Number Systems

A number system is a set of symbols (digits) that are assembled according to precise writing rules, making it possible to write, read and express numbers.

A base, in a positional number system, is the number of symbols (digits) that are used to represent numbers.

A base |10| number system (decimal numeration) uses |10| digits from |0| to |9,| while a base |2| number system (binary numeration) uses only |2| digits, zero |(0)| and one |(1).|

Writing numbers as we know it uses Indo-Arabic digits, namely |0,| |1,| |2,| |3,| |4,| |5,| |6,| |7,| |8,| |9.| In theory, this form of writing only applies to writing numbers in base |10.|

However, to make things easier to understand, these same digits will be used throughout this concept sheet. To distinguish between the different bases used, we will use a subscript for reference purposes: ||576_\boldsymbol{\color{#333fb1}{12}}=576\ \text{base}\ \boldsymbol{\color{#333fb1}{12}}||

Since many different forms of number writing existed around |3500| BC, humans at that time had to use the means at their disposal to count. In this case, they used the fingers of their left hand (5) and the phalanges of their right hand (12). So, 1 finger of the left hand was equal to all 12 phalanges of the right hand.

For example, ||\begin{align}\text{left index raised}+{1^\text{st}\ \text{right index phalange}}&=4\times\underbrace{\boldsymbol{\color{#333FB1}{12}}^1}_{\text{place value}}+10\times\underbrace{\boldsymbol{\color{#333FB1}{12}}^0}_{\text{place value}}\\&=4\times\boldsymbol{\color{#333FB1}{12}}+10\times\boldsymbol{\color{#333FB1}{1}}\\&=48+10\\&=58\end{align}||

​Just like with base |10| numeration, we can use the place value position system and the expanded form of the numbers to correctly interpret the value of a number in base |12.|

As well, this way of counting using fingers and phalanges is also related to the sexagesimal system, in base |60.|

Taking all |5| fingers of the left hand and all |12| phalanges of the right hand into consideration, there are a total of |60| possibilities: |(5 \times 12 =60).|

In today's society, this base is found in our units of time: seconds, minutes and hours.

How many seconds are there in 5 hours 17 minutes and 34 seconds? ||\begin{align}(5\ \text{h}\ 17\ \text{min} \ 34 \ \text{s})​_{\boldsymbol{\color{#333FB1}{60}}}&=5\times\underbrace{\boldsymbol{\color{#333FB1}{60}}^2}_{\text{place value}}+17\times\underbrace{\boldsymbol{\color{#333FB1}{60}}^1}_{\text{place value}}+34\times\underbrace{\boldsymbol{\color{#333FB1}{60}}^0}_{\text{place value}}\\&=5\times\boldsymbol{\color{#333FB1}{60\times60}}+17\times\boldsymbol{\color{#333FB1}{60}}+34\times\boldsymbol{\color{#333FB1}{1}}\\&=18\ 000+1020+34\\&=19\ 054\ \text{s}\end{align}||

When large quantities had to be represented, 10 fingers and 2 hands were not enough. So, between the 11^th and 14^th century BC, a new system emerged in China–the decimal system, or base 10.

Accustomed to counting with 10 fingers, it seemed natural to use 10. However, the introduction and the official use of the number 0 would not happen until the 5^th century AD.

Along with the arrival of the value zero came the idea of grouping elements to better represent them. A group of 10 ones became a ten; a group of 10 tens became a hundred, and so on.

The Mayan people also established their calendar in the 5^th century using base 20. Why? Not only did they use their 10 fingers, but also their 10 toes!

Since this is not a base 10 system, the symbols used to represent various quantities are not the digits as we know them.

Using today’s digits, determine the number that corresponds to this image.

​||\begin{align}\text{This number}​​_{\boldsymbol{\color{#333FB1}{20}}}&=11\times\underbrace{\boldsymbol{\color{#333FB1}{20}}^2}_{\text{place value}}+6\times\underbrace{\boldsymbol{\color{#333FB1}{20}}^1}_{\text{place value}}+16\times\underbrace{\boldsymbol{\color{#333FB1}{20}}^0}_{\text{place value}}\\&=11\times\boldsymbol{\color{#333FB1}{20\times20}}+6\times\boldsymbol{\color{#333FB1}{20}}+16\times\boldsymbol{\color{#333FB1}{1}}\\&=4400+120+16\\&=4536_{10}\ \text{is its decimal value}\end{align}||

Unlike the other bases, the base 2 binary system is quite recent. Though it was actually discovered in the late 17^th century, it has really only been heavily used since the early 20^th century.

This way of coding information is the only method understood by the various electronic components of a computer. Generally associated with bits (from the term “binary information digits”), this code has only two symbols: 0 and 1.

To standardize computer coding, these bits were combined into groups of 8 to form bytes. Each byte is associated with a digit, a letter, a punctuation symbol, etc. In other words, each byte is associated with a specific character in the Latin alphabet. Here is an overview of the table of characters used by the ISO/IEC 8859-1 code, namely the “computer alphabet” according to byte coding.​

Since each character has its own unique byte, a decimal value can be derived. Moreover, this decimal value often corresponds to universal shortcut keys.

