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Topos Uranos
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Algebraic fractions and partial fractions

Summary

This lesson extends the calculus of numerical fractions to quotients of polynomials. It defines algebraic fractions and their domain, proves that cancelling a common factor yields an equality only on the original domain, which explains the removed points of their graphs, and characterizes the equality of two fractions by means of the identity principle. It studies addition with the least common multiple of the denominators, multiplication, division and compound fractions, and it solves equations with fractions and with radicals, paying attention to the solutions that appear when denominators are cleared or both sides are squared. Finally, it proves that a proper fraction whose denominator has simple roots decomposes in a unique way into partial fractions, with coefficients computed by covering up a factor, extends the decomposition to repeated factors, with a proof of its uniqueness of its own, and to irreducible quadratic factors, and applies it to the computation of telescoping sums.

Learning objectives

  1. Determine the domain of an algebraic fraction, simplify it correctly and recognize the removed points of its graph.
  2. Add, multiply and divide algebraic fractions and reduce compound fractions, using the least common multiple of the denominators.
  3. Solve equations with fractions and with radicals, discarding the excluded values and the solutions added by squaring.
  4. Decompose a fraction into partial fractions, justify the decomposition and apply it to telescoping sums.

Algebraic fractions

An algebraic fraction is a quotient of two polynomials, PQ\frac{P}{Q}, with QQ nonzero. It defines the rational function that assigns to each real number xx the value P(x)Q(x)\frac{P(x)}{Q(x)}, of the type studied in the lesson on algebraic functions, and its domain consists of the real numbers that do not make the denominator vanish; the others are called excluded values. Since a nonzero polynomial of degree nn has at most nn roots, according to the lesson on the algebra of polynomials, the excluded values are always finite in number, and to find them one factors the denominator with the techniques of the lesson on the roots of higher-degree polynomials. In what follows, an equality between algebraic fractions is understood to hold at every real number at which both of its sides are defined, and when the domains of the two sides do not coincide this is stated explicitly.

Simplification

Numerical fractions are simplified by dividing numerator and denominator by a common factor, as seen in the lesson on the rational numbers. Algebraic fractions are handled in the same way, but the common factor is now a polynomial, which may vanish at some points; it follows that the simplified fraction may be defined where the original one is not.

LemmaSimplification

Let PP, QQ and CC be polynomials, with QQ and CC nonzero. At every real number that does not make QCQC vanish, the quotient PCQC\frac{PC}{QC} takes the same value as PQ\frac{P}{Q}. The domain of PCQC\frac{PC}{QC} is that of PQ\frac{P}{Q} without the roots of CC; therefore, the two fractions are the same function only if the second is restricted to the domain of the first.

PCQC=PQ,Q(x)≠0∧C(x)≠0\frac{PC}{QC} = \frac{P}{Q}, \qquad Q(x) \neq 0 \wedge C(x) \neq 0

The proof uses the lemma on the sign of a product, from the lesson on algebraic functions, and the expanding lemma, from the lesson on the rational numbers, which holds equally for real numbers. The figure follows the proof with P=x−3P = x - 3, Q=x−2Q = x - 2 and C=x+3C = x + 3.

Proof

  1. Q(x0)C(x0)≠0↔Q(x0)≠0∧C(x0)≠0\dato{d}{Q(x_0) C(x_0) \neq 0 \leftrightarrow \resaltar{Q(x_0) \neq 0 \wedge C(x_0) \neq 0}}

    Let x0x_0 be a real number that does not make QCQC vanish. A product of real numbers is zero if and only if one of its factors is; hence neither QQ nor CC vanishes at x0x_0.

  2. P(x0) cQ(x0) c=P(x0)Q(x0),c=C(x0)≠0\frac{P(x_0) \, \resaltar{c}}{Q(x_0) \, \resaltar{c}} = \frac{P(x_0)}{Q(x_0)}, \qquad c = C(x_0) \neq 0

    The number c=C(x0)c = C(x_0) is not zero, and by the expanding lemma multiplying both terms of a fraction by it does not change its value.

  3. dom⁡(PCQC)={x∈dom⁡(PQ):C(x)≠0}\operatorname{dom}\left( \frac{PC}{QC} \right) = \left\{ x \in \operatorname{dom}\left( \frac{P}{Q} \right) : \resaltar{C(x) \neq 0} \right\}

    The same equivalence describes the domain: a real number is in that of PCQC\frac{PC}{QC} exactly when it is in that of PQ\frac{P}{Q} and does not make CC vanish.

  4. C(r)=0∧Q(r)≠0→r∈dom⁡(PQ)∧r∉dom⁡(PCQC)C(r) = 0 \wedge Q(r) \neq 0 \rightarrow r \in \operatorname{dom}\left( \frac{P}{Q} \right) \wedge \resaltar{r \notin \operatorname{dom}\left( \frac{PC}{QC} \right)}

    At a root rr of CC that is not a root of QQ, the simplified fraction is defined and the original one is not: the graph of the original is that of the simplified fraction with that point removed, which is drawn as an open circle. In the figure, r=−3r = -3.

What the original function does near a removed point will be studied in the lesson on sign charts and inequalities.

Only common factors are cancelled, and that is why simplification requires factoring the numerator and the denominator first: a common summand is not a factor, and striking it out produces an error. Thus, striking out the xx in x+2x+4\frac{x + 2}{x + 4} would give the constant value 24\frac{2}{4}, whereas the fraction is worth 23\frac{2}{3} at x=2x = 2 and 35\frac{3}{5} at x=1x = 1.

