1. ## Difficult proving algebra

Prove that there are infinitely many pairs of real numbers x and y such that x does not equal y, and x^y = y^x

2. ## Re: Difficult proving algebra

Originally Posted by yugimutoshung
Prove that there are infinitely many pairs of real numbers x and y such that x does not equal y, and x^y = y^x
Consider x = 2 and y = 4...

-Dan

3. ## Re: Difficult proving algebra

prove it algebraically, not numerically

4. ## Re: Difficult proving algebra

Just trying to give you a starting point.

-Dan

5. ## Re: Difficult proving algebra

Originally Posted by yugimutoshung
Prove that there are infinitely many pairs of real numbers x and y such that x does not equal y, and x^y = y^x
Hi, am just another student here so take this for only a possibility..

Since you only need to prove there are an infinite number of pairs of real numbers (x,y) that make x^y = x^y true why not first restrict the domain to real positive numbers just to keep cases to a minimum.

let x,y,z be positive real numbers

let y = z^(1/x), with x, z not zero, then z = y^x

so z = y^x = x^y

Let z be a fixed value, then for that value of z the pairs (x, y) = (x, x^(1/x) satisfy z = x^y = y^x,

but those pairs are infinite in number since there are an infinite number of positive real numbers x,

so it is then that there are an infinite number of pairs that satisfy x^y = y^x for a fixed z let alone a variable z.

I only comment because nothing has been happening on the post so I figure that I am not getting in the way.

And again, I am only presenting an idea and I might easily be wrong.

.

7. ## Re: Difficult proving algebra

Originally Posted by topsquark
Consider x = 2 and y = 4...

-Dan
I found (2,4) too but how do I go from there?

8. ## Re: Difficult proving algebra

hint:

Is the function $f(x) = \dfrac{\log(x)}{x}$ one-to-one? Why is this relevant?

9. ## Re: Difficult proving algebra

Originally Posted by Deveno
hint:

Is the function $f(x) = \dfrac{\log(x)}{x}$ one-to-one? Why is this relevant?
I think it's one-to-one... but I have no idea regarding its relevance

10. ## Re: Difficult proving algebra

Have you tried graphing it?

Suppose it wasn't 1-1, wouldn't that mean we had:

$\dfrac{\log(x)}{x} = \dfrac{\log(y)}{y}$ for some pair $x \neq y$?

11. ## Re: Difficult proving algebra

Originally Posted by Deveno
Have you tried graphing it?

Suppose it wasn't 1-1, wouldn't that mean we had:

$\dfrac{\log(x)}{x} = \dfrac{\log(y)}{y}$ for some pair $x \neq y$?
Yeap I tried. Yes, and then...?

12. ## Re: Difficult proving algebra

If $x^y = y^x$, then take log of both sides:

$y\log x = x\log y$

Divide both sides by $xy$ to get:

$\dfrac{\log x}{x} = \dfrac{\log y}{y}$

So, $x^y = y^x$ if and only if the equation Deveno gave holds. He is saying try to prove it is not one-to-one. You have an example already (2,4).

Are you allowed to use Calculus? That could help.

Let $f(x) = \dfrac{\log x}{x}$. Then $f'(x) = \dfrac{1-\log x}{x^2}$. There is a critical point at $x=e$. On $(0,e)$, the function is increasing. On $(e,\infty)$, the function is decreasing.

Using L'Hospital's Rule, $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \dfrac{1}{x} = 0$. $f(1) = 0$. Since the function is continuous, it possesses the intermediate value property. So, for every value $y \in \left(0,\dfrac{1}{e}\right)$, there exists two values for $x$ such that $f(x) = y$. One will have $1 and the other will have $x>e$.

Note: I have given you some things to think about, but alone, this is not strictly a proof. I skipped a few steps.