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Topos Uranos

The rational numbers: fractions and their operations

Summary

This lesson constructs the rational numbers from the integers. It defines a fraction as a pair of integers whose second term is nonzero, proves that the equality of cross products behaves like an equality and that the sum and the product do not depend on the fractions chosen to represent each number, and proves that every rational number has one and only one fraction in lowest terms. It then studies calculation with fractions (the common denominator, simplification before multiplying, compound fractions and the most frequent errors) and the order of the rational numbers, with its two characteristic properties: density and the Archimedean property.

Learning objectives

  1. Define the rational numbers as fractions of integers identified by the equality of their cross products, and prove that the sum and the product are well defined.
  2. Prove the existence and uniqueness of the fraction in lowest terms, and use it to simplify and to decide whether two fractions represent the same number.
  3. Operate efficiently with fractions: reduce to a common denominator with the least common multiple, simplify before multiplying, work out compound fractions and recognize the usual errors.
  4. Compare rational numbers and apply in proofs the properties of the order, density and the Archimedean property.

Fractions and their equivalence

In the lesson on number systems, the rational numbers were presented as the fractions of integers with nonzero denominator, and it was pointed out that a fraction cannot be defined as a division, because division is precisely what the integers do not allow. This lesson carries out the construction announced there. Its starting point is the integers of the lesson on the integers, with their sum, their product, their order and a property that will be used at every turn: the product of two nonzero integers is nonzero. It follows that a nonzero factor can be cancelled, because if ab=acab = ac with a≠0a \neq 0, then a(b−c)=0a(b - c) = 0, and therefore b−c=0b - c = 0.

Definition (fraction). A fraction is an ordered pair (p,q)(p, q) of integers with q≠0q \neq 0, written pq\frac{p}{q}; the integer pp is its numerator and qq its denominator. The fractions pq\frac{p}{q} and rs\frac{r}{s} are said to be equivalent when their cross products coincide, that is, when ps=qrps = qr.

Thus 23\frac{2}{3} and 46\frac{4}{6} are equivalent, because 2⋅6=3⋅42 \cdot 6 = 3 \cdot 4, whereas 23\frac{2}{3} and 34\frac{3}{4} are not, because 2⋅4≠3⋅32 \cdot 4 \neq 3 \cdot 3. The definition is not arbitrary. If fractions are to be quotients, the equality pq=rs\frac{p}{q} = \frac{r}{s} must be equivalent, after multiplying both sides by qsqs, to the equality ps=qrps = qr; the definition takes as its starting point this consequence, which contains only products of integers and therefore makes sense before division exists. Now, for equivalence to be able to play the role of an equality, it must behave like one: it must be an equivalence relation, in the sense of the lesson on sets, relations and functions.

TheoremThe equivalence of fractions

The equivalence of fractions is reflexive, symmetric and transitive: every fraction is equivalent to itself; if pq\frac{p}{q} is equivalent to rs\frac{r}{s}, then rs\frac{r}{s} is equivalent to pq\frac{p}{q}; and if pq\frac{p}{q} is equivalent to rs\frac{r}{s} and rs\frac{r}{s} is equivalent to tu\frac{t}{u}, then pq\frac{p}{q} is equivalent to tu\frac{t}{u}. The first two properties follow from the commutativity of the product; the third requires cancelling a factor, and it is the only one in which the condition that the denominators be nonzero plays a role. The figure draws pq\frac{p}{q} as the point with coordinates (q,p)(q, p): equivalent fractions appear aligned with the origin.

Proof

  1. p⋅q=q⋅p\resaltar{p \cdot q} = \resaltar{q \cdot p}

    Reflexivity: by the commutativity of the product, the cross products of a fraction with itself coincide.

  2. ps=qr→rq=sp\resaltar{ps = qr} \rightarrow \resaltar{rq = sp}

    Symmetry: if ps=qrps = qr, the commutativity of the product and the symmetry of equality give the same condition with the roles of the fractions interchanged.

  3. ps=qr,ru=st\dato{h1}{ps = qr}, \quad \dato{h2}{ru = st}

    Transitivity: we start from the two hypotheses, and we want to reach pu=qtpu = qt. In the figure, 23\frac{2}{3}, 46\frac{4}{6} and 69\frac{6}{9}.

