Rational Numbers

Why Extend the Integers?

The integers make subtraction possible, but division can still lead outside the number system. For example, no integer multiplied by \(2\) gives \(1\). The rational numbers fill this gap by including ratios of integers with nonzero denominators. We write \(\mathbb Q\) for the set of rational numbers.

A ratio such as \(2/3\) has many equivalent descriptions: \(4/6\), \(-2/(-3)\), and \(10/15\) all describe the same number. To construct the rationals precisely, we begin with pairs \((a,b)\) of integers, where \(b\ne 0\), and specify when two pairs represent the same number. As in the construction of the integers, the numbers will be equivalence classes rather than individual pairs.

We use the familiar arithmetic and order laws for the integers, including commutativity and associativity, distributivity, cancellation, and the fact that a product of nonzero integers is nonzero. In particular, if \(k\ne0\) and \(kx=ky\) for integers \(x,y\), then \(x=y\).

The denominator must be nonzero because division by zero is not defined. The equality test \(ad=bc\) uses only integer multiplication. For instance, \(2/3\) and \(4/6\) represent the same rational number because \(2\cdot6=3\cdot4=12\). The notation \(a/b\) is convenient, but the precise object is the class \([(a,b)]\).

Equivalent Pairs and Fraction Equality

Proof. For reflexivity, let \((a,b)\) be any allowed pair. Since \(ab=ba\), the equality test gives \((a,b)\sim(a,b)\).

For symmetry, suppose \((a,b)\sim(c,d)\). Then \(ad=bc\). Reversing the equality gives \(cb=da\), so \((c,d)\sim(a,b)\).

For transitivity, suppose \((a,b)\sim(c,d)\) and \((c,d)\sim(e,f)\). Thus \[ ad=bc\qquad\text{and}\qquad cf=de. \] Multiplying the first equality by \(f\) gives \(adf=bcf\). Using \(cf=de\), the right side becomes \(bde\). Therefore \(adf=bde\), or \(d(af)=d(be)\). Since \(d\ne0\), cancellation gives \(af=be\). Hence \((a,b)\sim(e,f)\). All three properties hold, so \(\sim\) is an equivalence relation. \(\square\)

The theorem guarantees that the equivalence classes fit together consistently: every pair is equivalent to itself, equivalence can be reversed, and equivalence can be passed along a chain. The resulting equality criterion is

Arithmetic on Rational Numbers

The familiar fraction rules arise from operations on the representing pairs. Addition uses a common denominator, while multiplication multiplies corresponding entries. We must ensure that changing an equivalent representative does not change the result; otherwise these formulas would define operations on pairs, but not on rational numbers.

Both formulas produce allowed pairs: \(b\ne0\) and \(d\ne0\), so the integer product \(bd\) is nonzero. For the reciprocal, \(a\ne0\) ensures that its denominator is nonzero. The following result verifies that addition and multiplication do not depend on which equivalent pairs are used.

Proof. The hypotheses give \(ab'=a'b\) and \(cd'=c'd\). For the sums, the pair formed from the first representatives is \((ad+bc,bd)\), and the pair formed from the second is \((a'd'+b'c',b'd')\). Their cross-products are equal, because \[ (ad+bc)b'd'=ab'dd'+bb'cd' \] and \[ (a'd'+b'c')bd=a'bdd'+bb'c'd. \] The first terms agree by \(ab'=a'b\), and the second terms agree by \(cd'=c'd\). Thus the two sum pairs are equivalent.

For the products, the pairs are \((ac,bd)\) and \((a'c',b'd')\). Their cross-products satisfy \[ (ac)b'd'=(ab')(cd')=(a'b)(c'd)=(a'c')bd. \] Therefore the product pairs are equivalent as well. The formulas define operations on rational-number classes, as claimed. \(\square\)

Every integer \(n\) is represented in \(\mathbb Q\) by \(n/1\). This representation preserves addition and multiplication: the formulas give \[ \frac{m}{1}+\frac{n}{1}=\frac{m+n}{1}, \qquad \frac{m}{1}\cdot\frac{n}{1}=\frac{mn}{1}. \] It also preserves equality, since \(m/1=n/1\) exactly when \(m\cdot1=1\cdot n\), which means \(m=n\). We can therefore regard the integers as rational numbers without changing their arithmetic.

Ordering the Rational Numbers

To compare fractions by cross-products, their denominators should first be positive. Every rational number has a representative with positive denominator: if \(b<0\), replace \(a/b\) by \((-a)/(-b)\), which represents the same class because \(a(-b)=b(-a)\). Thus we can compare representatives \(a/b\) and \(c/d\) with \(b,d>0\).

The comparison does not depend on the positive-denominator representatives. If \(a/b=a'/b'\) and \(c/d=c'/d'\), where all four denominators are positive, then \(ab'=a'b\) and \(cd'=c'd\). In comparing \(a/b\) with \(c/d\), the inequality \(ad<bc\) can be multiplied by the positive integer \(b'd'\), preserving its direction. Substituting the two equalities gives \[ a'd'bd<b b'c'd. \] Since \(bd>0\), cancellation in the integer order yields \(a'd'<b'c'\). The reverse argument gives the converse, so the comparison is representative-independent.

Reciprocals and the Role of Nonzero Denominators

A nonzero rational number has a reciprocal. If \(a/b\ne0\), then \(a\ne0\), and \(b/a\) is an allowed rational representation. Their product is \[ \frac{a}{b}\cdot\frac{b}{a}=\frac{ab}{ba}=\frac{1}{1}, \] because \(ab=ba\) and both denominators are nonzero. This is why dividing by a nonzero integer can be performed in \(\mathbb Q\): for example, \(1/2\) is the reciprocal of the integer \(2=2/1\).

Zero has no reciprocal. Any pair representing zero has numerator zero: if \(a/b=0/1\), then the equality criterion gives \(a\cdot1=b\cdot0=0\), so \(a=0\). Reversing such a pair would place zero in the denominator, which is forbidden. Thus the requirement of a nonzero denominator is essential, not a technical detail.