Question

Let $\mathbf{\square }$ be a symmetric and transitive relation on a set A. What is wrong with the following “proof” that$\mathbf{\square }$ is reflexive:role="math" localid="1659504197825" $\mathit{a}\mathbf{\square }\mathit{b}$ impliesrole="math" localid="1659504216047" $\mathit{b}\mathbf{\square }\mathit{a}$ by symmetry; then$\mathit{a}\mathbf{\square }\mathit{b}$ and$\mathit{b}\mathbf{\square }\mathit{a}$ imply$\mathit{a}\mathbf{\square }\mathit{a}$ by transitivity. [Also see Exercise 8(f).]

Short answer

The given proof is not true.

Step-by-step solution

To prove □ is symmetric

Let$\square$ be a symmetric and transitive relation on a set A.

We need to find where the proof is wrong.

Assume that the proof given is true.

Define the relation$a\square b$ by,

$a-b=c\ne 0..........\left(1\right)$.

Then$\square$ is symmetric.

Thus,

$$\begin{gathered} a\square b\Rightarrow b\square a \\ \therefore a-b\ne 0\Rightarrow b-a\ne 0..........\left(\because by\left(1\right)\right) \end{gathered}$$

To prove that □ is transitive and verify the proof

Now,$\square$ is transitive.

Therefore,

$$\begin{gathered} a\square b,b\square a \\ \Rightarrow a\square a \\ \therefore a\square b\Rightarrow a-b\ne 0, \\ b\square a\Rightarrow b-a\ne 0 \end{gathered}$$

Then $a\square a\Rightarrow a-a\ne 0$.

But a - a = 0,$\forall a$ which is a contradiction.

Thus, our assumption is wrong.

Hence, the given proof is not true.