Chapter 1 — Fundamental Concepts¶
1.2 — Exponents and Radicals¶
Algebra becomes powerful when we abbreviate repetition.
In §1.1 we established the rules of addition and multiplication in \(\mathbb{R}\). Now we take one more step. Just as, among natural numbers, multiplication abbreviates a repeated sum, exponentiation abbreviates a repeated product. This section tells how that simple idea extends, without losing coherence at any step, from natural exponents to integer and rational ones. Along the way appear the radicals, which undo exponentiation, and a question that §1.1 left open: why \(\sqrt{2}\) exists.
§1. Powers with integer exponents¶
Let \(a \in \mathbb{R}\) and \(n \in \mathbb{N}\). The power \(a^n\) is the product of \(n\) factors equal to \(a\):
The number \(a\) is called the base and \(n\) the exponent. With a single factor, \(a^1 = a\). The ellipsis in the formula has the same character as the one in \(\mathbb{N}\) in §1.1. Rigorously, \(a^n\) is defined step by step, with \(a^1 = a\) and \(a^{n+1} = a^n \cdot a\), and §10.2 proves that this definition is legitimate. This definition counts factors and says nothing, as yet, about zero or negative exponents.
Those cases are not defined at whim. We define them in the only way that preserves a rule that natural powers already satisfy: if \(m > n\) and \(a \neq 0\), then \(a^m / a^n = a^{m-n}\), because the \(n\) factors of the denominator cancel with \(n\) of the \(m\) in the numerator. If we want the rule to hold also when \(m = n\), the left side is \(a^n / a^n = 1\) and the right side is \(a^0\). There is no choice but to define
If we now want the rule to hold with \(m = 0\), the left side is \(a^0 / a^n = 1/a^n\) and the right side is \(a^{-n}\). This forces us to define the negative exponent as
Thus \(a^{-n}\) is exactly the multiplicative inverse of \(a^n\), in the sense of §1.1. Since the inverse of a product is the product of the inverses, also \(a^{-n} = (1/a)^n\).
The expression \(0^0\) is left out of these definitions, because the argument that led to \(a^0 = 1\) divides by \(a\). In Calculus, "\(0^0\)" appears as an indeterminate form: expressions that approach that form can approach different values. In algebraic contexts, on the other hand, the convention \(0^0 = 1\) is convenient, and that is how the binomial theorem of §10.5 uses it. In this section we do not need it, and we leave \(0^0\) undefined.
Reminder. \(a^0 = 1\) and \(a^{-n} = 1/a^n\), always with \(a \neq 0\). The negative exponent inverts the power. \(0^0\) is not defined here.
§2. The laws of exponents¶
For real bases \(a\) and \(b\), nonzero when some exponent is zero or negative, and integer exponents \(m\) and \(n\), five laws hold:
| Law | Formula |
|---|---|
| Product of equal bases | \(a^m \cdot a^n = a^{m+n}\) |
| Quotient of equal bases | \(a^m / a^n = a^{m-n}\) |
| Power of a power | \((a^m)^n = a^{mn}\) |
| Power of a product | \((ab)^n = a^n b^n\) |
| Power of a quotient | \((a/b)^n = a^n/b^n\) |
With natural exponents, each law is a matter of counting factors. Multiplying \(a^m\) by \(a^n\) means writing \(m\) copies of \(a\) followed by \(n\) more copies, that is, \(m + n\) copies. Raising \(a^m\) to the \(n\) means repeating the block of \(m\) copies \(n\) times, which gives \(mn\) copies. And in \((ab)^n\), commutativity lets us rearrange the \(n\) factors \(ab\) into \(n\) factors \(a\) followed by \(n\) factors \(b\).
For arbitrary integer exponents we must check the cases that counting does not cover. We do so for the first law, and the others reduce to it.
Product of equal bases. If \(m, n \ge 0\), it is the counting above, with \(a^0 = 1\). If \(m \ge 0 > n\) and \(m + n \ge 0\), then \(m = (m+n) + (-n)\) is a sum of two naturals or zeros, and by the previous case \(a^m = a^{m+n} a^{-n}\). Multiplying by \(a^n\), which is the inverse of \(a^{-n}\), gives \(a^m a^n = a^{m+n}\). If \(m \ge 0 > n\) and \(m + n < 0\), the previous case applied to \(-n = m + (-(m+n))\) gives \(a^{-n} = a^m a^{-(m+n)}\), and taking inverses, \(a^n = a^{-m} a^{m+n}\), whence \(a^m a^n = a^{m+n}\). If \(m, n < 0\), then \(a^m a^n = \dfrac{1}{a^{-m}a^{-n}} = \dfrac{1}{a^{-m-n}} = a^{m+n}\). The case \(m < 0 \le n\) is symmetric.
