Proving Transitivity

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Consider a relation defined by $Z$ where $(a,b) = 2a^2 + b^2 -3ab = 0$

Is the relation $R_1$ reflexive? symmetric? transitive? Is it an equivalence relation?

I have said that it is reflexive as $$2(a)^2 + a^2 - 3(a)(a) = 0$$ $$3a^2 - 3a^2 = 0$$ $$0= 0$$

I said it wasn't symmetric and I proved it via an example where the element is (2,1)

$$ 2(2)^2 +(1)^2 -3(2)(1) ≠ 2(1)^2 + (2)^2 - 3(1)(2)$$ $$8+1-6 ≠ 2+ 4 -6$$ $$3≠0$$

I don't really know how to prove transitivity. I understand it that if (a,b) ∈ $R_1$ and (b,c) ∈ $R_1$ then (a,c) ∈ $R_1$.

Also, I understand that this relation cannot be an equivalence relation as it is not reflexive $and$ symmetric $and$ transitive.

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The relation of interest is not transitive, because $R_1(1,2)$ and $R_1(2,4)$ are true, but $R_1(1,4)$ is false.

By the way, a good specification of $R_1$ would be explicit about the values that the arguments can take. For example, you could use set-builder notation:

$$R_1 = \{(a,b) \in \mathbb{Z} \times \mathbb{Z} \mid 2a^2+b^2−3ab=0\}$$

Also, if you're wondering how to tackle these problems in future, try using a computer program to find integer-valued counterexamples. My method was basically to put the formula into a spreadsheet, and toy with the numbers for 30 seconds. And viola, out popped a counterexample.