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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.

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  • $\begingroup$ That is, you are having trouble saying why the relation $R$ is symmetric,reflexive and transitive? Once you have described the properties, it is only a question of substituting into the definition and checking whether it holds. $\endgroup$ – Teresa Lisbon Jan 18 '18 at 23:05
  • $\begingroup$ I think you have the letters a), b), c), and d) mixed up in your question. You seem to know the answer to a) and c), (not d), and want help for b) and d) (not a and d). $\endgroup$ – Steve Kass Jan 18 '18 at 23:07
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    $\begingroup$ @астонвіллаолофмэллбэрг The problem is, I know how to check those properties for simple relations (i.e. checking that $xRy \rightarrow yRx$), but I don't understand how to do it in this example, because I don't really understand the relation. What does "Let $R⊆S^2$ be defined by: $(a,b)R(c,d)↔ad=bc$" actually mean? How is it input into the simple reflexive ($xRx$), symmetric ($xRy \rightarrow yRx$) and transitive ($xRy, yRz \rightarrow xRz$) formulas I already know? $\endgroup$ – Shannon Jan 18 '18 at 23:07
  • $\begingroup$ @SteveKass Thank you, updated. $\endgroup$ – Shannon Jan 18 '18 at 23:08
  • $\begingroup$ Strictly speaking (d) is not well written. The elements of $R$ are pairs of pairs $((a,b),(c,d))$, and only those pairs where $ad=bc$. The function $f$ should be $f((a,b),(c,d))=a/b$. $\endgroup$ – orole Jan 18 '18 at 23:11
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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 $\mathbb Q$

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  • $\begingroup$ Such a fantastic answer, thank you! One question though: why does the question specify that $R⊆S^2$ when it doesn't seem to be used throughout the question? $\endgroup$ – Shannon Jan 18 '18 at 23:27
  • $\begingroup$ In my proof of transativity, I factored out and dropped the $c.$ I could not do that if $c = 0$ however by the definition of $S^2, c \ne 0$ and is not a concern. I suppose I should have included this note at that step. $\endgroup$ – Doug M Jan 18 '18 at 23:33
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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.

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

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