---
tags: [binary, binary-encoding]
---

# Signed and unsigned numbers

## Summary

- To represent negative integers in binary we use signed numbers._Signed binary_
  includes negative integers, _unsigned binary_ does not.

- The principal methnod for encoding signed binary numbers is called **two's
  complement**.

## Related notes

In order to represent negative integers alonside positive integers a natural
approach is to divide the available
[encoding space / word length](Binary_encoding.md) into two subsets: one for
representing non-negative integers and one for representing negative integers.

The primary method for doing this is to use _two's complement_. This method
makes signed numbers possible in a way that complicates the hardware
implementation of the binary arithmetic operations as little as possible.

## Two's complement

Signed numbers can be implemented in binary in a number of ways. The differences
come down to how you choose to encode the negative integers. A common method is
to use "two's complement".

> The two's complement of a given binary number is its negative equivalent

For example the two's complement of $0101$ (binary 5) is $1011$. There is a
simple algorithm at work to generate the complement for 4-bit number:

1. Take the unsigned number, and flip the bits. In other words: invert the
   values, so $0$ becomes $1$ and $1$ becomes $0$.
2. Add one

![](/img/unsigned-to-signed.png)

To translate a signed number to an unsigned number you flip them back and still
add one:

![](/img/signed-to-unsigned.png)

### Formal expresssion: $2^n - x$

The heuristic account of deriving two's complement above can be formalised as
follows:

> in a binary system that uses a word size of $n$ bits, the two's complement
> binary code that represents negative $x$ is taken to be the code that
> represents $2^n - x$

Let's demonstrate this, again using -5 as the value we wish to encode.

Applying the formula to a 4-bit system, negative 5 can be derived as follows:

$$
 2^4 -5
$$

$$
 16 -5 = 11
$$

$$
11 = 1011
$$

So basically we carry out the subtraction against the word length and then
convert the decimal result to binary.

We can also confirm the output by noting that when the binary form of the number
and its negative are added the sum will be `0000` if we ignore the overflow bit:

$$
  1011 + 0101 = 0000
$$

(This makes sense if we recall that earlier we derived the complement by
inverting the bits.)

### Advantages

The chief advantage of the two's complement technique of signing numbers is that
its circuit implementation is no different from the adding of two unsigned
numbers. Once the signing algorithm is applied the addition can be passed
through an [adder](Half_adder_and_full_adder.md) component without any special
handling or additional hardware.

Let's demonstrate this with the following addition:

$$
    7 + -3 = 4
$$

First we convert $7$ to binary: $0111$.

Then we convert $-3$ to unsigned binary, by converting $3$ to its two's
complement

$$
0011 \rArr 1100 \rArr 1101
$$

Then we simply add the binary numbers regardless of whether they happen to be
positive or negative integers in decimal:

$$
0111 \\
1101 \\
====\\
0100
$$

Which is 4. This means the calculation above would be identical whether we were
calculating $7 + -3$ or $7 + 13$.

The ease by which we conduct signed arithmetic with standard hardware contrasts
with alternative approaches to signing numbers. An example of another approach
is **signed magnitude representation**. A basic implemetation of this would be
to say that for a given bit-length (6, 16, 32...) if the
[most significant bit](Half_adder_and_full_adder.md#binary-arithmetic) is a 0
then the number is positive. If it is 1 then it is negative.

This works but it requires extra complexity to in a system's design to account
for the bit that has a special meaning. Adder components would need to be
modified to account for it.

## Considerations

A limitation of signed numbers via two's complement is that it reduces the total
informational capacity of a 4-bit number. Instead 16 permutations of bits giving
you sixteen integers you instead have 8 integers and 8 of their negative
equivalents. So if you wanted to represent integers greater than decimal 8 you
would need to increase the bit length.