Faceted Curriculum Project/B3/Math Facet

Mathematical Introduction
Most rules of exponents arise from thinking about exponentiation as repeated multiplication. For example, when $$n$$ is a natural number, and $$x$$ is a nonzero real number, the following is the definition of $$x^n$$:

$$x^n = x \cdots x,$$ where $$x$$ occurs $$n$$ times on the right side.

To be rigorous, the definition of $$x^n$$ is given recursively, when $$n$$ is a natural number and $$x$$ is a nonzero real number by the following:
 * $$x^0$$ = 1.
 * $$x^n = x \cdot x^{n-1}$$ when $$0 < n$$

Note that $$x^n$$ is nonzero, when $$x$$ is nonzero as required above.

Adding and Subtracting Exponents
We begin with "rules" involving addition and subtraction of exponents. These rules can be proven from the definition of exponents. Moreover, these rules suggest a useful way of extending the definition of exponents to allow negative exponents.

The Rule for Adding Exponents
Theorem: If $$x$$ is a nonzero real number, and $$m,n$$ are natural numbers, then:

$$x^{m+n} = x^m \cdot x^n $$.

Proof: We prove this by induction on $$m+n$$. For the base step of induction, we note that if $$m,n$$ are natural numbers, and $$m+n=0$$, then $$m=n=0$$. In this case, the definition of exponents yields:

$$x^{0+0} = x^0 = 1 = 1 \cdot 1 = x^0 x^0$$.

For the inductive step, suppose that $$m+n > 0$$, and the theorem has been proven in all smaller cases. Then, we find:

Q.E.D.
 * $$x^{m+n} = x^{m+n-1} \cdot x$$ (by the definition of exponents).
 * $$x^{m+n} = (x^m x^{n-1}) x $$ (by induction).
 * $$x^{m+n} = x^m (x^{n-1} x) $$ (by the associative property of multiplication).
 * $$x^{m+n} = x^m x^n $$ (by the definition of exponents).

Corollary: If $$x$$ is a nonzero real number, and $$m,n$$ are natural numbers, and $$m \geq n$$, then:

$$x^{m-n} = x^m / x^n$$.

Proof: Since $$ m \geq n$$, we find that $$m-n$$ is a natural number. It follows from the theorem that:
 * $$x^{(m-n) + n} = x^m$$.
 * Hence $$x^{m-n} = x^m / x^n $$, by dividing both sides by the nonzero number $$ x^n $$.

Q.E.D.

Digression on Exponents and Multiplicative Inverses
Theorem: If $$x$$ is a nonzero real number, and $$n$$ is a natural number, then $$(1/x)^n = 1/(x^n)$$. Equivalently, $$(1/x)^n x^n = 1$$.

Proof: We prove this by induction on $$n$$. When $$n=0$$, the conclusion of the theorem follows from the fact that $$1=1/1$$.

Now, we apply induction and assume $$n>0$$. In this case, we find: The Theorem follows.
 * $$(1/x)^n x^n = (1/x)^{n-1} (1/x) x^{n-1} x $$.
 * $$(1/x)^n x^n = (1/x) x $$, by induction.
 * $$(1/x)^n x^n = 1 $$.

Q.E.D.

Extension to Negative Exponents
From the previous results, we can extend our definition to allow negative exponents:

Theorem: There is a unique way to extend the function $$ (x,n) \mapsto x^n$$, from its initial domain $$ (\mathbb{R} - \{ 0 \}) \times (\mathbb{N}) $$ to the domain $$(\mathbb{R} - \{ 0 \}) \times (\mathbb{Z})$$, in such a way that:

$$ x^{m+n} = x^m \cdot x^n $$ for all nonzero real numbers $$x$$, and all integers $$m,n$$.

Proof: First, we desribe an extension of exponentiation to allow negative exponents. If $$x$$ is a nonzero real number, and $$n$$ is a negative integer, then $$-n$$ is a positive integer. Moreover, since $$x$$ is nonzero, it has a (unique) multiplicative inverse: $$1/x$$. In this situation, we define:

$$x^n = (1/x)^{-n}$$.

This extends the domain of the function as desired. It remains to check that the "law of exponents" is satisfied in this extended domain, i.e., that $$ x^{m+n} = x^m \cdot x^n $$ in the extended domain. There are four cases to consider, using the symmetry between $$m$$ and $$n$$:


 * If both $$m$$ and $$n$$ are positive, the "law of exponents" is satisfied since we are working within the original domain.
 * If both $$m$$ and $$n$$ are negative, then we check:
 * $$x^{m+n} = (1/x)^{m+n} = 1/(x^{m+n}) = 1/(x^m x^n) = 1/(x^m) 1/(x^n) = (1/x)^m (1/x)^n = x^{-m} x^{-n}.$$
 * If $$m$$ is positive, and $$n$$ is negative, and $$m\geq(-n)$$ then we check:
 * $$x^{m+n} = x^{m-(-n)} = x^m / x^{-n} = x^m (1/x)^{-n} = x^m x^n $$, using the previous Theorem and Corollary.
 * If $$m$$ is positive, and $$n$$ is negative, and $$m<(-n)$$, then a similar argument suffices.

Q.E.D.

Exponentiation Distributes over Multiplication
Next, we discuss a rule, which involves the interaction between multiplication and exponentiation. It can be summarized by the sentence: "Exponentiation distributes over multiplication", in analogy to the more familiar distributive law:  "Multiplication distributes over addition".

The Distributive Law for Exponentiation
Theorem: Suppose that $$x,y$$ are nonzero real numbers, and $$n$$ is an integer. Then $$ (x \cdot y)^n = x^n \cdot y^n $$.