The other four. The quotient is the product with \(a^{-n}\) in place of \(a^n\): \(a^m / a^n = a^m a^{-n} = a^{m-n}\). For the power of a power with \(n \ge 0\), we apply the product law \(n\) times: \((a^m)^n = a^{m + \dots + m} = a^{mn}\). If \(n = -k < 0\), then \((a^m)^{-k} = 1/(a^m)^k = 1/a^{mk} = a^{-mk}\). For the power of a product with \(n = -k < 0\), we have \((ab)^{-k} = \dfrac{1}{a^k b^k} = a^{-k} b^{-k}\). The quotient law is the product law applied to \(a\) and \(1/b\).
There is a constant temptation that is best disarmed from the start: exponentiation does not distribute over addition or subtraction. The equality \((a+b)^n = a^n + b^n\) is not an identity when \(n \ge 2\): it may hold for particular values, such as \(b = 0\), but it fails in general. The case \(n = 2\) is the most useful to remember:
The term \(2ab\) is precisely what is lost by "distributing" the exponent. With \(a = b = 1\), the left side is \(4\) and \(a^2 + b^2\) is \(2\).
Reminder. The exponent distributes over products and quotients, never over sums or differences: \((a+b)^2 = a^2 + 2ab + b^2\).
§3. \(n\)th roots¶
Exponentiation answers the question "what is \(b\) raised to the \(n\)?". The inverse question, "what number, raised to the \(n\), gives \(a\)?", leads to root extraction. Before giving the answer a name, we must make sure that it exists and that it is unique. We begin with uniqueness, which follows from a property of order.
Lemma. Let \(n \in \mathbb{N}\). If \(0 \le x < y\), then \(x^n < y^n\).
Proof. Multiplying \(x < y\) by \(x \ge 0\) and by \(y > 0\) gives \(x^2 \le xy < y^2\). Repeating the argument, multiplying \(x^k < y^k\) by \(x\) and by \(y\), we pass from each exponent to the next, up to \(n\). \(\blacksquare\)
Consequently, two nonnegative numbers with the same \(n\)th power are equal: if they were different, the smaller would have the smaller power.
Existence requires something that algebra alone does not provide: the supremum axiom of §1.1. This is the moment to fulfill the promise made there.
Theorem. Let \(n \in \mathbb{N}\) with \(n \ge 2\) and \(a \ge 0\). There is a unique real number \(b \ge 0\) such that \(b^n = a\).
Proof. Uniqueness is the lemma. For existence, if \(a = 0\) then \(b = 0\) works, so we assume \(a > 0\). We will use an inequality: if \(0 \le x \le y\), then
To see it, note that \(y^n - x^n = (y - x)(y^{n-1} + y^{n-2}x + \dots + x^{n-1})\). When the product on the right is expanded, every term appears once with a plus sign and once with a minus sign, except \(y^n\) and \(-x^n\). The second factor has \(n\) terms, and each is at most \(y^{n-1}\) because \(x \le y\).
Let \(T = \{x \ge 0 \mid x^n \le a\}\). It is nonempty, because \(0 \in T\), and it is bounded above by \(1 + a\): if \(x > 1 + a\), then \(x > 1\), so \(x^n \ge x > a\). By the supremum axiom, \(T\) has a supremum; call it \(s\). We will prove that \(s^n = a\) by ruling out the other two possibilities.
If we had \(s^n < a\), take \(h = \min\left\{\dfrac{1}{2},\ \dfrac{a - s^n}{2n(s+1)^{n-1}}\right\}\), which is positive and less than \(1\). By the inequality with \(x = s\) and \(y = s + h \le s + 1\),
Hence \((s+h)^n < a\), so \(s + h \in T\) and it is greater than \(s\), contradicting the fact that \(s\) is an upper bound.
If we had \(s^n > a\), then \(s > 0\). Take \(h = \min\left\{\dfrac{s}{2},\ \dfrac{s^n - a}{2n s^{n-1}}\right\}\), which is positive and less than \(s\). By the inequality with \(x = s - h\) and \(y = s\),
Hence \((s-h)^n > a\). Every \(x \in T\) satisfies \(x^n \le a < (s-h)^n\), and by the lemma \(x < s - h\). Thus \(s - h\) is an upper bound of \(T\) less than \(s\), contradicting the fact that \(s\) is the least. With both possibilities ruled out, \(s^n = a\). \(\blacksquare\)
With \(n = 2\) and \(a = 2\), the theorem says that there is a unique positive real number whose square is \(2\): it is the number \(\sqrt{2}\) of §1.1, and its existence rests on the supremum axiom.
Definition. Let \(n \in \mathbb{N}\) with \(n \ge 2\) and \(a \ge 0\). The \(n\)th root of \(a\), written \(\sqrt[n]{a}\), is the unique real number \(b \ge 0\) with \(b^n = a\). In the expression \(\sqrt[n]{a}\), the sign \(\sqrt{\ }\) is the radical, \(n\) is the index and \(a\) is the radicand. With index \(2\) the index is not written: \(\sqrt{a}\) is the square root.
