# Thread: Determine is relation is transitive

1. ## Determine is relation is transitive

Problem:

$nRm$ in $\mathbb{Z}$ if $nm > 0$. 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 $p$ as a third variable for the transitive test. I said that if $nRp$ and $pRm$ were true, then $np>0$ and $mp>0$.
$(nRp \land pRm) \to (np>0 \land pm>0)$
Since $np>0$ as proven in the first step by assuming $nRp$, and likewise for $mp$, n, p and m are all not zero.
$(np>0) \to (n \neq 0 \land p \neq 0)$
$(mp>0) \to (m \neq 0 \land p \neq 0)$
And therefore $nRp$ and $pRm$ are both true. Referring to the definition of a transitive relation that says $(xRy \land yRz) \to (xRz)$, 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.

2. ## Re: Determine is relation is transitive

Originally Posted by tangibleLime
Problem:
$nRm$ in $\mathbb{Z}$ if $nm > 0$. Determine if this is an equivalence relation.
Is this true $0R0~?$
If no then why do any more?

3. ## Re: Determine is relation is transitive

Originally Posted by Plato
Is this true $0R0~?$
If no then why do any more?
It might be a useful exercise...

4. ## Re: Determine is relation is transitive

Originally Posted by TheChaz
It might be a useful exercise...
But that is not what was posted.
I think that it is important to deal only what was actually posted.
Otherwise, this forum could dissolve into wild guessing.

5. ## Re: Determine is relation is transitive

Originally Posted by Plato
But that is not what was posted.
I think that it is important to deal only what was actually posted.
Otherwise, this forum could dissolve into wild guessing.
...or into speculation about the future of the forum

I think the exercise would be useful.

6. ## Re: Determine is relation is transitive

Oooh, I think I was assuming that we were supposed to assume that nm > 0 was true. Now I see that it's not necessarily true... and if it's false, then something like 0R0 can arise... correct?

7. ## Re: Determine is relation is transitive

There is a time to make such an assumption.

Usually, we prove that - FOR ALL elements "x" in the set - xRx.
This is not true, since 0R0 <=> (0*0 > 0)

Since reflexivity fails, there is no need to continue.

However, suppose that we are dealing with the integers (or reals) except (set difference) 0.

Then it is indeed true that for all nonzero integers "x", xRx since x*x > 0

For symmetry, we need to prove that
(xRy) => (yRx)
And HERE we DO assume the hypothesis (that x is related to y) and show that the conclusion (yRx) is true
Assume xRy <=> x*y > 0
Well, since integers commute in multiplication, this is equivalent to (y*x > 0) <=> yRx as desired.

For transitive, we would assume that xRy AND that yRz
xy > 0 AND yz > 0
We multiply these together to get
xy*yz > 0
x(y^2)z > 0
Since y^2 is positive, so is xz. QED

8. ## Re: Determine is relation is transitive

Originally Posted by tangibleLime
Oooh, I think I was assuming that we were supposed to assume that nm > 0 was true.
Without saying what n and m are, this statement is meaningless. It is true for some n, m and false for others. Usually when people say that some statement P(n, m) is true without specifying n and m, they mean "For all n and m it is the case that P(n, m) is true." But if you mean this, then the relation R is total, i.e., nRm holds for all n and m. Then it is automatically reflexive, symmetric and transitive (e.g., transitivity holds just because the conclusion is true regardless of the premise).

Moral: always define the objects you use (like n and m above).

9. ## Re: Determine is relation is transitive

Originally Posted by emakarov
Without saying what n and m are, this statement is meaningless.
Moral: always define the objects you use (like n and m above)
Actually it was defined in the OP.
Originally Posted by tangibleLime
Problem:
$nRm$ in $\color{blue}\mathbb{Z}$ if $nm > 0$. Determine if this is an equivalence relation.

10. ## Re: Determine is relation is transitive

I am still not sure which n and m in Z the OP had in mind when he/she said "I was assuming that we were supposed to assume that nm > 0." Either "nm > 0" is not a proposition and cannot be assumed, or it stands for "for all n and m, nm > 0," i.e., "for all n and m, nRm." In the latter case, as I said, the problem is trivial since a total relation is an equivalence relation.