What is the decimal value of this character?

||\begin{align}11001000_{\boldsymbol{\color{#333FB1}{2}}}&=1\times\underbrace{\boldsymbol{\color{#333FB1}{2}}^7}_{\text{place value}}+1\times\underbrace{\boldsymbol{\color{#333FB1}{2}}^6}_{\text{place value}}+1\times\underbrace{\boldsymbol{\color{#333FB1}{2}}^3}_{\text{place value}}\\&=1\times\boldsymbol{\color{#333FB1}{2\times2\times2\times2\times2\times2\times2}}+1\times\boldsymbol{\color{#333FB1}{2\times2\times2\times2\times2\times2}}+1\times\boldsymbol{\color{#333FB1}{2\times2\times2}}\\&=128+64+8\\&=200_{10}\ \text{is its place value position}\end{align}||

By definition, all numbers from |0| to |15| inclusively can be used to form hexadecimal code. However, this could be a problem since there could only be one character per position. The following naming system was adopted.||\begin{align}​0&=0&4&=4&8&=8&12&=C\\1&=1&5&=5&9&=9&13&=D\\2&=2&6&=6&10&=A&14&=E\\3&=3&7&=7&11&=B&15&=F\end{align}||What is the decimal value of this character?

||\begin{align}\text{CA}​​_{\boldsymbol{\color{#333FB1}{16}}}&=C\times\underbrace{\boldsymbol{\color{#333FB1}{16}}^1}_{\text{place value}}+A\times\underbrace{\boldsymbol{\color{#333FB1}{16}}^0}_{\text{place value}}\\&=12\times\underbrace{\boldsymbol{\color{#333FB1}{16}}^1}_{\text{place value}}+10\times\underbrace{\boldsymbol{\color{#333FB1}{16}}^0}_{\text{place value}}\\&=12\times\boldsymbol{\color{#333FB1}{16}}+10\times\boldsymbol{\color{#333FB1}{1}}\\&=192+10\\&=202_{10}\ \text{is its decimal value}\end{align}||

From Base 10 to Another Base

How can we write 345 in base 12?

1. Determine the place value position of the identified base.

||\begin{align}1^\text{st}\ \text{position}​​_{\boldsymbol{\color{#333FB1}{12}}}&=12^0=1\\2^\text{nd}\ \text{position}​​_{\boldsymbol{\color{#333FB1}{12}}}&=12^1=12\\3^\text{rd}\ \text{position}​​_{\boldsymbol{\color{#333FB1}{12}}}&=12^2=144\\4^\text{th}\ \text{position}​​_{\boldsymbol{\color{#333FB1}{12}}}&=12^3=1728\end{align}||

2. Subtract, starting with the largest possible position.

||\begin{align}&345-(144+144)&&3^\text{rd}\ \text{place}\\ \Rightarrow\ &57-(12+12+12+12)&&2^\text{nd}\ \text{place}\\\Rightarrow\ &9-(1+1+1+1+1+1+1+1+1)&&1^\text{st}\ \text{place}\end{align}||

3. Match a quantity to each position.

||\begin{align}345_{10}&=\underbrace{144 + 144}_{3^\text{rd}\ \text{place}}+\underbrace{12+12+12+12}_{2^\text{nd}\ \text{place}}+\underbrace{1+1+1+1+1+1+1+1+1}_{1^\text{st}\ \text{place}}\\&=2\times\underbrace{144 }_{\text{place value}}+4\times\underbrace{12}_{\text{place value}}+9\times\underbrace{1}_{\text{place value}}\\&=2\times\underbrace{\boldsymbol{\color{#333FB1}{12}}^2}_{\text{place value}}+4\times\underbrace{\boldsymbol{\color{#333FB1}{12}}^1}_{\text{place value}}+9\times\underbrace{\boldsymbol{\color{#333FB1}{12}}^0}_{\text{place value}}\\&=249_{\boldsymbol{\color{#333FB1}{12}}}\end{align}||

Converting From Any Base to Base 10

How can we write |{346_8}| in base 10?

1. Expand according to the value of each place.

||346_\boldsymbol{\color{#333FB1}{8}}=3\times\boldsymbol{\color{#333FB1}{8}}^2+4\times\boldsymbol{\color{#333FB1}{8}}^1+6\times\boldsymbol{\color{#333FB1}{8}}^0||

2. Calculate the total.

||\begin{alignat}{13}&=3\times64&&+4\times8&&+6\times1\\&=192&&+32&&+6\\&=230_{\boldsymbol{\color{#333FB1}{10}}}\end{alignat}||

Here are the symbols used to change from writing in base |10| to writing in base |12:|

|0,| |1,| |2,| |3,| |4,| |5,| |6,| |7,| |8,| |9,| |A| and |B.|

Here, |A| stands for “ten” and |B| for “eleven.”

For example, to write |22_{10}| in base |12,| |22_{10}| becomes |1A| in base |12.| ||\begin{align}22_{10}&=1\times12^1+10\times12^0\\&=1A_{12}\ \text{(1 dozen and 10 ones)}\end{align}||