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The question is:

Let $A = \{\text{cat, dog, mouse, bird}\}$ and let $R$ be the binary relation on $A$ given by $R = \{(x,y) \mid \text{$x$ and $y$ have no letter in common}\}$.

(a) Draw $R$.
(b) State whether or not $R$ is an equivalence relation, explaining your answer.

I have done both parts of the question, and wanted to confirm if the answers and my reasoning behind the answers are correct, and would be acceptable for an exam answer.

(a)

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(b)

No, it is not an equivalence relation, as the relation is not reflexive. An example contradiction would be $R(\text{dog, dog})$, which is not reflexive, because they have letters in common.

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  • I find it perfect. – ajotatxe Jan 18 '18 at 19:42
  • Basically correct. I would reword to say “$R(dog,dog)$, which is not true, because dog and dog have (all three) letters in common.” – Steve Kass Jan 18 '18 at 19:42
  • @SteveKass I find irrelevant that all letters are common. A single common letter prevents the relationship. – ajotatxe Jan 18 '18 at 19:45
  • Yes, of course. I think the parenthetical comment could be enlightening to a reader learning the material for the first time. If the aim of the proof is to convince a more seasoned audience, it could be left out. That's my opinion as a teacher, however. – Steve Kass Jan 18 '18 at 23:05
  • One may note that it also cannot be repaired simply by extending $R$ to include all pairs $(x,x)$, either, as $R$ is also not transitive: $R(\text{mouse}, \text{cat})$ and $R(\text{cat}, \text{dog})$, but it is not true that $R(\text{mouse}, \text{dog})$. However, this is extra, simply noting it isn't reflexive is enough. – Paul Sinclair Jan 19 '18 at 00:47

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