What if the radicand is negative? Here the index makes a difference. If \(n\) is even, \(b^n = (b^{n/2})^2 \ge 0\) for every real \(b\), so no real number raised to \(n\) gives a negative number. Negative numbers have no root of even index in \(\mathbb{R}\): \(\sqrt{-4}\) is not a real number. If \(n\) is odd, on the other hand, \((-b)^n = -b^n\), and the number \(-\sqrt[n]{-a}\) satisfies \(\big(-\sqrt[n]{-a}\big)^n = -(-a) = a\). For this reason, for odd \(n\) and \(a < 0\), we define \(\sqrt[n]{a} = -\sqrt[n]{-a}\). For example, \(\sqrt[3]{-8} = -2\). This root is also unique, because for odd \(n\) the function \(x \mapsto x^n\) is strictly increasing on all of \(\mathbb{R}\): if \(x < 0 \le y\), then \(x^n < 0 \le y^n\), and if \(x < y < 0\), the lemma applied to \(0 < -y < -x\) gives \(-y^n < -x^n\).
The even-index restriction has a consequence that often goes unnoticed and that connects with the absolute value of §1.1.
Proposition. For every \(a \in \mathbb{R}\) and every even \(n\), \(\sqrt[n]{a^n} = |a|\).
Proof. The number \(|a|\) is nonnegative and, since \(n\) is even, \(|a|^n = a^n\): if \(a \ge 0\) this is immediate, and if \(a < 0\), then \(|a|^n = (-a)^n = a^n\). By the uniqueness of the root, \(\sqrt[n]{a^n} = |a|\). \(\blacksquare\)
In particular, \(\sqrt{a^2} = |a|\), and not \(a\). A numerical example exposes the trap: \(\sqrt{(-3)^2} = \sqrt{9} = 3\), which is \(|-3|\) and not \(-3\).
Reminder. For \(a \ge 0\), \(\sqrt[n]{a}\) is the unique \(b \ge 0\) with \(b^n = a\); it exists by the supremum axiom. With even index, the radicand cannot be negative and the result is always \(\ge 0\). With odd index, \(\sqrt[n]{a} = -\sqrt[n]{-a}\) for \(a < 0\). For every \(a\), \(\sqrt{a^2} = |a|\).
§4. Rational exponents¶
Roots allow us to extend exponentiation to rational exponents. Here we must proceed carefully, because the extension works without surprises only when the base is positive. That is why we begin with \(a > 0\).
Definition. Let \(a > 0\), \(m \in \mathbb{Z}\) and \(n \in \mathbb{N}\). We define
With \(n = 1\), understanding that \(\sqrt[1]{a} = a\), we recover the integer power. The number \(a^{m/n}\) is positive, and it can also be computed as \(\sqrt[n]{a^m}\). Indeed, \(\big((\sqrt[n]{a})^m\big)^n = \big((\sqrt[n]{a})^n\big)^m = a^m\) by the laws of part 2, and the uniqueness of the root gives \((\sqrt[n]{a})^m = \sqrt[n]{a^m}\).
Before going on, we must solve a problem that the notation hides. The same rational number can be written in infinitely many ways: \(\frac{1}{2}\), \(\frac{2}{4}\) and \(\frac{50}{100}\) are the same number. The definition starts from one of those forms, \(\frac{m}{n}\), without saying what happens if we use another. For the symbol \(a^{m/n}\) to make sense, two ways of writing the same exponent must give the same real number. This is not automatic; it must be proved.
Proposition (the rational exponent is well defined). Let \(a > 0\). If \(\frac{m}{n} = \frac{p}{q}\), that is, if \(mq = np\), then \(a^{m/n} = a^{p/q}\).
Proof. Let \(b = a^{m/n}\) and \(c = a^{p/q}\), both positive. We raise each to the integer power \(nq\) and use the laws of part 2:
Since \(mq = np\), we get \(b^{nq} = c^{nq}\), and since \(b\) and \(c\) are positive, the lemma of part 3 gives \(b = c\). \(\blacksquare\)
With the definition secured, the five laws of part 2 hold for rational exponents when the bases are positive. For rational \(r\) and \(s\) we write both with a common denominator, \(r = \frac{m}{n}\) and \(s = \frac{p}{n}\). Then the product is the integer law applied to the base \(\sqrt[n]{a}\):
For the power of a product, \(\sqrt[n]{a}\sqrt[n]{b}\) is positive and its \(n\)th power is \(ab\), so \(\sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b}\) by uniqueness; raising to the \(m\) gives \((ab)^{m/n} = a^{m/n} b^{m/n}\). For the power of a power, let \(d = (a^{m/n})^{p/q}\). Then \(d^q = (a^{m/n})^p = a^{mp/n}\), so \(d^{qn} = a^{mp}\), and since \(d > 0\), uniqueness gives \(d = a^{mp/(nq)}\). The quotient reduces to the product, as in the integer case.
