In my first post in the Constructing the Reals series, I explained how to create the natural numbers, such as 0, 7, and 23, using only sets, as well as define addition and multiplication. Now, using those tools, we can construct the (non-negative) rationals, like 11/8, 78/7, and 11/1. To avoid repetition, I’ll leave out the word non-negative in the rest of this post.

But first, we have to define an ordered pair, which is exactly what it sounds like. Formally, the ordered pair (a, b) is the set {{a}, {a, b}}. Using this definition, we distinguish the first part of the pair as the object in both sets and the second element as the object in only one set. We can now define the (non-negative) rational number |x/y|, with x and y natural numbers and y ? 0, as the set of ordered pairs of natural numbers (a, b) with� b ? 0 such that ay = bx. Obviously, this definition includes the ordered pair (x, y), since xy = yx. The reason for the bars is that I intend to use the general notation x/y for all rational numbers; they should not be confused with the absolute value operator.

Much like with the non-negative integers, now that we have the rationals, we need to define what you can do with them. Equality again is defined as being the same set; it’s not too hard to show then that |a/b| = |x/y| if and only if ay = bx. For addition and multiplication we basically pull the definition from how we ordinarily work with rationals: |a/b| < |x/y| if and only if ay < bx, |a/b| + |x/y| = |(ay + bx)/by|, and |a/b||x/y| = |(ax)/(by)|.� But we can now define a new operation that didn’t exist in the natural numbers: division. We just define it as usual: |a/b| / |x/y| = |a/b||y/x| = |ay/bx|. Again, these have the properties we usually associate with them.

Now that we have the rationals, the question arises: how do we place each one of the natural numbers as a rational? It turns out that the answer is simple: each natural number x is identified with the rational number |x/1|. If we do this, then 0 maps to |0/1|, which has the property that |a/b| + |0/1| = |(a1 + b0)/(b1)| = |a/b|: |0/1| is the additive identity. Similarly, 1 maps to |1/1|, which satisfies |a/b||1/1| = |(a1)/(b1)| = |a/b|; it’s the multiplicative identity. If you’re familiar with group theory, then the non-negative rationals as we’ve defined them form a group under multiplication with |0/1| excluded: closure is obvious, associativity follows from the associativity of natural number multiplication, since (|a/b||c/d|)|e/f| = |(ab)c/(de)f| = |a(bc)/x(yz)| = |a/b|(|c/d||e/f|), we just showed that |1/1| is an identity, and the inverse of |a/b| is |b/a|: |a/b||b/a| = |ab/ba| = |1/1|.

Strictly speaking, the standard method to construct the rationals from the natural numbers is to define an equivalence relation, ?, between ordered pairs (x, y) of natural numbers (again with y nonzero), and define rational numbers as equivalence classes of these ordered pairs (the equivalence class of an ordered pair (x, y) is the set of all pairs (a, b) with (a, b) ? (x, y)).� But that definition is essentially identical to the one I used. I will use this definition for negative numbers, to give you an idea of what it’s like, but that will have to wait until the next post! Your homework is to prove that multiplication of rational numbers distributes over addition.