Number Bases

Lessons

Senior 3 · Mathematics

Number Bases

5 lessons

Lesson 5 of 5

Base two, base twelve and equations in bases

25 min

A clock that tells the time in base two

The front of the railway station in St. Gallen, Switzerland, carries a giant clock with three rows of lamps. Each row is a binary numeral: the top row gives the hours, the middle row the minutes and the bottom row the seconds. A lit lamp is the digit 1 and a dark lamp is 0. In this lesson you read it, meet base twelve, and finish the unit by solving equations in which the unknown is a digit or the base itself.

Definition

Bit: a binary digit, 0 or 1. Byte: a group of 8 bits.

Notation: bits are usually grouped in fours when written, for example $0110\;1001_2$ .

Example: one byte can hold any numeral from $00000000_{2} = 0$ to $11111111_{2} = 255$ : 256 different values.

Important

Rule. With $n$ bits you can write $2^{n}$ different numbers, from 0 up to $2^{n} - 1$ .

Why: each bit has 2 choices, so $n$ bits give $2 \times 2 \times \cdots \times 2 = 2^n$ patterns. The largest, all ones, is one less than $1\underbrace{00\ldots0}_{n}\,{}_2 = 2^n$ .

Computers use base two because an electronic switch has two states that are easy to tell apart, on and off, so every number, letter, colour and sound is stored as bits.

Worked example 1: reading the station clock

In the photograph the lamps are: hours $01001$ , minutes $011001$ , seconds $101110$ . What time is shown?

Solution

\begin{aligned} 01001_2 &= 8 + 1 = 9 \\ 011001_2 &= 16 + 8 + 1 = 25 \\ 101110_2 &= 32 + 8 + 4 + 2 = 46 \end{aligned}

The time is 9:25:46, as the photographer recorded. Five bits are enough for the hours, because $2^5 = 32 \gt 23$ ; six bits are needed for minutes and seconds, because $2^{5} = 32$ is less than 60 but $2^{6} = 64$ is not.

Definition

Duodecimal (base twelve): the place-value system that groups in twelves. It needs twelve digits, so after 0 to 9 it uses two extra symbols for ten and eleven.

Notation: in this unit $\mathrm{A} = 10$ and $\mathrm{B} = 11$ , as in hexadecimal. Some books use $\mathrm{T}$ and $\mathrm{\mathrm{E}}$ (for "ten" and "eleven") instead.

Example: $2\mathrm{B}7_{12} = 2 \times 144 + 11 \times 12 + 7 = 288 + 132 + 7 = 427$ .

Why twelve?

Twelve can be divided exactly by 2, 3, 4 and 6, while ten can be divided only by 2 and 5. That is why twelve appears in so many old measures: 12 months in a year, 12 hours on a clock face, eggs and pencils sold by the dozen (12) and the gross ( $12^{2} = 144$ ). A trader who counts "5 gross, 3 dozen and 8" is already writing a base-twelve numeral: $538_{12} = 5 \times 144 + 3 \times 12 + 8 = 764$ .

Base ten	10	11	12	13	24	100	144
Base twelve	A	B	10	11	20	84	100

Worked example 2: working in base twelve

(a) Write 1000 in base twelve. (b) Work out $7\mathrm{A}_{12} + 5\mathrm{B}_{12}$ .

Solution

(a) $1000 \div 12 = 83$ remainder 4; $83 \div 12 = 6$ remainder 11 ( $= \mathrm{B}$ ); $6 \div 12 = 0$ remainder 6. So $1000 = 6\mathrm{B}4_{12}$ . Check: $6 \times 144 + 11 \times 12 + 4 = 864 + 132 + 4 = 1000$ . ✓

(b) Ones: $\mathrm{A} + \mathrm{B} = 10 + 11 = 21 = 1 \times 12 + 9$ : write 9, carry 1. Twelves: $7 + 5 + 1 = 13 = 1 \times 12 + 1$ : write 1, carry 1. Answer $119_{12}$ . Check: $94 + 71 = 165 = 144 + 12 + 9$ . ✓

Equations in number bases

Some questions give a numeral with a missing digit, or in an unknown base, and ask you to find it. Write each side in base ten using place values; this gives an ordinary equation.

