Base b representation

• Interpreting a string of digits in base b
• Every number can be written uniquely in any base
• Converting to base b

Working in base b

Base b representation is a way to write numbers using the digits {0, 1, …, (b − 1)}.

Common bases:

• you use base 10 (decimal) every day (digits are {0, 1, . . . , 9})

• base 2 (binary) uses digits {0, 1}. It is convenient for digital logic, a digit (called a bit) can be represented using a single wire: the wire has high voltage for 1, low for 0. Binary numbers are often designated by a trailing b: for example 1101b.

• base 16 (hexadecimal) uses the digits {0, 1, 2, . . . , 9, A, B, C, D, E, F}. it is useful becausee a single digit can be represented using 4 bits. Hex numbers are often written with a prefix of "0x": for example 0xFC39.

• base 8 (octal) uses the digits {0, 1, 2, . . . , 7}, and is occasionally used when 3-bit numbers are useful.

A string of digits in base b, written (a_na_n − 1. . . a₃a₂a₁a₀)_b, represents the number a₀b⁰ + a₁b¹ + a₂b² + ⋯ + a_nbⁿ.

Arithmetic in base b

There are two ways to work with numbers in base b. The hard way: convert to base 10, apply the algorithms you already know for addition and multiplication, convert back.

The easy way: use the algorithms you already know for addition, multiplication, division, but remember that (10)_b stands for b and not 10.

We did examples with long addition. Long multiplication and division work the same way.

Writing a number in base b

Theorem: for any n ≥ 1 and b ≥ 2, you can write n in base b. That is, there are digits a₀, a₁, ..., a_k such that (a_ka_k − 1. . . a₂a₁a₀)_b = n.

Note: as with the division algorithm, the proof of this theorem contains the algorithm used to construct the base-b representation.

Proof: by strong induction on n. In the base case, we can choose a₀ = 1. Then since b > 1, b > a₀, and n = a₀ = (a₀)_b.

For the inductive step, assume that any number k < n can be written in base b. We wish to write n in base b. To do so, use Euclidean division to write n = qb + r. Let a₀ = r. By the inductive hypothesis, we can write q in base b: q = (a_lʹa_l − 1ʹ. . . a₂ʹa₁ʹa₀ʹ) (we have to check that q < n, but this is true).

It turns out that these digits of q are also the higher digits of n. That is, a₁ = aʹ₀ and a₂ = aʹ₁ and so on. To check this, we know:

q = ∑ _i = 0^la_iʹbⁱ

We also know that n = qb + r, so

n = qb + r = b(∑ _i = 0^la_iʹbⁱ) + r = a₀ + ∑ _i = 0^la_iʹb^i + 1 = a₀ + ∑ _i = 0^la_i + 1b^i + 1

By changing i to k − 1 in the summation this becomes

n = a₀ + ∑ _k = 1^l + 1a_kb^k = ∑ _k = 0^l + 1a_kb^k

which shows that (a_i) is the base b representation of n.

Note: we stated (but did not prove) that the base b representation is unique (just as quotient and remainder are), this is a good exercise. It follows directly from the uniqueness of the quotient and remainder.