Why isn't reflexivity redundant in the definition of equivalence relation? An equivalence relation is defined by three properties: reflexivity, symmetry and transitivity.
Doesn't symmetry and transitivity implies reflexivity? Consider the following argument.
For any $a$ and $b$,
$a R b$ implies $b R a$ by symmetry. Using transitivity, we have $a R a$.
Source: Exercise 8.46, P195 of Mathematical Proofs, 2nd (not 3rd) ed. by Chartrand et al
 A: Actually, without the reflexivity condition,   the empty relation would count as an equivalence relation, which is non-ideal.
Your argument used the hypothesis that for each $a$, there exists $b$ such that $aRb$ holds. If this is true, then symmetry and transitivity imply reflexivity, but this is not true in general.
A: No.
The missing condition is sometimes called  'seriality' -- for any x there must be an y such that x R y.
If you add seriality to the symmetry and transitivity you get a reflexive relation again.
A: You can get kind of rid of reflexivity.  Assume the $R$ is a symmetric relation which satisfies the property of "Drittengleichheit": 
$xRz\land yRz\Rightarrow xRy$. In this case  $R$ is a equivalence relation; you can easily deduce transitivity and reflexibility of $R$.
Do you notice the difference between "Drittengleichheit" and transitivity?
A: To answer Muno's question, $R$ on $S$ is not transitive: $x=z=1$ and $y=3$ is a counterexample (any $y$ other than $1$ will do). $x+y=y+z>3$ but $x+z<3$.  
silvascientist: $R$ satisfies Drittengleichheit if the implication is true; it is vacuously true if $xRz$ and $yRz$ is false. This does not mean that $xRy$ is true  (similarly $xRy$ and $xRy$ false does not mean $xRx$ is true, simply that it vacuously satisfies the property) for the empty relation on a non-empty set. The same applies if $x$ isn't related to anything, so in neither case is $R$ reflexive. 
A: Consider the set $S =\{a,b,c\}$, with $a, b$, and $c$ distinct, and the relation $$R = \{(a,c),(c,a),(a,a),(c,c)\} \subset S \times S$$
It's symmetric and transitive but it's not reflexive.
Added 3/4/2021
The argument $aRb \implies bRa \implies aRa$ is compelling except for one thing. What if there is no $b$ such that $aRb$?
A: Reflexivity always requires for the binary relation to include every element of the set, which it may not even if transitivity and symmetry are both valid for the relation. However, it may be redundant if every element of the set is said to have a relative or the relation is redefined to be only on the subset of current set leaving out the excluded members for whome the relation doesn't map to anywhere. 
Although sometimes the definition of the relation itself might restrict the reflexivity, for example: the pair (a,b) where 'a' is b's brother.
