Quickie #13

**Soroban** · January 18th 2007, 05:39 AM

This is a classic (very old).
If you've seen it, please don't post the answer.

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

Prove it or provide a counterexample.

**ThePerfectHacker** · January 18th 2007, 06:28 AM

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

**AfterShock** · January 18th 2007, 07:15 AM

Originally Posted by Soroban

This is a classic (very old).
If you've seen it, please don't post the answer.

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

**topsquark** · January 18th 2007, 11:31 AM

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

**ThePerfectHacker** · January 18th 2007, 02:27 PM

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.

**topsquark** · January 18th 2007, 02:34 PM

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

**Soroban** · January 18th 2007, 06:14 PM

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 . . .

**ThePerfectHacker** · January 18th 2007, 06:20 PM

I found a different counterexample.

$e$ is irrational.

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

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

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