Confused on equivalence relationships with the equation $y=x^2$ Help me get my head on straight, please!  Isn't the equality sign an equivalence relationship by definition?  If so, then if a relationship R is defined as $y=x^2$ (I don't believe which universe matters to the question could be Reals or Integers take your pick), then why doesn't it seem to be an equivalence relationship?  If I had to form partitions from this relationship wouldn't this have to be an equivalence class?
Is $y=x^2$ reflexive?
Isn't the answer No.  Counter-example $(2,2) \not \in$ the relationship.
Is $y=x^2$ symmetric?
Once again no.  Counter-example $(4,2) \in$ R but $(2,4) \not \in$ R.
Is $y=x^2$ transitive?
And no. $(16,4) \in$ R and $(4,2) \in$ R, but $(16,2) \not \in$ R.
Thank you.
 A: The equal sign in $y= x^2$ does not say that $y$ is the same thing as $x^2$. It says that you are thinking about a real valued function of a real variable (namely, "square the input"), using $x$ for a typical number in the domain and $y$ for the corresponding value in the range. This is a nice convention if you want to think about graphing the function. It's not particularly useful when you think of the function as a set of ordered pairs. That set of ordered pairs is not an equivalence relation.
So the underlying problem is that mathematicians sometimes use $=$ in a possibly confusing way. When the equality sign is used to say two things are the same then it is denoting an equivalence relation. You have to pay attention to the context.

Long edit in response to comment.
Here's a formal way to make sense of the $=$ sign in the colloquial (but well understood if you don't worry too much) $y=x^2$. 
The first step is to say what $x$ and $y$ might actually be. Calling them "variables" begs the question. I suggest that a good way to define them is as the projection functions from $\mathbb{R} \times \mathbb{R}$ to the first and second factors $\mathbb{R}$ in that cartesian product.  Then the function $x^2$ is the pointwise product of $x$ with itself: $x^2(t,s) = x(t,s)^2 = t^2$.
Now let $f: \mathbb{R} \to \mathbb{R}$ be the function that squares its input. If you think of a function as a set of ordered pairs then $f$ is its graph:
$$
f = \{(t,s) \in \mathbb{R} \times \mathbb{R} \ | \ s = t^2 \}.
$$
Finally, restrict the functions $y$ and $x^2$ to the set $f$. Those two restriction are literally equal: they are the same function. As a set of ordered pairs, both are
$$
\{ ((t,s),c) \in f \times \mathbb{R} \ | \  c = s \}.
$$
I chose not to offer this clumsy formalism as an answer since I didn't think it would help the OP with his question.
This analysis is a special case of an idea I learned from Andy Gleason years ago about a good way to think of "variables" in a more general context -  for example, the classic problem about a freely falling body. There the fundamental equation is $s = 16t^2$ (in elementary calculus) or $d^2s/dt^2 = a$ (in differential equations or physics). What formal meaning might the variables $s$ and $t$ have here? Andy said you should think of them as real valued functions on the abstract state space of the physical system. Then the laws of physics are identities satisfied by these functions:  when the variables are combined in particular ways they define the same function on state space - in this example, our friend $s = 16t^2$.
