Chapter 1 — Fundamental Concepts¶
1.4 — Fractional Expressions¶
A quotient of polynomials inherits the rules of a numerical fraction, but it also inherits its caution: not every cancellation is legitimate.
In §1.3 we learned to factor. Now we use that tool in its most immediate application: quotients of polynomials. Everything §1.1 established about fractions of real numbers holds here almost unchanged. There is a single essential difference, and it shapes the whole section: the denominator is no longer a fixed number but an expression that depends on \(x\). The condition that it not vanish, which with numbers ruled out a single case, now rules out in general a set of values of \(x\), and that set must be kept in view at every step.
§1. Rational expressions and their domain¶
A rational expression is a quotient of two polynomials,
where \(Q\) is not the zero polynomial. For each real number we put in place of \(x\), the numerator and the denominator become numbers, and the quotient makes sense only if the resulting denominator is not zero. We call the domain of the expression the set of reals for which it is defined:
The formula reads: the domain is the set of real \(x\) such that \(Q(x)\) is different from zero. The vertical bar, as in §1.1, separates the variable from the condition it must satisfy. The same idea applies to any algebraic expression: its domain is the set of reals for which all of its operations are defined.
To find the domain, we factor the denominator. The lemma of §1.3 says that a product of reals is zero only if one of the factors is; and if a factor is zero, so is the product. Therefore the values that make \(Q\) vanish are exactly those that make one of its factors vanish. For example, in
the denominator vanishes at \(x = 3\) and at \(x = -2\), and at no other value. The domain is \(\mathbb{R} \setminus \{-2, 3\}\); the symbol \(\setminus\) indicates that we remove from \(\mathbb{R}\) the numbers listed in braces. An irreducible quadratic factor contributes no exclusions: by §1.3, a trinomial with negative discriminant vanishes at no real number. Thus the domain of \(\dfrac{1}{x^2 + 1}\) is all of \(\mathbb{R}\).
Reminder. The domain of a rational expression is determined by the denominator, not the numerator. A value is excluded if it makes \(Q(x)\) vanish, regardless of what happens to \(P(x)\).
§2. Simplification¶
Simplifying a rational expression consists in factoring the numerator and the denominator completely and canceling the factors they share. The rule that allows it is
and it rests on a fact about the real numbers.
Proposition. If \(a, b, c \in \mathbb{R}\), with \(b \neq 0\) and \(c \neq 0\), then \(\dfrac{ab}{cb} = \dfrac{a}{c}\).
Proof. Since \(b \neq 0\) and \(c \neq 0\), the lemma of §1.3 gives \(cb \neq 0\), so the left-hand side makes sense. Moreover, \((cb)(c^{-1}b^{-1}) = (cc^{-1})(bb^{-1}) = 1\), so the inverse of \(cb\) is \(c^{-1}b^{-1}\). Then \(\dfrac{ab}{cb} = ab \cdot c^{-1}b^{-1} = a c^{-1} (b b^{-1}) = ac^{-1} = \dfrac{a}{c}\), using the associativity and commutativity of §1.1. \(\blacksquare\)
For each \(x\) in the domain of the original expression, the polynomials \(P\), \(Q\) and \(R\) take real values \(a\), \(c\) and \(b\) that satisfy the hypotheses, and the proposition gives the equality. Outside that set the proposition says nothing, and therein lies the delicate point. Let us look at an example:
The expression on the left is not defined at \(x = 1\), because its denominator equals \(0\). The one on the right is, and it equals \(\frac{2}{3}\). The two expressions agree on the whole domain of the first, but they are not the same expression, because their domains differ. The simplified expression does not remember where it came from, and so whoever simplifies must carry the exclusion along explicitly, as we did in parentheses.
We say that a rational expression is simplified when its numerator and its denominator share no factor of degree greater than or equal to \(1\).
Reminder. Canceling a common factor preserves the value of the expression only where the original expression was defined. The values that made the canceled factor vanish remain excluded, even though the simplified expression no longer shows it.
§3. Operations with rational expressions¶
The product, quotient, sum and difference of rational expressions follow the rules for fractions of real numbers from §1.1. If \(A\), \(B\), \(C\) and \(D\) are polynomials,
The justification is the same as in the previous subsection: for each admissible \(x\), the four polynomials take real values, and for real numbers these rules are already proved. What "admissible" means depends on the operation. In the product and the sum, the zeros of \(B\) and \(D\) are excluded. In the quotient we must also exclude the zeros of \(C\), because dividing by \(\frac{C}{D}\) requires that fraction not to be zero, and in the formula \(C\) moves to the denominator.
