PG2.2 The exponent rules, proved

Simulate

Check the claim with numbers

Lesson 1.4.1 found its exponent rules by counting factors in examples. Before proving them, test three of them with exact whole numbers. The cell tries every base $a$ and $b$ from $-3$ to $3$ , zero included, and exponents $m$ , $n$ from 1 to 6.

checks = fails = 0 for a in range(-3, 4): for b in range(-3, 4): for m in range(1, 7): for n in range(1, 7): checks += 1 product = a**m * a**n == a**(m + n) power = (a**m)**n == a**(m * n) of_product = (a * b)**n == a**n * b**n fails += not (product and power and of_product) print(checks, "checks,", fails, "failures")

⌘/Ctrl + Enter runs the cell

All 1,764 checks pass. Rules that look similar do not. With $a = 2$ , $b = 3$ , $m = 3$ and $n = 2$ :

a, b, m, n = 2, 3, 3, 2 print("a^m * a^n =", a**m * a**n, " a^(m+n) =", a**(m + n), " a^(mn) =", a**(m * n)) print("(a^m)^n =", (a**m)**n, " a^(mn) =", a**(m * n), " a^(m+n) =", a**(m + n)) print("(a+b)^n =", (a + b)**n, " a^n + b^n =", a**n + b**n)

⌘/Ctrl + Enter runs the cell

A product of powers adds the exponents (32, not 64), and a power of a power multiplies them (64, not 32). A power of a sum is not the sum of the powers: 25 against 13.

Lesson 1.4.1 also gave $a^0 = 1$ and $a^{-k} = \frac{1}{a^k}$ . The cell below builds each power from those definitions alone, by repeated multiplication, and tests the quotient rule $\frac{a^m}{a^n} = a^{m-n}$ .

from fractions import Fraction def power(a, k): """a to a whole-number power k, from the definitions only.""" if k > 0: result = Fraction(1) for _ in range(k): result *= a return result if k == 0: return Fraction(1) return 1 / power(a, -k) fails = 0 for a in [Fraction(2), Fraction(-3), Fraction(1, 2)]: for m in range(1, 8): for n in range(1, 8): fails += power(a, m) / power(a, n) != power(a, m - n) print("quotient rule failures:", fails, "of", 3 * 7 * 7) try: power(Fraction(0), 2) / power(Fraction(0), 3) except ZeroDivisionError: print("a = 0: dividing by 0^3 is impossible")

⌘/Ctrl + Enter runs the cell

No failures in 147 cases, for exponents $m - n$ from $-6$ to $6$ . A base of 0 breaks the rule, because it divides by zero.

Conjecture

a

All 1,764 checks and all 147 cases pass. The rules are about every number $a$ and every whole number $m$ and $n$ , so they still need a proof.

Proof from a definition

Every rule here follows from one definition. A proof uses it in both directions: to unpack a power into factors, and to pack factors back into a power.

Rule: A whole-number power

a^n = \underbrace{a \cdot a \cdots a}_{n \text{ factors}} \quad (n \ge 1)

The base $a$ can be any number. The exponent $n$ counts the factors.

The proofs also use two facts about multiplication. Brackets in a product can be moved, so a long product needs none. Factors can be swapped, so they can be put in any order.

Counting factors is a direct proof (lesson PG2.1). Stage PG3 makes "and so on" in a product fully formal, with induction.

Format 1

Reorder the steps

Here is the proof of the product rule $a^m \cdot a^n = a^{m+n}$ , with its lines shuffled. Put them in order.

Brackets in a product can be moved, so this is a single product of $m + n$ factors of $a$ .
So $a^m \cdot a^n$ is a product of $m$ factors of $a$ , followed by $n$ more.
By the definition again, that product is $a^{m+n}$ .
Let $a$ be a number and $m, n \ge 1$ whole numbers. By the definition, $a^m$ is a product of $m$ factors, each equal to $a$ .
Likewise, $a^n$ is a product of $n$ factors, each equal to $a$ .

The same counting proves the next two rules. A power of a power is $n$ copies of $a^m$ , and each copy is $m$ factors of $a$ :

\begin{gathered} (a^m)^n = \underbrace{a^m \cdots a^m}_{n \text{ copies}} \\ = a^{m + m + \cdots + m} = a^{mn} \end{gathered}

The middle step is the product rule, used once for each copy after the first. A power of a product is $n$ copies of $ab$ , and swapping factors puts the $a$ 's first:

\begin{gathered} (ab)^n = \underbrace{(ab) \cdots (ab)}_{n \text{ copies}} \\ = \underbrace{a \cdots a}_{n} \cdot \underbrace{b \cdots b}_{n} = a^n b^n \end{gathered}

Format 2

Fill the gap

The quotient rule, in three cases. Fill each gap as you would a quick answer, such as m+n or 1/a^2, then check.

a \ne 0

How to type answers

Powers with ^, brackets round a power that is a sum: r^k, r^(k+1).
Fractions with / and brackets: (1-r)/(1+r). Multiply with * or side by side: 2k+1.
Press Enter in any box to check them all.

Format 3

Find the error

This proof claims that $(a + b)^n = a^n + b^n$ . The simulation found $(2 + 3)^2 = 25$ but $2^2 + 3^2 = 13$ , so one line must be wrong. Select it, then say why.

a^n

The exponent rules, proved

Check the claim with numbers

Proof from a definition

Rule: A whole-number power

Reorder the steps

Fill the gap

Find the error

Summary card

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