Equivalence Relation Proof? I can't seem to prove this equivalence relation statement (Tools of the Trade, Paul Sally): 
Let $R$ be a relation on $X$ that satisfies:
(a) For all a that are elements of $X$, $(a,a)$ is an element of $R$;
(b) For all $a,b,c$ that are elements of $X$, if $(a,b)$, $(b,c)$ that are elements of $R$, then $(c,a)$ is an element of $R$
It just doesn't seem true to me that $R$ MUST be an equivalence relation. 
 A: Property (a) already says that $R$ is reflexive, so in order to prove that $R$ is an equivalence relation, you need only show that $R$ is symmetric and transitive. For symmetry, suppose that $\langle x,y\rangle\in R$; you want to show that $\langle y,x\rangle\in R$. From (a) you know that $\langle x,x\rangle\in R$, and you’re assuming that $\langle x,y\rangle\in R$, so (c), applied with $a=b=x$ and $c=y$, tells you that $\langle y,x\rangle\in R$. This shows that $R$ is symmetric.
Now see if you can use symmetry together with property (b) to show that $R$ is transitive.
A: I take it the problem is to show that $R$ is an equivalence relation?
Say $(a,b)\in R$. Now (a) shows that $(a,a)\in R$. So you have $(a,a),(a,b)\in R$, hence (b) shows $(b,a)\in R$.
That's one of the three parts of the definition, and now that we know $(a,b)\in R$ implies $(b,a)\in R$, the third part follows from (b).
A: A relation is an equivalence relation if it is reflexive, symmetric and transitive. Clearly (a) is reflexivity and (b) is transitivity, so we just have to check symmetry. $(b,b) \in R$ by (a), and then if $(a,b)$ and $(b,b)$ are in $R$, so is $(b,a)$ by (b). Hence for every $a$ and $b$, if $(a,b) \in R$, $(b,a) \in R$, which is symmetry.
A: $(a,a)$ and $(a,b)$ and b) gives $(b,a)$ so symmetry is proven.
$(a,b)$ and $(b,c)$ and b) gives $(c,a)$. Using a) gives $(a,c)$ so transitivity is proven.
A: *

*reflexive : (given) all $\left(a,a\right)$ belongs to $R$.

*symmetric : if$\left(a,b\right)$ belongs to $R$, we know that $\left(a,a\right)$ also belongs to $R$, so by the second definition $\left(a,a\right)$ and $\left(a,b\right)$ belongs to $R$ implies $\left(b,a\right)$ belongs to $R$. so it is symmetric. [$\left(a,b\right)$ belongs to $R$ implies $\left(b,a\right)$ belongs to $R$]

*transitive : if$\left(a,b\right)$ , $\left(b,c\right)$ belongs to $R$ then by second definition $\left(c,a\right)$ belongs to $R$ . also we proved the $R$ is symmetric so if$\left(c,a\right)$ belongs to $R$ then $\left(a,c\right)$ also belongs to $R$.  [$\left(a,b\right)$ and $\left(b,c\right)$ belongs to $R$ implies $\left(a,c\right)$ belongs to $R$.]

