Writing Numbers in Other Bases

We write numbers in base 10: $2024$ means $2 \cdot 10^{3} + 0 \cdot 10^{2} + 2 \cdot 10 + 4$. Nothing is special about ten except our fingers. Computers use base 2, programmers use base 16, and the division algorithm is exactly the tool for moving between bases.

Base $b$. Fix an integer $b \ge 2$. Every positive integer $n$ can be written in exactly one way as $$n = d_k b^{k} + d_{k-1} b^{k-1} + \cdots + d_1 b + d_0,$$ with digits $0 \le d_i \le b - 1$ and $d_k \neq 0$. We write $n = (d_k d_{k-1} \dots d_0)_b$, or $d_k \dots d_0$ with a subscript $b$. Base 16 needs six extra digits: A = 10, B = 11, …, F = 15.

Why it works

Divide $n$ by $b$: every term except $d_0$ is a multiple of $b$, so $$n = b\,(d_k b^{k-1} + \cdots + d_1) + d_0, \qquad 0 \le d_0 < b.$$ By the uniqueness in the division algorithm, $d_0$ is the remainder of $n$ divided by $b$, and the quotient is the number whose digits are $d_k \dots d_1$. Repeat on the quotient and the remainders peel off the digits one at a time, last digit first. That same uniqueness is why the digits are unique.

Worked example: Decimal to binary

Write $44$ in base 2.

$44 = 2 \cdot 22 + 0$
$22 = 2 \cdot 11 + 0$
$11 = 2 \cdot 5 + 1$
$5 = 2 \cdot 2 + 1$
$2 = 2 \cdot 1 + 0$, and $1 = 2 \cdot 0 + 1$. Stop when the quotient is $0$.
The remainders were $0, 0, 1, 1, 0, 1$. Read them from the last to the first: $101100$.
Check: $1 \cdot 32 + 0 \cdot 16 + 1 \cdot 8 + 1 \cdot 4 + 0 \cdot 2 + 0 = 44$.

Answer: $44 = 101100_2$.

Worked example: Decimal to base 7

Write $2024$ in base 7.

$2024 = 7 \cdot 289 + 1$
$289 = 7 \cdot 41 + 2$
$41 = 7 \cdot 5 + 6$
$5 = 7 \cdot 0 + 5$
Read the remainders upward: $5621_7$. Check: $5 \cdot 343 + 6 \cdot 49 + 2 \cdot 7 + 1 = 1715 + 294 + 14 + 1 = 2024$.

Answer: $2024 = 5621_7$.

Worked example: Hexadecimal to decimal

Write $\mathrm{2F3}_{16}$ in base 10.

The digits are $2$, $\mathrm{F} = 15$ and $3$, standing for $16^{2}$, $16^{1}$ and $16^{0}$.
$2 \cdot 256 + 15 \cdot 16 + 3 = 512 + 240 + 3 = 755$.

Answer: $755$.

Place values in base 2. Each hex digit stands for 4 binary digits, since $16 = 2^{4}$.
Power of 2	$2^{0}$	$2^{1}$	$2^{2}$	$2^{3}$	$2^{4}$	$2^{5}$	$2^{6}$	$2^{7}$
Value	1	2	4	8	16	32	64	128

Common mistake

Writing the remainders in the order you found them. For $44$ that gives $001101_2 = 13$, not $44$: the first remainder is the ones digit, which goes on the right. Always check by expanding the answer back into base 10.

Base 16 is popular because it is a shorthand for base 2: split the binary digits into groups of four from the right and replace each group by one hex digit. For example $101100_2 = 0010\ 1100_2 = \mathrm{2C}_{16}$, since $0010_2 = 2$ and $1100_2 = 12 = \mathrm{C}$. Check: $2 \cdot 16 + 12 = 44$.

Binary will come back in fast powers, where the binary digits of an exponent tell you which squares to multiply.

Key idea

To write $n$ in base $b$, divide by $b$ again and again; the remainders are the digits, from the ones digit up. To go back, multiply each digit by its power of $b$ and add.

Writing Numbers in Other Bases

Practice

Builds on

Leads to

Sources and further reading