Important

Rule. To solve an equation in number bases: (1) write every numeral in base ten, with the unknown base $b$ or digit $x$ in the place values; (2) solve the equation; (3) keep only answers that make sense: a base must be a whole number greater than every digit used in it, and a digit must be a whole number from 0 to $b - 1$ .

Why the check: the equation in base ten does not know that digits must be smaller than the base, so it can give an answer that no real numeral could have.

Worked example 3: an unknown base, linear

(a) Find $b$ if $23_{b} = 17$ . (b) Find $b$ if $34_{b} = 31_{8}$ . (c) Is there a base $b$ with $52_{b} = 17$ ?

Solution

(a) $2 b + 3 = 17$ , so $b = 7$ . The digits 2 and 3 are less than 7, so $b = 7$ is valid. Check: $23_{7} = 14 + 3 = 17$ . ✓

(b) $31_8 = 3 \times 8 + 1 = 25$ , so $3 b + 4 = 25$ and $b = 7$ . Valid (digits 3 and 4 are less than 7).

(c) $5 b + 2 = 17$ gives $b = 3$ . But the numeral uses the digit 5, which is not allowed in base 3. So there is no such base.

Worked example 4: an unknown base, quadratic

Find $b$ if $132_{b} = 56$ .

Solution $132_b = 1 \times b^2 + 3 \times b + 2$ , so

b^2 + 3b + 2 = 56 \;\Rightarrow\; b^2 + 3b - 54 = 0 \;\Rightarrow\; (b + 9)(b - 6) = 0.

$b = - 9$ is rejected (a base is a positive whole number), so $b = 6$ , which is greater than every digit (1, 3, 2). Check: $36 + 18 + 2 = 56$ . ✓ This is the factorising method from the unit on quadratic equations, used in a new setting.

Worked example 5: a missing digit

Find the digit $x$ if $2 x 3_{5} = 73$ .

Solution $2 \times 25 + 5x + 3 = 73$ , so $53 + 5 x = 73$ , $5 x = 20$ and $x = 4$ . A base-five digit may be 0 to 4, so $x = 4$ is valid: $243_{5} = 73$ , as in the first lesson of this unit. ✓

Warning

Solving is not the end. An answer such as $b = 3$ for a numeral containing the digit 5, $b = \frac{11}{4}$ , or $b = - 9$ must be rejected. If no answer survives the check, say clearly that no such base exists.

Tip

The smallest possible base is a quick first check. Before solving, note the largest digit in the numeral: the base must be at least one more than it. In $132_{b}$ , $b \ge 4$ , so any answer below 4 is wrong at once.

Numberphile: Base 12 (9 min). Dr James Grime makes the case for counting in twelves. Watch for the reasons he gives, such as twelve having more factors than ten, and compare them with the ones in this lesson.

✦

Practice

The last three questions are harder.

How many different numbers can be written with 5 bits? What is the largest?
Convert $11111111_{2}$ to base ten.
Convert $3\mathrm{A}9_{12}$ to base ten.
Convert 500 to base twelve.
Find $b$ if $35_{b} = 23$ .
Find the digit $x$ if $3 x 2_{5} = 97$ .
Find $b$ if $121_{b} = 49$ .
Find $b$ if $213_{b} = 58$ .

Show answersHide answers

$2^{5} = 32$ numbers, from 0 to 31 ( $11111_{2}$ ).
$255$ ( $2^{8} - 1$ )
$3 \times 144 + 10 \times 12 + 9 = 561$
$500 = 3 \times 144 + 5 \times 12 + 8$ , so $358_{12}$ .
$3 b + 5 = 23$ , $b = 6$ (digits 3 and 5 are less than 6).
$75 + 5 x + 2 = 97$ , $x = 4$ .
$b^{2} + 2 b + 1 = 49$ , $(b + 1)^{2} = 49$ , $b + 1 = 7$ (reject $- 7$ ), so $b = 6$ .
$2 b^{2} + b + 3 = 58$ , $2 b^{2} + b - 55 = 0$ , $\Delta = 1 + 440 = 441$ , $b = \frac{-1 \pm 21}{4}$ : $b = 5$ (reject $- 5.5$ ). Check: $50 + 5 + 3 = 58$ .

1A clock that tells the time in base two

2Why twelve?

3Equations in number bases

A clock that tells the time in base two

Why twelve?

Equations in number bases