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A relation $R$ is defined as $(x,y)\in R \implies x^y=y^x$ for $x,y\in\mathbb{I}-\{0\}$, where $\mathbb{I}$ is the set of integers. Find whether the relation $R$ is transitive or not.

Let $x^y=y^x$ and $y^z=z^y$

We have to prove that $x^z=z^x$

Now, $y\ln x=x \ln y$ and $z\ln y=y \ln z$

$\implies y \ln x=\frac{y\ln z}{z}\cdot x$ or $z \ln x=x \ln z$ or $x^z=z^x$

So it seems that this relation is transitive. Here's the catch, I can't find even one triplet where this transitive relation holds. And here's another proof which claims that the relation is not transitive.

Since $x^y=y^x$ we can say that $y=x^{\frac{y}{x}}$

$\implies (x^{\frac{y}{x}})^z=z^y$ or $x^{yz}=z^{xy}$ which says that it isn't transitive.

Which one is correct and why$?$

Any help is greatly appreciated.

Edited. Sorry for the typo.

Bill Dubuque
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Blue Cat Blues
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    I don't understand your second proof. Your former proof is correct. The reason you can't find any tuples where transitivity holds is because there are no tuples where $xRy$ and $yRz$ at all. You might say it is vacuously transitive. Every number greater than $1$ has another number it is paired with, which is the only number it is related to besides itself. Numbers $\le 1$ are only related to themselves. So nothing is related to two different numbers other than itself. – Zoe Allen Apr 19 '24 at 22:23
  • Wait my comment was if the relation was defined on the positive reals. For the positive integers, $4$ and $2$ are the only two distinct numbers related to each other. Again making it vacuously transitive. – Zoe Allen Apr 19 '24 at 22:26
  • I don’t think the second proof is valid because the implication is not correct. It should either say $z^y$ or $z^{x^{\frac{y}{x}}}$ on the RHS. – Egor Larionov Apr 19 '24 at 22:28

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So, the biggest underlying problem here is that the solutions to the equation $x^y = y^x$ are only at $x=y \in \mathbb{Z}_{\ne 0}$ and $(x,y) = (2,4)$ and $(x,y) = (4,2)$. That is, $$ R = \{(2,4),(4,2)\} \cup \{(n,n) \mid n \in \mathbb{Z}_{\ne 0}\} $$ So ultimately, there is no nontrivial pairing $(x,y),(y,z) \in R$ to work with. If $y \in \{2,4\}$, then you have pairs of the type $$ (x,2),(2,z) $$ let's say. But then $x,z \in \{2,4\}$, and all possible pairs $(x,z)$ from that set work, so no violations can occur there. It's similar for $y=4$. And if $y$ is anything else, then $x$ and $z$ are forced to share the same value as $y$ to have $(x,y)$ and $(y,z)$ in $R$ in the first place.

So ultimately, yes, the relation is transitive.

Solving $x^y=y^x$ is ultimately a well-known problem, c.f. some past discussion on MSE here or Wikipedia here.

With your first proof, I take umbrage with the use of the logarithm since that implicitly assumes your numbers are positive; $\ln(-1)$, for instance, is not well-defined in the reals.

The second proof sketches me out for similar reasons, by making assumptions on exponent laws and how they work for negative numbers. (What if $y/x$ was a fraction, for instance?) I don't think this proof arrived at a convincing enough conclusion to determine (non)transitivity either.

PrincessEev
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