# Is this relation symmetric? I think it is but my professor claims it isn't.

Is the relation $$R = \{(a,b),(a,c),(b,a),(b,c),(c,a),(c,b),(d,d)\}$$ symmetric?

My professor claims that if $$(d,d)$$ was not included, it would be symmetric, but the inclusion of $$(d,d)$$ ruins it because $$d$$ has to connect to another element of the relation.

• Two of the defining properties of equivalence relations are that they are both reflexive and symmetric, which would be useless if they were mutually exclusive :-P Sep 24 '21 at 4:12
• Hope your professor is willing to get corrected in this case! Sep 24 '21 at 8:39
• You could tell your professor that $(d,d)\in R$ "connects" to itself, ie, for the element $(\color{red}d,\color{green}d)$ in $R$, we have the element $(\color{green}d,\color{red}d)$ in $R$. There is no restriction that it must "connect" to a distinct element in $R$. As an example, the relation $R=\{(a,a)\}$ on the singleton set $\{a\}$ is an equivalence relation: reflexive, symmetric and transitive. Sep 24 '21 at 19:16

A very simple way to see if a relation $$R$$ is symmetric is to check its inverse relation $$R^{-1}$$, where $$(x,y)\in R\Leftrightarrow (y,x)\in R^{-1}$$

Since for the given relation, $$R=R^{-1}$$, it is symmetric.

$$\color{green}{\text{YOU ARE CORRECT.}}$$

Yes, that relation is symmetric. The definition of symmetry does not require each element to be connected to some other element; $$R$$ is symmetric iff for every $$x,y$$ such that $$(x,y)\in R$$ it is also the case that $$(y,x)\in R$$.

One way to see if a relation is symmetric is to draw the table what is in relation to what:

$$\begin{array}{r|c|c|c|c}R&a&b&c&d\\\hline a&F&T&T&F\\b&T&F&T&F\\c&T&T&F&F\\d&F&F&F&T\end{array}$$

and just see if the table is symmetrical across the main diagonal.

And - in this case it is. Thus, the relation is symmetric. Having $$(d,d)$$ in it has nothing to do with it, it would be symmetric either way (i.e. whether the entry for $$(d,d)$$ is "true" or "false").