Number Bases « CoDefined

The number base of a number is the number of unique digits possible in that number. eg:

Binary - Base-2 contains 2 digits: 1 and 0
Hexadecimal - Base-16 contains 16 digits: 0-F
Decimal - Base-10 contains 10 digits: 0-9
Octal - Base-8 contains 8 digits: 0-7

Counting in binary, decimal and hexadecimal

In each number system, we count from 0 up to the maximum value in a single digit. eg in decimal:

When we reach a maximum in one digit, we start a new column of digits and reset the first column.

                        .
                        .
                        .
                        8
                        9 - Max digit
Second column set to -  10
 one                    11
                        12
                        13
                        .
                        .
                        .

Place values

In decimal, the place values increase by a factor of 10 as new columns are added.

Units of binary

1 binary digit = 1 bit 8 bits = 1 byte 4 bits = 1 nybble $\gets$ Joke unit 2 bits = crumb $\gets$ Joke unit

Multiples of bytes and bits

The unit multipliers of bits and bytes follow the S.I. convention:

Kilo = 1,000 = k
Mega = 1,000,000 = M
Giga = 1,000,000,000 = G
Terra = 1,000,000,000,000 = T
Peta = 1,000,000,000,000,000 = P
Exa = 1,000,000,000,000,000,000 = E
Zetta = 1,000,000,000,000,000,000,000 = Z
Yotta = 1,000,000,000,000,000,000,000,000 = Y

These prefixes are in common use but are not the best units to use for binary data. Becuase of this, we have preferred units:

(ki) Kibi = $1024$ = $2^1$
(Mi) Mebi = $1024^2$ = $2^{20}$
(Gi) Gibi = $1024^3$ = $2^{30}$

Bytes and hexadecimal

1 byte is 2 nybbles.

Hexadecimal uses the digits 0-9 A-F to represent 16 different values. A binary nybble can contain 16 different values so it maps neatly to hexadecimal.

Decimal	Hexadecimal	Binary
0	0	0000
7	7	0111
12	C	1100

To convert hex to decimal

6f to decimal $6\times16^1+F\times16^0 = 6\times16+F\\ 6\times16 + 15 = 111_{10}$

Note: It is unlikely but possible that you might need to convert a multi-digit hex number to decimal. eg: $6\times16^2+F\times16^1+C\times16^0=1788_{10}$

Number conversion exercises

Decimal	Hexadecimal	Binary
243	F3	11110011
97	61	01100001
123	7B	01111011
195	C3	11000011
189	BD	10111101
251	FB	11111011
26	1A	00011010
90	5A	01011010

More about binary numbers

In $1001 111$ , the first $1$ is the most significant bit (MSB), and the last $1$ is the least significant bit (LSB). This example is also an unsigned binary integer. This means that the kind of number this is is never specified.

Note The type of binary number needs to be specified if you are using anything other than an unsigned binary integer, because pure binary numbers are ambiguous.

Addition of unsigned binary integers

$\begin{align} &01101010& \\ &\underline{00101001}+&\\ &\underline{10010011}&\\ &^{1\,1\,\;\,1}& \end{align}$

Overflow If 2 numbers are added together and the sum is bigger than the spaaaaaace allocated, an overflow will occur.

Multiplying unsigned binary integers

eg:

$10111\\ \; \;\;\;\;\;\;\;\;\;11 \times \\ \overline{101110 }\;\;\\ \,\underline{10111} \\ 1000101\;\;\;$

Numbers with a fractional part

Often, calculations produce numbers that have fractional parts, such as $5\div3$ . The decimal system is very similar to the binary system.: Decimal:

$100$	$10$	$1$	$.$	$\frac{1}{10}$	$\frac{1}{100}$
$2$	$5$	$6$	$.$	$7$	$1$

Binary:

$16$	$8$	$4$	$2$	$1$	$.$	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{8}$
$1$	$1$	$0$	$1$	$0$	$.$	$1$	$0$	$1$

Note In the binary example, it was assumed that the 3 LSBs were the fractional part becuase in binary, there is no symbol for “ $.$ ” and so the position of the “ $.$ ” must be assumed. This assumption is known as fixed point coding.

We can also use 2’s compliment with this fixed point coding. eg:

$-16$	$8$	$4$	$2$	$1$	$.$	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{8}$
$1$	$1$	$0$	$1$	$0$	$.$	$1$	$0$	$1$

Floating point form

Often, scientific notation is used to write numbers. for example: $A\times10^8$

A similar notation is used when 2’s compliment is used to to represent signed numbers that range from small to large: M $\times$ Base $^E$

Floating points in binary

Floating point gets around the limitations of fixed point by using a format similar to scientific notation. eg: $3.141\times10^3$

The “ $3.141$ ” is known as the Mantissa and the “ $\times10^3$ ” is known as the exponent. So say there was a 32 bit system, the mantissa would be 24 bits and the exponent would be 8 bits.

