Is x irrational if x^x = 2?

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In summary, to show that 2 has no rational roots, you can assume that it does and use the fact that the square of an odd number is always odd to eventually reach a contradiction. Additionally, you can extend this proof to show that 2 to any power greater than 1 has no rational roots. Also, if x^x = 2, then x must be irrational because for any rational number x, the expression x = 2^{1/x} has 2 rational roots, which contradicts our assumption that x is rational.
  • #1
Icebreaker
How do I show that 2 has no rational roots?
 
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  • #2
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.
 
  • #3
Write square root of 2 = p / q, where p and q are integers.

Where p and q are coprime integers...
 
  • #4
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.
 
  • #5
No, I meant ANY rational root. That is,

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

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

[tex]2a^n=b^n[/tex]

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

Or does it, let me think...
 
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  • #6
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.
 
  • #7
Yup, that's what I thought too. Thanks everyone.

On a side note, [tex]x^x = 2[/tex] implies that x cannot be rational because [tex]x = 2^{1/x}[/tex] and 2 has 2 rational roots, as we've shown. Is that correct?
 
  • #8
Icebreaker said:
Yup, that's what I thought too. Thanks everyone.

On a side note, [tex]x^x = 2[/tex] implies that x cannot be rational because [tex]x = 2^{1/x}[/tex] and 2 has 2 rational roots, as we've shown. Is that correct?
Our assumption before was that 21/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.
 
  • #9
Icebreaker said:
Yup, that's what I thought too. Thanks everyone.

On a side note, [tex]x^x = 2[/tex] implies that x cannot be rational because [tex]x = 2^{1/x}[/tex] 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 = 2q

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, 2q 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.
 

1. Is there a way to prove that x is irrational if x^x = 2?

Yes, it is possible to prove that x is irrational using a proof by contradiction. This involves assuming that x is rational and then showing that it leads to a contradiction, thus proving that x must be irrational.

2. Can x be both rational and irrational if x^x = 2?

No, x cannot be both rational and irrational because these two types of numbers are mutually exclusive. If x is rational, it can be expressed as a ratio of two integers. If x is irrational, it cannot be expressed as a ratio and has an infinite number of non-repeating decimals.

3. Does the value of x have any relationship with the number 2 if x^x = 2?

Yes, the value of x is directly related to the number 2 in this equation. The value of x represents the root of the exponent, so in this case, x is the square root of 2. This means that x is approximately 1.41421356.

4. Are there any other numbers besides x that satisfy the equation x^x = 2?

No, there are no other numbers besides x that satisfy this equation. The only solution is x = √2, which has been proven to be irrational. This means that there are no other rational or irrational numbers that can satisfy this equation.

5. What is the significance of the equation x^x = 2?

The significance of this equation lies in the fact that it has a unique solution, which is an irrational number. This equation also demonstrates the relationship between exponentiation and irrational numbers, as the root of the exponent leads to an irrational solution.

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