# Thread: provide a counter example.

1. ## provide a counter example.

I have been assigned this fro my proofs class, and i haven't been able to figure it out after several tries.

"There do not exist rational numbers x and y such that X^y is positive and Y^x is negative."

does anyone have any idea? Thanks!!

2. Try -3 and -2?

3. Ya, but they have to be rational..

4. Originally Posted by meg0529
Ya, but they have to be rational..
-2 and -3 are rational, and so are $\displaystyle (-2)^{-3}=-1/8 \in \mathbb{Q}$ and $\displaystyle (-3)^{-2}=1/9 \in \mathbb{Q}$...

On the other hand, $\displaystyle y= \pm 1$ for $\displaystyle x,y,x^y, y^x \in \mathbb{Z}$. This is because, assuming. $\displaystyle x^y < 0 \Rightarrow x<0 \Rightarrow y^x = 1/{y^{x'}}$ with $\displaystyle x'>0$. Thus, $\displaystyle 0 < y^{x'} \leq 1 \Rightarrow 0 < y \leq 1 \Rightarrow y = 1$ as is an integer. The fact that $\displaystyle x^y > 0$ gives $\displaystyle y=-1$ is attained through symmetry.

Clearly, $\displaystyle 1^{-i} = 1$ and $\displaystyle {(-i)}^1 = -i$ for $\displaystyle i \in \mathbb{N}$.

5. Oh! okay,

Thank you both! I'm so used to rational being something w/o a 1 on the bottom forgot that 2,3,4.. are rational >.<

that makes things easier! again thanks so much! I been trying this for 2 hours already!