Recap of 2’s compliment If it is positive, convert to binary normally. If it is negative, convert to binary, invert the bits and add 1.

$\overset{\mathsf{Mantissa}}{\overset{-2^{0}}{0}\ \overset{•}{}\ \overset{2^{-1}}{1}\ \overset{2^{-2}}{1}\ \overset{2^{-3}}{0}\ \overset{2^{-4}}{1}\ \overset{2^{-5}}{0}}\ \ \ \ \ \ \ \ \overset{\mathsf{Exponent}}{\overset{-2^3}{0}\ \overset{2^2}{1}\ \overset{2^1}{0}\ \overset{2^0}{1}}$ This can be written as $0.11010\times10^{0101}$ . The exponent evaluates to $+5_{10}$ This means that the binary point is moved 5 places to the right: $0.\overset{↷}{1}\overset{↷}{1}\overset{↷}{0}\overset{↷}{1}\overset{↷}{0}_2 = 011010.0_2 \mathsf{\ or\ just\ }011010_2$ Which evaluates to $16 + 8 + 2 = 26_{10}$

It is also easy to convert back: You simply convert the decimal number to binary. You then work out how many places you need to move the binary point to move it to before the MSB. The “ $.$ ” is implied after the MSB.

Negative mantissas can also be used:

$\overset{\mathsf{Mantissa}}{\overset{-2^{0}}{1}\ \overset{2^{-1}}{1}\ \overset{2^{-2}}{0}\ \overset{2^{-3}}{0}\ \overset{2^{-4}}{1}\ \overset{2^{-5}}{1}}\ \ \ \ \ \ \ \ \overset{\mathsf{Exponent}}{\overset{-2^3}{0}\ \overset{2^2}{0}\ \overset{2^1}{1}\ \overset{2^0}{1}}$

So the point is implied after the MSB:

$\overset{\mathsf{Mantissa}}{\overset{-2^{0}}{1}\ \overset{•}{}\ \overset{2^{-1}}{1}\ \overset{2^{-2}}{0}\ \overset{2^{-3}}{0}\ \overset{2^{-4}}{1}\ \overset{2^{-5}}{1}}\ \ \ \ \ \ \ \ \overset{\mathsf{Exponent}}{\overset{-2^3}{0}\ \overset{2^2}{0}\ \overset{2^1}{1}\ \overset{2^0}{1}}$

The exponent moves the binary point $11_2$ or $3_{10}$ positions to the right:

$\overset{\mathsf{Mantissa}}{\overset{-2^{3}}{1}\ \overset{2^{2}}{1}\ \overset{2^{1}}{0}\ \overset{2^{0}}{0}\ \overset{•}{}\ \overset{2^{-1}}{1}\ \overset{2^{-2}}{1}}\ \ \ \ \ \ \ \ \overset{\mathsf{Exponent}}{\overset{-2^3}{0}\ \overset{2^2}{0}\ \overset{2^1}{1}\ \overset{2^0}{1}}$

Which equates to $1100.11_2$ or $-3.25_{10}$

Converting from unsigned decimal numbers to unsigned binary

With integers this is easy:

$15_{10} \to 1111_2$

And with some decimal:

$15.625_{10} \to 1111.101_2\ \ \ =\ \ \ .\overset{0.5}{1}\ \overset{0.25}{0}\ \overset{0.125}{0}$

But with others:

$15.6_{10} \to 1111.1010\dots_2$

Which has to be rounded and when converted back $\neq15.6$ .

Note Converting decimal numbers to fixed point unsigned binary introduces errors.

Research task

Find out how signed binary integers can be stored using 2’s compliment.

Two’s complement is a clever way of storing integers so that common math problems are very simple to implement.

To understand, you have to think of the numbers in binary.

It basically says,

for zero, use all 0’s.
for positive integers, start counting up, with a maximum of 2^{(number of bits - 1)}-1.
for negative integers, do exactly the same thing, but switch the role of 0’s and 1’s (so instead of starting with 0000, start with 1111 - that’s the “complement” part).

Let’s try it with a mini-byte of 4 bits (we’ll call it a [nibble][1] - 1/2 a byte).

0000 - zero
0001 - one
0010 - two
0011 - three
0100 to 0111 - four to seven

That’s as far as we can go in positives. 2³-1 = 7.

For negatives:

1111 - negative one
1110 - negative two
1101 - negative three
1100 to 1000 - negative four to negative eight

Note that you get one extra value for negatives (1000 = -8) that you don’t for positives. This is because 0000 is used for zero.

Doing this, the first bit gets the role of the “sign” bit, since it is always ‘1’ for negative numbers, and ‘0’ for non-negatives (zero and positive).