Do irrational numbers have equivalence classes the way rational numbers do? Rational numbers are defined as equivalence classes of ordered pairs (less formally, "fractions") of integers, where $m_1n_2=m_2n_1$. This equivalence relation justifies the common practice of multiplying the numerator and denominator of a rational number by a common factor, $x$, to get another member of the equivalence class.
However, irrational numbers are often written like rational numbers (e.g., $\frac{\sqrt3}{2}$ often seen in trig), and it's common practice to multiply the numerator and denominator by a common factor, x, to get what's presumably an equivalent (irrational) number (e.g., $\frac{2\sqrt3}{4}$).
But if we do this, aren't we treating an irrational number like $\frac{\sqrt3}{2}$ as if it were a rational number? Where's the justification for this?

We're not treating $\frac{\sqrt{3}}{2}$ as a rational number. The notation $\frac{a}{b}$ is not reserved for rational numbers. For $a, b \in \mathbb{R}$, $b \neq 0$, $\frac{a}{b}$ is a perfectly well-defined number (it is the unique solution to $bx = a$). If $a$ and $b$ are integers ($b$ non-zero), then such a number is rational. 
This is in response to the question in the title. One way we can obtain the real numbers (in particular, the irrationals) from the rationals is to consider Cauchy sequences of rational numbers - these are sequences which should converge, but don't necessarily. We say two sequences are equivalent if they should converge to the same thing. We then take the equivalence classes of such sequences to be the elements of $\mathbb{R}$, much like we take equivalence classes of elements of $\mathbb{Z}\times\mathbb{Z}^*$ under the appropriate equivalence relation to be the elements of $\mathbb{Q}$. So an irrational number can be considered as an equivalence class of rational Cauchy sequences. For more details on this contruction, see the corresponding Wikipedia section.