  4. (ps)u=q(ru)=q(st)(ps)u = q(\resaltar{ru}) = q(\resaltar{st})

    We multiply the first equality by uu and replace ruru by stst, according to the second.

  5. s(pu)=s(qt)→s(pu−qt)=0\resaltar{s(pu) = s(qt)} \rightarrow \resaltar{s(pu - qt) = 0}

    By associativity and commutativity, the factor ss appears on both sides; we move everything to the first side and take it out as a common factor.

  6. pu−qt=0→pu=qt\resaltar{pu - qt = 0} \rightarrow \resaltar{pu = qt}

    Since s≠0s \neq 0 and the product of two nonzero integers is nonzero, the other factor is zero: the outer fractions are equivalent.

If this is so, the general results of the lesson on sets, relations and functions hold for the equivalence of fractions, and there is no need to repeat them: by the theorem on equivalences and partitions, the classes of equivalent fractions form a partition of the set of fractions, so that each fraction belongs to exactly one class and two different classes have no fraction in common. Each class is a number.

Definition (rational number). A rational number is the equivalence class of a fraction, that is, the set of all fractions equivalent to it, and any one of them represents it; in the notation of the lesson on sets, relations and functions, the rational number represented by pq\frac{p}{q} is the class C(pq)C\left( \frac{p}{q} \right), and here it is written simply pq\frac{p}{q}. The set of rational numbers, which is the quotient set of the fractions by their equivalence, is denoted by Q\mathbb{Q}. One writes pq=rs\frac{p}{q} = \frac{r}{s} when both fractions represent the same rational number, that is, when they are equivalent, as the lemma on equivalence classes proved there states.
pq=rs↔ps=qr\frac{p}{q} = \frac{r}{s} \leftrightarrow ps = qr

Each integer pp is identified with the rational number represented by p1\frac{p}{1}. The identification does not confuse different integers, because p1=r1\frac{p}{1} = \frac{r}{1} is equivalent to p⋅1=1⋅rp \cdot 1 = 1 \cdot r, that is, to p=rp = r; in this sense, Z⊆Q\mathbb{Z} \subseteq \mathbb{Q}. It will be checked later that the operations on rational numbers, applied to the integers, give the usual ones. The practical rule most often used to change the fraction without changing the number is the following.

LemmaExpanding and reducing

For all integers pp, qq and kk, with q≠0q \neq 0 and k≠0k \neq 0, we have kpkq=pq\frac{kp}{kq} = \frac{p}{q}. In particular, −p−q=pq\frac{-p}{-q} = \frac{p}{q}, so that every rational number is represented by a fraction with natural denominator. Read from left to right, the equality cancels a common factor; read from right to left, it expands the fraction. The figure shows its meaning: if each part of the whole is divided into kk equal parts, the same portion is described with kk times as many parts.

Proof

  1. pq=23\frac{p}{q} = \frac{2}{3}

    In the figure, 23\frac{2}{3} divides the whole into q=3q = 3 equal parts and takes p=2p = 2 of them.

  2. kpkq=2⋅22⋅3=46\frac{kp}{kq} = \frac{\resaltar{2 \cdot 2}}{\resaltar{2 \cdot 3}} = \resaltar{\frac{4}{6}}

    If each part is divided into k=2k = 2 equal parts, the whole is divided into kqkq parts, and the portion taken contains kpkp of them.

  3. 4⋅24⋅3=812\frac{\resaltar{4 \cdot 2}}{\resaltar{4 \cdot 3}} = \resaltar{\frac{8}{12}}

    With k=4k = 4, the same portion is described with twelve parts.

  4. (kp)⋅q=k(pq)=(kq)⋅p(kp) \cdot q = \resaltar{k(pq)} = \resaltar{(kq) \cdot p}

    The equivalence does not depend on the drawing: the cross products coincide, by the associativity and commutativity of the product. The condition k≠0k \neq 0 only ensures that kqkq is an admissible denominator.

  5. −p−q=pq,2−3=−23\frac{-p}{-q} = \frac{p}{q}, \quad \frac{2}{-3} = \frac{\resaltar{-2}}{\resaltar{3}}

    With k=−1k = -1 we obtain the change of sign of both terms; if the denominator is negative, this change makes it natural.