Showing that a relation is reflexive, symmetric and transitive

Question

I am attempting the following question:

Let $S = \Bbb Z$ x $(\Bbb Z $ \ $\{0\})$, be pairs of integers where the second coordinate is non-zero. Let the relation $R \subseteq S ^2$ be defined by:

$(a,b)R(c,d) \leftrightarrow ad = bc$

(a) Describe the following properties of relations: reflexive, symmetric and transitive.

(b) Show that $R$ has these properties.

(c) What is the name for relations satisfying the properties in (a).

(d) Is the function $f: R\times R \rightarrow \Bbb Q$ defined by $f(a,b) = > a/b$ bijective?

I know how to decribe the properties of relations, so question (a) shouldn't be an issue.

I believe the answer to question (c) is "an equivalence relation".

However, I am completely lost as to where I start with questions (b) and (d). I have been trying to understand relations for weeks, but just can't seem to wrap my head around it.

Any help explaining (b) and (d) would be greatly appreciated.

Answer 1

What does this relation actually mean?

It really is all about the equivalence of fractions. $\frac {a}{b} = \frac dc \implies ac = bd$

Now, instead of writing these fractions in the more familiar form we show them as and ordered pair.

a,b)

reflexive:

$(a,b)R(a,b) \iff ab = ba$

symmetric:

$(a,b)R(c,d) \implies (c,d)R(a,b)$

$ad = bc \implies cb =da$

Transitive:

$(a,b)R(c,d)$ and $(c,d)R(p,q) \implies (a,b)R(p,q)$

$ad = bc$ and $cq = dp$

$adp = bcp\\ acq = bcp\\ aq = bp$

d)

Is the function $f(a,b) = \frac ab$ bijective?

It is not injective as $(1,2)$ and $(2,4)$ both map onto the same element of $\mathbb Q$

Answer 2

The problem is, I know how to check those properties for simple relations (i.e. checking that $xRy→yRx$), but I don't understand how to do it in this example, because I don't really understand the relation.

$\def\R{\operatorname R}\R$ is symmetric if $\forall s\in S\;\forall t\in S\; [s\R t\to t\R s]$.   Now $S=\Bbb Z\times(\Bbb Z\setminus \{0\})$, and $\forall (a,b)\in S\;\forall (c,d)\in S\;[(a,b)\R(c,d)\leftrightarrow (ad=cb)]$ so that would be $\forall (a,b)\in S\;\forall (c,d)\in S\;[(ad=cb)\to (cb=ad)]$.   Because this is true (via the symmetry of equality), therefore $\R$ is symmetric.

Use similar principle to check for Reflexivity and Transitivity.

$\\[3ex]$

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Hint: You might find it helpful that $(ad=cb)\leftrightarrow (a/b=c/d)$ if $b\neq0, d\neq 0$.