To add or subtract, the denominator \(BD\) always works, but it rarely leads to a simplified result. It is better to use the least common denominator: we factor each denominator into irreducible factors (§1.3) and take the product of all the factors that appear, each with the highest exponent with which it occurs. For example, to compute
we factor \(x^2 - 1 = (x - 1)(x + 1)\). The least common denominator is \((x - 1)(x + 1)\), and it suffices to multiply the second fraction, top and bottom, by \(x - 1\):
Multiplying top and bottom by \(x - 1\) is the simplification rule of §2 used in reverse, and so it requires \(x \neq 1\), which was already excluded anyway.
Reminder. In a quotient of rational expressions, the values that make the numerator of the divisor vanish are also excluded.
§4. Complex fractions¶
A complex fraction is a fraction whose numerator, whose denominator, or both, themselves contain fractions. There are two ways to reduce it to a simple fraction, and both lead to the same result. The first is to work separately on the numerator and the denominator until each is a single fraction, and only then apply the quotient rule, \(\dfrac{A/B}{C/D} = \dfrac{AD}{BC}\), valid where \(B\), \(C\) and \(D\) do not vanish. The second is to multiply the main numerator and the main denominator by the least common denominator of all the inner fractions, which disappear in one stroke; again, this is the rule of §2 used in reverse.
Let us look at an example with two variables, \(x\) and \(h\). By the first route,
The starting expression requires \(x \neq 0\), \(x + h \neq 0\) and \(h \neq 0\). The last step cancels the factor \(h\), and the proposition of §2 allows it precisely because \(h \neq 0\). Here the warning of §2 reappears: the result \(\frac{-1}{x(x + h)}\) makes sense for \(h = 0\), but the original expression does not, so the equality holds only for \(h \neq 0\). This quotient is not just any example. It compares how much \(\frac{1}{x}\) changes when \(x\) is shifted by \(h\), and Calculus studies what happens to it as \(h\) approaches \(0\). That the simplification removes \(h\) from the denominator is what makes that question possible.
Let us also note what was canceled: the factor \(h\), which multiplies the whole numerator and the whole denominator. A summand is never canceled.
Reminder. Only factors that multiply the entire numerator and the entire denominator are canceled. The equality \(\dfrac{a + b}{a} = 1 + b\) is not an identity; the correct statement, for \(a \neq 0\), is \(\dfrac{a + b}{a} = \dfrac{a}{a} + \dfrac{b}{a} = 1 + \dfrac{b}{a}\).
§5. Fractional expressions with radicals¶
Fractional expressions are not limited to quotients of polynomials. They may contain radicals, and then the domain has a second source of exclusions: by §1.2, a radical of even index is defined only for nonnegative radicands. For example, in \(\dfrac{\sqrt{x}}{x - 1}\) we must require \(x \ge 0\) because of the root and \(x \neq 1\) because of the denominator. In the interval notation of §1.1, the domain is \([0, 1) \cup (1, \infty)\).
In §1.2 we rationalized denominators by multiplying by the conjugate. Sometimes the opposite is convenient, rationalizing the numerator, and the tool is the same: the difference of squares from §1.3. Let us return to the quotient of the previous subsection, now with a root:
Let us review the conditions. The starting expression requires \(x \ge 0\), \(x + h \ge 0\) and \(h \neq 0\). In the first step we multiplied top and bottom by \(\sqrt{x + h} + \sqrt{x}\), and for that this number must not be zero. The conditions guarantee it: a sum of two nonnegative roots equals \(0\) only if both are \(0\), that is, if \(x = 0\) and \(x + h = 0\), which would give \(h = 0\), an excluded value. In the second step we used \(\left(\sqrt{u}\right)^2 = u\) for \(u \ge 0\), and in the last we canceled the factor \(h\), which is not zero. As in the previous subsection, the result makes sense for \(h = 0\) and the original expression does not.
Reminder. In an expression with radicals of even index, the domain excludes the values that make a denominator vanish and also those that make a radicand negative.
With the handling of algebraic and fractional expressions, the chapter now has everything it needs to take up the solution of equations, the subject of the next section.