Proof: We begin by proving this theorem when $$ n $$ is a natural number, by induction on $$ n $$. When $$ n = 0$$, we find: When $$ n > 0 $$, we apply induction, and find: This proof by induction yields the theorem, when $$ n $$ is a natural number.
 * $$(x \cdot y)^n = (x \cdot y)^0 = 1 = 1 \cdot 1 = x^0 \cdot y^0 $$.
 * $$(x \cdot y)^n = (x \cdot y)^{n-1} (x \cdot y) $$, by the definition of exponents.
 * $$(x \cdot y)^n = x^{n-1} y^{n-1} x y $$, by induction.
 * $$(x \cdot y)^n = x^{n-1} x y^{n-1} y $$, by the commutativity of multiplication.
 * $$(x \cdot y)^n = x^n y^n $$, by the definition of exponents.

Now, suppose that $$ n < 0 $$. In this case, we find that:
 * $$ (x \cdot y)^n = (1/(x \cdot y))^{-n} = (1/x \cdot 1/y)^{-n} = (1/x)^{-n} \cdot (1/y)^{-n} = x^n y^n. $$

Q.E.D.

Multiplying and Dividing Exponents
Next, we discuss a rule for multiplying and dividing exponents. Once again, this rule can be proven using the definition of exponents. Furthermore, it suggests a natural way of extending exponentiation to allow rational exponents, and incorporate the theory of roots within.

The Rule for Multiplying Exponents
Theorem: Suppose that $$x$$ is a nonzero real number, and $$m,n$$ are natural numbers. Then $$x^{mn} = (x^m)^n$$.

Proof: We prove this theorem by induction on the natural number $$m \cdot n$$. When $$m \cdot n = 0$$, it must be true that $$ m = 0$$ or $$ n = 0 $$. When $$m=0$$, we find that: When $$n=0$$, the same argument applies, yielding the base step of the induction.
 * $$x^{mn} = x^0 = 1 = 1^n = (x^0)^n = (x^m)^n$$.

Now, suppose that $$m \cdot n = N > 0$$. Then, both $$m$$ and $$n$$ are positive. It follows that: Q.E.D.
 * $$x^{mn} = x^{(m-1)n + n} = x^{(m-1)n} x^n$$, using the rules for adding exponents.
 * $$x^{mn} = (x^{m-1})^n x^n $$, by induction (since $$(m-1)n < mn$$).
 * $$x^{mn} = (x^{m-1} \cdot x)^n $$, since exponentiation distributes over multiplication.
 * $$x^{mn} = (x^m)^n $$, by the definition of exponentiation.

Extension to Rational Exponents, Positive Base
Lemma:If $$x$$ is a positive real number, and $$b$$ is a positive integer, there exists a unique positive real number $$y$$ such that $$y^b = x$$. We call $$y$$ the "unique positive b-th root of $$x$$".

Proof: Consider the function $$f(y) = y^b$$, when $$y$$ is a positive real number, and $$b$$ is a positive integer  In this domain, the function is strictly increasing, and satisfies the intermediate value theorem.

Since $$f(y)$$ approaches zero, as $$y$$ approaches zero from the right, there exists a real number $$u$$ such that $$f(u) < x$$ Similarly, $$f(y)$$ becomes arbitrarily large, as $$y$$ approaches infinity. It follows that there exists a real number $$v$$ such that $$f(v) > x$$.

By the intermediate value theorem, it follows that there exists a real number $$y$$ such that $$f(y) = y^b = x$$. By the monotonicity of $$f$$, this $$y$$ is unique as well.

Q.E.D.

Theorem: There is a unique function $$(x,n) \mapsto x^n$$, whose domain consists of those pairs $$(x,n)$$ such that $$0 < x \in \mathbb{R}$$, and $$n$$ is rational, whose restriction to integer values of $$n$$ is the function we have already defined, and which still satisfies the rule: $$ x^{mn} = (x^m)^n$$ for all $$x,m,n$$ within the domain.

Proof: First, we describe such an extension. In other words, we define $$x^{a/b}$$, when $$x$$ is a positive real number, and $$a,b$$ are integers (with $$b > 0$$). By the previous Lemma, there exists a unique positive real number $$y$$ such that $$y^b = x$$.

Define $$x^{a/b} = y^a$$.

First, we must check that this is well-defined, in the sense that using equivalent fractions as exponents yields equal results.

Now, we have extended the function as desired. We need to check that the new function still satisfies the multiplication rule for exponents, and is the unique extension with this property.

Summary
Exponentiation is a function, sending pair of numbers $$(x,n)$$ to a real number $$x^n$$. Initially, this function is defined only on the domain consisting of nonzero real numbers $$x$$ and natural numbers $$n$$.

This function satisfies a number of rules, including: These rules are valid for $$x,m,n$$ in appropriate domains.
 * Addition of Exponents: $$x^{m+n} = x^m x^n$$.
 * Subtraction of Exponents: $$x^{m-n} = x^m/x^n$$.
 * Distributivity of Exponents over Multiplication: $$(xy)^m = x^m y^m$$.
 * Distributivity of Exponents over Division: $$(x/y)^m = x^m / y^m$$.
 * Multiplication of Exponents: $$x^{mn} = (x^m)^n$$.

The theorems proven in the previous section tell us that the function $$(x,n) \mapsto x^n$$ extends to a domain including integer values of $$n$$, and rational values of $$n$$ (when $$x > 0$$), in such a way that the five rules listed above still hold.