Just as repeated addition defines multiplication, so repeated multiplication defines the operation of exponentiation (or raising to a power). We write r.r.r.....r, where there are n occurrences of r, as r^n (or more traditionally rn). The result of the operation r^n is called a power, with r as base and n as exponent (or index).
Some calculators have a key labelled for calculation of powers: instead of the operator sign ^ it usually shows xy.
In contrast to addition and multiplication, exponentiation is non-commutative. We only have r^n = n^r in the special cases when r = n or when r = 2 and n = 4.
Rules for manipulation of exponents.
Product of powers: (a^m).(a^n) = a^(m+n).
Power of product: (a.b)^m = (a^m).(b^m).
Ratio of powers: (a^m)/(a^n) = a^(mHn) provided m>n so that a^m is a multiple of a^n.
Power of ratio: (a/b)^m = (a^m)/(b^m).
Power of power: (a^m)^n = a^(m.n).
For any nonzero number n: 0^n = 0, n^0 = 1
For n=0 these rules yield 0^0 = 1 and 0^0 = 0 which are incompatible, so it is usual to leave the expression 0^0 undefined, i.e. indeterminate.
For any number n: n^1 = n, 1^n = 1. The second power, r^2 is often called the square of r, and r^3 is called the cube of r, these terms coming from applications in geometry. Higher powers are read as ‘r to the nth’.
Exponentiation table (base ten)
| ^ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 0 | ? | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 4 | 8 | 16 | 32 | 64 | 128 | 256 | 512 | 1024 |
| 3 | 1 | 3 | 9 | 27 | 81 | 243 | 729 | 2187 | 6561 | 19683 | 59049 |
| 4 | 1 | 4 | 16 | 64 | 256 | 1024 | 4096 | 16384 | 65536 | 262144 | * |
| 5 | 1 | 5 | 25 | 125 | 625 | 3125 | 15625 | 78125 | 390625 | * | * |
| 6 | 1 | 6 | 36 | 216 | 1296 | 7776 | 46656 | 279936 | * | * | * |
| 7 | 1 | 7 | 49 | 343 | 2401 | 16807 | 117649 | 823543 | * | * | * |
| 8 | 1 | 8 | 64 | 512 | 4096 | 32768 | 262144 | * | * | * | * |
| 9 | 1 | 9 | 81 | 729 | 6561 | 59049 | 531441 | * | * | * | * |
| 10 | 1 | 10 | 100 | 1000 | 10000 | 100000 | * | * | * | * | * |
Theorem (Sequence Rule): The number of ways of choosing a sequence of m elements from a set of n elements is n^m. Proof: There are m choices for each element. So the total number of choices is m.m. ... .m where there are n occurrences of m. This is the definition of n^m.
Example: People staying at a hotel can choose either fish or eggs or bacon or sausages for breakfast. In how many ways can a resident arrange breakfasts for a week? Answer: 4^7 = 16384, there being 4 choices each day.
Example: How many code words of 4 letters can be formed from the 26 letters of the alphabet? Answer 26^4 = 456976. If an alphabet has k letters how many distinct sequences of m letters are possible? The answer is the power k^m.
Example: If beads are available in k colours and m beads are put on a string, how many distinct strings are possible?
The powers of 2 in base ten begin: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576 ... (see the 2-row of the power table). Here are some examples where powers of 2 arise.
Theorem (Counting Subsets): The number of subsets of a set of n elements (including the null and full sets) is 2^n. Proof: A subset either includes or excludes a given element. So there are 2 choices in each of n cases. Hence 2^n choices altogether. In particular applications the null set is often excluded, making the total 2^n – 1, or both the null and the full set are excluded, making 2^n – 2.
Theorem (Partitions): The number of ways of expressing n as a sum of smaller numbers is 2^(n–1) – 1 for n>0. Proof: One way of expressing n is as 1+1+1+...+1 where there are n occurrences of 1. The number of + signs used is n–1. Any other way of expressing n is equivalent to retaining a subset of the + signs and summing the intervening 1s. The number of subsets is, as proved above 2^(n–1), but we must deduct one from this total since the null case gives us n = n, which is not a sum of smaller numbers.
Examples: The first nonzero cases are: 2 = 1+1 (1 expression); 3 = 1+1+1 = 1+2 = 2+1 (3 expressions); 4 = 1+1+1+1 = 1+1+2 = 1+2+1 = 2+1+1 = 2+2 = 1+3 = 3+1 (7 expressions).
Example: The number of ways of partitioning a word of n letters is 2^(n–1) – 1. For example ABCD partitions into A|BCD, AB|CD, ABC|D, AB|C|D, A|BC|D, A|B|CD, A|B|C|D. Each of the m–1 pairs of successive elements is either separated or not, and the case ABCD is not counted.
Example: A Morse Code Problem: In Morse code a letter is represented as one or more dots and dashes in sequence. How many dots and dashes must be used to represented all the letters of the alphabet, plus the ten digits? Answer. We can represent 2 with one sign, 4 with two signs, 8 with three signs, 16 with four signs, total 30. So for the remaining six letters or numerals we must use five dots or dashes.