# Irrational power of an irrational number

1. Mar 18, 2012

### analyzer

What kind of number is sqrt(2)^sqrt(2)?

I have noted sqrt(2)^sqrt(2) = 2^(sqrt(2)/2) = 2^(1/sqrt(2)), i.e. a rational number to an irrational power.

Now, 1/sqrt(2) is less than 1, but greater than zero. So, given that 2^x is an increasing function, 2^(1/sqrt(2)) is less than 2^1, but greater than 2^0.

Also, I tried the following. Let a,b be positive integers such that 2^(1/sqrt(2)) = a/b. What I want to do is reach a contradicition for otherwise I would assume the number in question is rational.

2. Mar 18, 2012

### HallsofIvy

Staff Emeritus
Your last statement is poor logic. If you can show $2^{1/\sqrt{2}}= a/b$ then, yes, by definition, it is rational. If you cannot show that you don't know what it is. The fact that you cannot show something doesn't mean it isn't true.

3. Mar 18, 2012

### analyzer

Thanks for replying. I see what you mean by poor logic. I feel silly. After posting the problem, I read a thread in which you answer the question of whether an irrational number to an irrational power can be rational. That makes my day as I thought I had to know what kind of number sqrt(2)^sqrt(2) is.

4. Mar 19, 2012

### analyzer

Anyways, the question remains: what kind of number is sqrt(2)^sqrt(2)?

1 < sqrt(2) < 2

Therefore,

1/2 < sqrt(2)/2 = 1/sqrt(2) < 1

Now, 1/sqrt(2) is irrational, i.e. some infinte nonrepeating decimal.

It has the form

0 + digit number/10 + digit number/100 + digit number/1000 + ...

Then, sqrt(2)^ sqrt(2) = 2^(1/ sqrt(2)) = 2^0 * 2^(digit number/10) *
2^(digit number/100) * 2^(digit number/1000) * ...

Now, what kind of number is 2^(digit number/(10^n)) for every positive

integer n?

First, 10^n > digit number. When digit number = 2, digit number/10^n

takes the form 1/(5*10^(n-1)).

Now, let 2^(1/(5*10^(n-1))) = a/b, with a, b positive integers (necessarily)

greater than 1:

2 = (a/b)^(5*10^(n-1))
2 * b^(5*10^(n-1)) = a^(5*10^(n-1))

But the last statement seems to be a contradiction. Both sides of the

equation represent nonprime natural numbers greater than one. Every

nonprime natural number greater than one can be expressed as a unique

product of primes. The right-hand side of the equation has 2 as factor a

different number of times from the left-hand side. According to this,

2^(1/(5*10^(n-1))) is an irrational number. Correct me if I am wrong.

At last, what I suspect, but have not proved of course, is that 2^(1/sqrt(2))

is the product of infinitely many irrational numbers. But, does this

5. Mar 20, 2012

### analyzer

A way to prove that sqrt(2) is an irrational number is the following.

Let a, b be natural numbers greater than 1. Now, I plug

sqrt(2) = a/b

As you will see, that statement leads to a contradiction. Square both sides of the equation:

2 = (a^2)/(b^2)

2(b^2) = a^2

The fundamental theorem of arithmetic: every nonprime number greater than one can be expressed as a unique product of primes.

Now, clearly, the right-hand side of the equation has the factor 2 a different number of times from that of the left-hand side. Namely, on the right-hand side, 2 appears as a factor an even number of times. But on the left-hand side, it appears as factor an odd number of times.

What went wrong? It happens to be that sqrt(2) is not a rational number (a/b). It is an irrational number.

I have to say that I took this proof from Varberg and Fleming's Algebra and Trigonometry with Analytic Geometry: A problem-solving approach (the Spanish language version).

I have used the same logic above to show that 2^(1/(5*10^(n-1))) is an irrational number. Maybe I omitted some details. So I will be more explicit:

Let a, b, n be positive integers. Then, plug

2^(1/(5*10^(n-1))) = a/b

Now, 0 < 1/(5*10^(n-1)) < 1

Given that 2^x, with x real, is an increasing function,

2^0 < 2^(1/(5*10^(n-1))) < 2^1

So, 2^(1/(5*10^(n-1))) lies somewhere between 2 and 1. Being that the case, it is not an integer. So b can't be equal to 1. Also, a is not equal to b; it must be greater than it. That's why a is not equal to 1.

Going back, I raise the following equation to the power of 5*10^(n-1):

2^(1/(5*10^(n-1))) = a/b

2 = (a/b)^(5*10^(n-1))

2*b^(5*10^(n-1)) = a^(5*10^(n-1))

On the left-hand side, there may be only one factor 2. And, if tha't's the case, the right-hand side also must have one factor 2, but that's not the case.

Now, the other possibility is that, on the left-hand side, 2 appears as factor a multiple-of -five-plus-one number of times, whereas on the right-hand side it appears a multiple-of-five number of times. Now, it seems to me that that's a contradiction.

The conclusion is that 2^(1/(5*10^(n-1))) is an irrational number.

I want to prove in the same fashion (on a case by case basis) what kinds of numbers the expression 2^(digit number/(10^(n))) yields (rational, irrational?). But, does this program lead anywhere concerning the question of what kind of number is sqrt(2)^sqrt(2)?

Remember,

sqrt(2)^sqrt(2) = 2^(1/sqrt(2)) = 2^0 * 2^(digit number/10) * 2^(digit number/100) * 2^(digit number/1000) * ...

I would also be grateful if someone can assess the reasoning I have followed on this post.

6. Mar 20, 2012

### Dick

You aren't going to be able to show a number is irrational by showing it's a limit of rational or irrational numbers. sqrt(2)^sqrt(2) is irrational. In fact, it's transcendental. It's a easy consequence of the Gelfond-Schneider theorem. I'm not sure there is any elementary way to show this.

Last edited: Mar 20, 2012