ExampleA simplification and its removed point

Simplify the fraction x2−9x2+x−6\frac{x^2 - 9}{x^2 + x - 6} and describe its graph. The figure marks the excluded values on the line and then draws the simplified fraction with the point that the original one lacks.

Proof

  1. x2−9x2+x−6=(x−3)(x+3)(x+3)(x−2)\frac{\resaltar{x^2 - 9}}{\resaltar{x^2 + x - 6}} = \frac{\resaltar{(x - 3)(x + 3)}}{\resaltar{(x + 3)(x - 2)}}

    We factor the numerator, a difference of squares, and the denominator, a trinomial with roots −3-3 and 22.

  2. (x+3)(x−2)≠0↔x≠−3∧x≠2\resaltar{(x + 3)(x - 2)} \neq 0 \leftrightarrow \dato{d}{\resaltar{x \neq -3 \wedge x \neq 2}}

    The excluded values are the roots of the denominator; the domain is the line without those two points.

  3. x2−9x2+x−6=x−3x−2,x≠−3\frac{x^2 - 9}{x^2 + x - 6} = \resaltar{\frac{x - 3}{x - 2}}, \qquad \resaltar{x \neq -3}

    By the simplification lemma, the factor x+3x + 3, which does not vanish on the domain, can be cancelled. The simplified fraction has a larger domain: it excludes only 22.

  4. −3−3−3−2=−6−5=65\frac{\resaltar{-3} - 3}{\resaltar{-3} - 2} = \frac{-6}{-5} = \resaltar{\frac{6}{5}}

    At −3-3 the simplified fraction is worth 65\frac{6}{5} and the original one is not defined: the graph of the original is the curve of the simplified fraction without the point with abscissa −3-3.

The equality of two fractions

Two numerical fractions pq\frac{p}{q} and rs\frac{r}{s} are equal if and only if ps=qrps = qr. For algebraic fractions an analogous criterion holds, whose proof rests on a fact about polynomials: an identity that holds at every real number except finitely many also holds at those.

LemmaIdentities except at finitely many points

If two polynomials take the same value at every real number except, at most, finitely many of them, they have the same coefficients and take the same value at every real number.

The proof is by contradiction, using the lemma on distinct roots from the lesson on the algebra of polynomials. The figure draws the difference of the two polynomials over the xx-axis, with two exceptional points.

Proof

  1. x∉{s1,…,sm}→D(x)=0\dato{d}{x \notin \{ s_1, \ldots, s_m \} \rightarrow \resaltar{D(x) = 0}}

    Let AA and BB be the two polynomials, D=A−BD = A - B their difference and s1,…,sms_1, \ldots, s_m the real numbers at which they may fail to coincide. The difference vanishes at all the others.

  2. t1,…,tk+1∉{s1,…,sm}→D(t1)=⋯=D(tk+1)=0t_1, \ldots, t_{k + 1} \notin \{ s_1, \ldots, s_m \} \rightarrow \resaltar{D(t_1) = \cdots = D(t_{k + 1}) = 0}

    If DD were not the zero polynomial, it would have a degree kk and, by the lemma on distinct roots, at most kk roots. But among the k+m+1k + m + 1 natural numbers 1,2,…,k+m+11, 2, \ldots, k + m + 1 there are at least k+1k + 1, which we call t1,…,tk+1t_1, \ldots, t_{k + 1}, that are not exceptional, and all of them are roots of DD: this is a contradiction.

  3. D=0→A(x)=B(x),x∈R\resaltar{D = 0} \rightarrow A(x) = B(x), \quad x \in \mathbb{R}

    Therefore DD is the zero polynomial: its coefficients are all zero, and AA and BB coincide coefficient by coefficient, as in the identity principle, and take the same value at every real number, also at s1,…,sms_1, \ldots, s_m.

TheoremEquality of algebraic fractions

Let PQ\frac{P}{Q} and RS\frac{R}{S} be two algebraic fractions. They take the same value at every real number that does not make QSQS vanish if and only if the polynomials PSPS and QRQR are equal.

PQ=RS↔PS=QR\frac{P}{Q} = \frac{R}{S} \leftrightarrow PS = QR

The figure follows the proof with the fractions x2−9x2+x−6\frac{x^2 - 9}{x^2 + x - 6} and x−3x−2\frac{x - 3}{x - 2} of the previous example.

Proof

  1. Q(x)S(x)≠0→P(x)S(x)=Q(x)R(x)\dato{m}{Q(x) S(x) \neq 0 \rightarrow \resaltar{P(x) S(x) = Q(x) R(x)}}

    Suppose that the two fractions coincide away from the roots of QSQS. Multiplying by Q(x)S(x)Q(x) S(x) gives an equality between the values of two polynomials, valid at every real number except finitely many, because QSQS is not zero and has finitely many roots.

  2. PS=QR\resaltar{PS = QR}

    By the lemma on identities except at finitely many points, PSPS and QRQR are the same polynomial. In the example, both products are x3−2x2−9x+18x^3 - 2x^2 - 9x + 18.

  3. PS=QR∧Q(x)S(x)≠0→P(x)Q(x)=R(x)S(x)PS = QR \wedge Q(x) S(x) \neq 0 \rightarrow \resaltar{\frac{P(x)}{Q(x)} = \frac{R(x)}{S(x)}}

    Conversely, if PS=QRPS = QR, it suffices to divide by Q(x)S(x)Q(x) S(x) at the real numbers that do not make it vanish, by the expanding lemma.

The theorem justifies checking a simplification by cross-multiplying. The lemma, in turn, is the key to partial fraction decomposition: it allows an identity to be evaluated at the roots of a denominator, where the fractions it comes from are not defined.