Question involving proving equivalence relations

Question

In order to prove if a relation is an equivalence relation, it needs to be show that is all of:

• Reflexive
• Symmetric
• Transitive

Whilst I am familiar with this, I am unsure how to approach the following set of questions:

State and explain whether each of the following relations $R$ is an equivalence relation.

1. The relation $R\subseteq(\mathbb{Z}\times\mathbb{Z} \ \backslash \ \{0\}) \times (\mathbb{Z}\times\mathbb{Z} \ \backslash \ \{0\}) $ is defined via $((a,b),(c,d))\in R$ if, and only if, $ad=bc$.
2. The relation $R\subseteq(\mathbb{Z}\times\mathbb{Z}) \times (\mathbb{Z}\times\mathbb{Z} ) $ is defined via $((a,b),(c,d))\in R$ if, and only if, $ad=bc$.

The main thing I am confused about is how to deal with relations involving cartesian products of cartesian products. To prove it is reflexive, am I supposed to prove all $$((a,a),(a,a))$$ exists within the relation, if and only if $a^2=a^2$? And I'm not sure how to begin with the others. Thanks for any tips.

Answer 1

Concerning the first binary relation, not that the elements of $\mathbb Z\times(\mathbb Z\setminus\{0\})$ are pairs $(a,b)$ (with $a,b\in\mathbb Z$ and $b\neq0$, not pairs $(a,a)$. Now:

• Reflexive: $\bigl((a,b),(a,b)\bigr)\in R$, because $ab=ab$.
• Symmetric: It is symmetric because the definition itself is symmetric.
• Transitive: Suppose that $\bigl((a,b),(c,d)\bigr),\bigl((c,d),(e,f)\bigr)\in R$. This means that $ad=bc$ and that $cf=de$. Do we have $\bigl((a,b),(e,f)\bigr)\in R$? In other words, is it true that $af=be$. Yes, because$$\frac ab=\frac cd=\frac ef.$$

Can you deal with the other one now? Hint: It is not an equivalence relation.

Answer 2

How to deal with cartesian products of cartesian products:

Let $\Omega := \mathbb{Z}\times\mathbb{Z}$ (or, respectively, $\mathbb{Z}\times\mathbb{Z}\setminus \{0\}$).

For reflexivity: You are supposed to prove (or disprove) $((a,b),(a,b))\in R$ for all $(a,b)\in\Omega$.

For symmetry: Prove/Disprove $((a,b),(c,d))\in R \iff ((c,d),(a,b))\in R$ for all $(a,b),(c,d)\in\Omega$.

For transitivity: Prove/Disprove $((a,b),(c,d))\in R$ and $((c,d),(e,f))\in R$ implies $((a,b),(e,f))\in R$ for all $(a,b),(c,d),(e,f)\in\Omega$.

José has already taken care of the three points above for your particular relation from 1., so I won't do that here.

Answer 3

You can get around the "cartesian products of cartesian products" confusion by writing the relation the way you write "$=$", so $$ (a,b) R (c,d) \iff ad = bc . $$ Now the only cartesian product involves the pairs that may or may not be related. So to show that the relation is reflexive you need to show that for every pair $(a,b)$, $$ (a,b) R (a,b). $$

$xRy$ rather than $(x,y) \in R$ is in fact how equivalence relations (and order relations) are usually written in practice. Similarly, functions are often written as $y = f(x)$ rather than $(x,y) \in f$ even though, formally, a function is just a relation with certain properties.