## Question:

The following is a ‘proof ’ that reflexivity is an unnecessary axiom for an equivalence relation.
Because of symmetry X ∼ Y implies Y ∼ X. Then by transitivity X ∼ Y and Y ∼ X imply X ∼ X. Thus symmetry and transitivity imply reflexivity, which therefore need not be separately required.
Demonstrate the flaw in this proof using the set consisting of all real numbers plus the number i. Show by investigating the following specific cases that, whether or not reflexivity actually holds, it cannot be deduced from symmetry and transitivity alone.
(a) X ∼ Y if X + Y is real.
(b) X ∼ Y if XY is real.

## Step-by-step

Let elements X and Y be drawn from the set S consisting of the real numbers together with i.
(a) For the definition X ∼ Y if X + Y is real, we have
(i) that
X ∼ Y ⇒ X + Y is real ⇒ Y + X is real ⇒ Y ∼ X,
i.e symmetry holds;
(ii) that if X ∼ Y then neither X nor Y can be i and, equally, if Y ∼ Z then neither Y nor Z can be i. It then follows that X + Z is real and X ∼ Z, i.e.
transitivity holds.
Thus both symmetry and transitivity hold, though it is obvious that $X \nsim X$ if X is i. Thus symmetry and transitivity do not necessarily imply reflexivity, showing that the ‘proof’ is flawed – in this case the proof fails when X is i because there is no distinct ‘Y ’ available, something assumed in the proof.
(b) For the definition X ∼ Y if XY is real, we have
(i) that
X ∼ Y ⇒ XY is real ⇒ YX is real ⇒ Y ∼ X,
i.e symmetry holds;
(ii) that if X ∼ Y then neither X nor Y is i. Equally, if Y ∼ Z then neither Y nor Z is i. It then follows that XZ is real and X ∼ Z, i.e. transitivity holds.
Thus both symmetry and transitivity hold and, setting Z equal to X, they do imply the reflexivity property for the real elements of S. However, they cannot establish it for the element i – though it happens to be true in this case as ${i}^{2}$ is real.