Do irrational numbers have equivalence classes the way rational numbers do?

Question

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?

Answer 1

The justification is that $\frac{a}{b}$ means $a*b^{-1}$ from an algebraic point of view ($\mathbb R$ is a division ring). So $\frac{ax}{bx}$ means $ax*(bx)^{-1}$ and the equality is a consequence of $x*x^{-1}=1$.

But it seems that you think about the construction of irrationals. The real numbers can be built with Cauchy sequences of $\mathbb Q$ and explains why $\mathbb R$ is again a division ring.(see more here or here)

And the two definitions of $\frac{a}{b}$ are equivalents on Q because $b*\frac1{b}=\frac{b}{b}=\frac11=1$ so $\frac1{b}=b^{-1}$ and $\frac{a}{b}=a*\frac1{b}=a*b^{-1}$

Answer 2

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.

Answer 3

You’ve mixed up a number of different concepts. It’s a fact of the way fractions (of any kind) behave that top and bottom can be multiplied by the same nonzero quantity ($\star$) without changing the value of the fraction. This has nothing specific to do with rational numbers, it’s good for fractions where top and bottom are functions, for instance. It’s really a reflection of the fact that division (represented by the horizontal line) is an undoing of the process of multiplication.

The commonest construction of the rationals is to set them up as equivalence classes of pairs (top, bottom) of nonzero whole numbers. The rules for arithmetic manipulation of these pairs are designed so that they accord with the behavior of quantities in any system where you can add, subtract, multiply, and divide (“field”). There could conceivably be other ways of defining the rationals without using the familiar construction that you’re referring to. As long as the ultimate reality turns out to be the same, the particular construction doesn’t matter.

Real numbers are usually constructed by a process of forming equivalence classes, just as the rationals are constructed as classes of pairs, but the process is very different, and indeed there are several equivalent but very different constructions, all giving rise to the same ultimate reality (the constructs turn out to be “isomorphic”).

($\star$) Strictly speaking, you can multiply only by quantities that have a reciprocal in your system (“units in the ring”).