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I am currently studying for my exams this summer. I have a hard time getting to grips with the following question on discrete mathematics:

Consider the relationship $T$ between ordered pairs of natural numbers such that $(a, b)$ is related to $(c, d):$

$$[(a, b) T (c, d)] \iff ad = bc$$

Is T an equivalence relation?

Community
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    You write down the definition of an equivalence relation. "A relation is an equivalence relation if... ". Then you check whether this relation fulfils all the required conditions. For example, we should have (a, b) equivalent to (a, b). How would you check that? – gnasher729 Apr 15 '17 at 21:32
  • An essential question: is $0$ a natural number for you? (For some people it is, for the others it isn’t.) – Lubin Apr 15 '17 at 22:12

2 Answers2

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So hopefully from your discrete mathematics class, you learned that an equivilance relation $\sim$ (or T if you prefer, the symbol is irrelevant) for some set $A$ satisfies the following properties:

  1. Identity: $\forall a \in A$, $a \sim a$.
  2. Symmetry: $\forall a,b \in A$, if $a \sim b$ then $b \sim a$
  3. Transitivity: $\forall a,b,c \in A$, if $a \sim b, b \sim c$ then $a \sim c$

In your case, $A$ is the set of pairs of numbers $(a,b)$.

Artimis Fowl
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To prove a relation ~ is an equivalence relation, we need to prove:

  • Reflexive: $a$~$a$
  • Symmetric: $a$~$b$ implies $b$~$a$
  • Transitive: $a$~$b$, $b$~$c$ implies $a$~$c$

If we define $(a,b)$~$(c,d) \iff ad=bc$, we then need to check all three properties above.

$(a,b)$~$(a,b)$ is true

$(a,b)$~$(c,d)\implies (c,d)$~$(a,b)$ is true

$(a,b)$~$(c,d)$, $(c,d)$~$(e,f)$ is not true, consider $(1,1),(0,0),(1,2)$

so it is not an equivalence relation on $\mathbb{N^+}\cup \{0\}$, as it is intransitive.

If however we only consider $\mathbb{N^+}$, it is an equivalence relation, as;

$\dfrac ab=\dfrac cd=\dfrac ef$.

JMP
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