Is it possible to prove reflexive, symmetric and transitive properties of equality and the transitive property of inequality? This may be a bit of a trivial question, but can one prove the reflexive, symmetric and transitive properties of equality and the transitive property of inequality of real numbers?(and if so, how? Is there a fairly straightforward, possibly algebraic method?) i.e. Is it possible to prove that $\forall\;a, b\in\mathbb{R}$
$$a=a,$$ $$\text{if } a=b \text{ then } b=a,$$  and $$\text{if } a=b \text{ and } b=c \text{ then } a=c.$$  Also, $$\text{if } a<b \text{ and } b<c \text{ then } a<b$$
The sources that I've read so far seem to have conflicting points of view. 

There is a construction whereby every real number is really a set of things. (For the purposes of proving the equality properties, it's not material what the construction is.) We then rely on basic properties of sets.
Now, for any given sets $x$ and $y$, if we can show they have the same elements, then they are equal. Let $a,b,c \in \mathbb{R}$. Now, treating these as sets, we know that $a$ has the same elements as itself, and thus $a = a$. 
Further, if $a = b$, then $a$ and $b$ have the same elements. But then $b$ and $a$ have the same elements, and hence, $b=a$. (And vice-versa.)
You can imagine the transitive argument.
To address the inequality, simply note that if $x < y$ if and only if there exists $z \in \mathbb{R}^+$ such that $x + z = y$. 
