Is x Irrational When x^x Equals 2?

[Icebreaker]

How do I show that 2 has no rational roots?

 

[krab]

First assume it does. Write square root of 2 = p / q, where p and q are integers. Then p^2 = 2 q^2, and so must be even... you can guess the rest. Eventually you come to a conclusion that contradicts one of your original assumptions.

 

[Muzza]

Write square root of 2 = p / q, where p and q are integers.

Where p and q are coprime integers...

 

[HallsofIvy]

Note: you will need the fact that the square of an odd number is always odd:
(2n+1)²= 4n²+ 4n+ 1= 2(2n²+ 2n) + 1.

 

[Icebreaker]

No, I meant ANY rational root. That is,

2^{\frac{1}{n}} is not rational for any positive integer n > 1.

I tried "extending" the square root of 2 proof, however, at some point,

2a^n=b^n

If n is even, then it works. But if n is odd, then the argument breaks down.

Or does it, let me think...

 

[AKG]

I think the same proof applies. Suppose (p/q)ⁿ = 2 where gcd(p,q) = 1. Then

pⁿ = 2qⁿ

This tells us that pⁿ is even, which tells us that p is even, which tells us that 2ⁿ|pⁿ. This in turn tells us that qⁿ is even, which in turn tells us that q is even. Both q and p are even, and thus aren't co-prime, contradicting our assumption that they were.

 

[Icebreaker]

Yup, that's what I thought too. Thanks everyone.

On a side note, x^x = 2 implies that x cannot be rational because x = 2^{1/x} and 2 has 2 rational roots, as we've shown. Is that correct?

 

[Zurtex]

Yup, that's what I thought too. Thanks everyone.

On a side note, x^x = 2 implies that x cannot be rational because x = 2^{1/x} and 2 has 2 rational roots, as we've shown. Is that correct?

Our assumption before was that 2^1/n where n was an integer greater than 1. However I don't think it is hard to extend if you let x = p/q.

 

[AKG]

Yup, that's what I thought too. Thanks everyone.

On a side note, x^x = 2 implies that x cannot be rational because x = 2^{1/x} and 2 has 2 rational roots, as we've shown. Is that correct?

If x is rational, then there are coprime p and q such that x = p/q. We get:

(p/q)^(p/q) = 2

(p/q)^p = 2^q

If p/q is not a whole number, then in general (p/q)ⁿ is not whole for natural n, and in particular when n = p. On the other hand, 2^q is of course whole, so we have a contradiction unless p/q is a whole number. But it's easy to check 1¹ is not 2, 2² is not 2, etc. So we get a contradiction regardless, so x is irrational.