Example. \(8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4\), and also \(\sqrt[3]{8^2} = \sqrt[3]{64} = 4\).
Why exclude negative bases? Because with them the definition ceases to be well defined. With odd index, \((-8)^{1/3} = \sqrt[3]{-8} = -2\) makes sense. But \(\frac{1}{3} = \frac{2}{6}\), and the form \(\frac{2}{6}\) would give \(\sqrt[6]{(-8)^2} = \sqrt[6]{64} = 2\). The same exponent would produce two different values. With negative bases we therefore use only radicals of odd index, \(\sqrt[n]{a}\), and not the fractional-exponent notation.
Reminder. For \(a > 0\): \(a^{m/n} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}\), it does not depend on how the fraction is written, and all the laws of exponents hold. With negative bases, the notation \(a^{m/n}\) is avoided.
§5. Simplifying radicals¶
The properties of radicals are the laws of exponents written in another notation. With nonnegative radicands, and \(b > 0\) in the quotient,
All three are proved in the same way: the right side is nonnegative and, raised to the corresponding index, gives the radicand; uniqueness of the root does the rest. In the third, for example, \(\big(\sqrt[mn]{a}\big)^{mn} = a\) says that \(\big(\sqrt[mn]{a}\big)^m\) is the \(n\)th root of \(a\), and then \(\sqrt[mn]{a}\) is the \(m\)th root of \(\sqrt[n]{a}\). Nesting two roots is thus equivalent to a single root whose index is the product of the indices.
The condition on the radicands matters. With even index and negative radicands, the first formula does not even make sense on the right side: \(\sqrt{(-2)(-8)} = \sqrt{16} = 4\), but \(\sqrt{-2}\) and \(\sqrt{-8}\) are not real numbers.
We say that a radical is simplified when three conditions hold. The first is that all the factors appearing in the radicand with exponent greater than or equal to the index have been extracted. The second, that the index is as small as possible. The third, that no radicals remain in any denominator. For example, since \(72 = 36 \cdot 2 = 6^2 \cdot 2\), we get \(\sqrt{72} = \sqrt{6^2}\sqrt{2} = 6\sqrt{2}\). Likewise, \(\sqrt[3]{54} = \sqrt[3]{3^3 \cdot 2} = 3\sqrt[3]{2}\). And since \(\sqrt[4]{9} = \sqrt{\sqrt{9}} = \sqrt{3}\), the index \(4\) reduces to \(2\). The third condition is the subject of the next part.
§6. Rationalizing denominators¶
To rationalize a fraction is to transform it into an equal one whose denominator has no radicals. The procedure depends on whether the denominator has one term or two.
A single radical. Suppose the denominator is \(\sqrt[n]{a^k}\), with \(a > 0\) and \(0 < k < n\). For the radical to disappear, we must complete the \(n\)th power of the radicand. That is why we multiply numerator and denominator by \(\sqrt[n]{a^{n-k}}\):
Examples. \(\dfrac{5}{\sqrt{3}} = \dfrac{5\sqrt{3}}{\sqrt{3}\sqrt{3}} = \dfrac{5\sqrt{3}}{3}\), and \(\dfrac{1}{\sqrt[3]{2}} = \dfrac{\sqrt[3]{4}}{\sqrt[3]{2}\sqrt[3]{4}} = \dfrac{\sqrt[3]{4}}{2}\).
Two terms. Suppose the denominator is \(\sqrt{a} + \sqrt{b}\) or \(\sqrt{a} - \sqrt{b}\), with \(a, b \ge 0\) and \(a \neq b\). Here the device is different: we multiply by the conjugate, the same expression with the middle sign changed. The difference of squares, \((x+y)(x-y) = x^2 - y^2\), obtained by expanding, makes the roots get squared and disappear:
The condition \(a \neq b\) guarantees that the denominator \(a - b\) is not zero. With \(\sqrt{a} - \sqrt{b}\) in the denominator we proceed in the same way, multiplying by \(\sqrt{a} + \sqrt{b}\). For example, \(\dfrac{1}{\sqrt{5} - \sqrt{2}} = \dfrac{\sqrt{5} + \sqrt{2}}{5 - 2} = \dfrac{\sqrt{5} + \sqrt{2}}{3}\).
Reminder. With a single radical, we multiply to complete the power of the index. With two terms, we multiply by the conjugate: \((x+y)(x-y) = x^2 - y^2\) has no roots.
This settles the handling of powers and roots in \(\mathbb{R}\) and the existence of roots. The next section applies these tools to algebraic expressions: polynomials, special products and factoring.