Problem:

$\displaystyle nRm$ in $\displaystyle \mathbb{Z}$ if $\displaystyle nm > 0$. Determine if this is an equivalence relation.

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I know that an equivalence relationship requires it to be reflexive, symmetric and transitive. I know how to do reflexive and symmetric, but I'm having some trouble with transitive. I think I may have gotten it, or maybe I'm close.

Here's my attempt:

First introduced $\displaystyle p$ as a third variable for the transitive test. I said that if $\displaystyle nRp$ and $\displaystyle pRm$ were true, then $\displaystyle np>0$ and $\displaystyle mp>0$.$\displaystyle (nRp \land pRm) \to (np>0 \land pm>0)$

Since $\displaystyle np>0$ as proven in the first step by assuming $\displaystyle nRp$, and likewise for $\displaystyle mp$, n, p and m are all not zero.$\displaystyle (np>0) \to (n \neq 0 \land p \neq 0)$

$\displaystyle (mp>0) \to (m \neq 0 \land p \neq 0)$

And therefore $\displaystyle nRp$ and $\displaystyle pRm$ are both true. Referring to the definition of a transitive relation that says $\displaystyle (xRy \land yRz) \to (xRz)$, I say that this IS a transitive relation because both of the required relations in the implication of the definition (replacing x with n, y with p and z with m) are true, and therefore xRz is true, or in my case, nRm, which it what I was out to prove.

But now that I look at it more closely... what if p was negative? Then this wouldn't be transitive. I think I'm assuming something I'm not supposed to assume.

I was looking at an example in my book, and they are proving that it's not transitive if it's simply nm >= 0. They said that if aRb and bRc, then ab>=0 and bc>=0. Thus acb^2 >= 0. What allowed them to combine those two expressions to get that b^2? If I know that, I think I can just say that (in my problem) p^2 must be > 0 because all numbers squared are positive except for zero, but I've already proved that it's not zero.

A push in the right direction would be perfect. Please don't just spit out the answer

Thanks.