sqrt(2)^sqrt(2)?

Question

Is sqrt(2)^sqrt(2) rational or irrational?

Answer 1

This is not an unsolved problem. (√2)^(√2) is irrational (in fact, transcendental) by the Gelfond-Schneider theorem (the exponent and base are both algebraic and the exponent is not a real rational number).

Answer 2

I believe this is an unsolved problem.

To the person that said that if a is rational and b is irrational, then a^b is irrational, you are wrong. For example, let a=2, b=log_2(3).
Then a^b = 3, but you can show that b must be irrational. (If b=p/q, then 2^p = 3^q, violating the rule that the prime factorization of a number is unique.)

This number is often used as a proof that if a and b are irrational, then a^b is can be rational. That's because if we set x to be:

sqrt(2)^sqrt(2)

then either this number is rational, or it is irrational, and

x^sqrt(2) = 2

is rational, so either way we have an example of a^b where a and b are irrational and a^b is rational.

Answer 3

sqrt(2)^sqrt(2) = [2^(1/2)]^sqrt2 = 2^[(sqrt2)/2]
= 2^(1/sqrt 2)
This is now a rational number raised to an irrational power. So I'd say it would have to be irrational, as I don't see how this number could be expressed in the form p/q, where p & q are integers.

Answer 4

I disagree with spell_check. In fact, this is not a simple problem. I did a quick net search and found this reference: http://www.cut-the-knot.org/do_you_know/irrat.shtml , which indicates the author, at least, does not know if it is rational.

Look at 2^x = 3, 2 is rational, 3 is rational, but x is not (log2(3)), so it is not true that a rational to an irrational must be irrational.

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I defer to Pascal. His reference is correct. (but I was right, it is not a simple problem)

Answer 5

Irrational.

(√2)^(√2) = 1.6325269

√2 raised to a real power will be rational only if that power is zero or a multiple of 2, either positive or negative.

Answer 6

[sqrt(2)]^[sqrt(2)] is the same as

[2^(1/2)]^[2^(1/2)]

Since we have a power to a power, we now have

2^([2^(1/2)]/2)

Let's deal with the inside of the exponent first; that is, this part:

[2^(1/2)]/2

We can change this, because we have powers of 2 and we're dividing; we can subtract the exponents.

2^(1/2 - 1)

This is equal to

2^(-1/2)

So now, we have

2^(2^(-1/2))

or

2^(1/[2^(1/2)])

or

2^(1/sqrt(2))

There is nothing to indicate this is rational.
Perhaps someone can have a better proof.

Answer 7

i'm guessing you advise you prefer a fence to go around the 800 sq. foot section diagonally. the difficulty is which you particularly decide on specific lengths for the perimeters. you notice, if one factor is 80 and the different is 10, the diagonal would be 80.sixty two, even regardless of the indisputable fact that it the perimeters have been 20 and 40, the diagonal would be 40 4.seventy two. it definitely relies upon on the lengths of the perimeters.

Answer 8

what am i missing here. isn't this expression the sq root of 2 times the sq root of 2? if it is the answer is 2 and it is rational

Answer 9

rational !!!