isn't a good counterexample without proving this fact.

-Dan

5. Originally Posted by topsquark
Interesting, but you are assuming that $\sqrt{2}^{\sqrt{2}}$ is irrational, else this isn't a counterexample. I have little doubt that it is, but this isn't a good counterexample without proving this fact.

-Dan
Funny thing is you do not need to. The assumption is that irrational raised to irrational is always irrational. Thus, he is using this assumption. And arrives at a contradiction.

6. Originally Posted by ThePerfectHacker
Funny thing is you do not need to. The assumption is that irrational raised to irrational is always irrational. Thus, he is using this assumption. And arrives at a contradiction.
Aaaaah! I get it now.

-Dan

7. The "classic" solution goes like this:

Theorem: . $(\text{irrational})^{\text{irrational}}$ can be rational.

Proof

$\text{Consider }a \:=\:\sqrt{2}^{\sqrt{2}}$

There are only two possibilities:

. . (1) $a$ is rational.

. . (2) $a$ is irrational.

If (1) $a$ is rational, the theroem is verified.

If (2) $a$ is irrational, consider: . $a^{\sqrt{2}}$
. . an irrational number raised to an irrational power.

Then we have: . $a^{\sqrt{2}} \:=\:\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} \:=\:\left(\sqrt{2}\right)^2\:=\:2$ . . . a rational number.

Either way, an irrational raised to an irrational power can be rational.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I've always found this proof amusing.

Note that we still don't know if $\sqrt{2}^{\sqrt{2}}$ is rational or irrational
. . but it doesn't matter . . .

8. I found a different counterexample.

$e$ is irrational.

$\ln 2$ is irrational (but do not know how to show it).

Then,
$e^{\ln 2}=2$