System of decimal, binary and hexadecimal numeration

In school we learn that there are $10$ symbols or digits that are used to write all the numbers; they are $0, 1, 2, 3, 4, 5, 6, 7, 8$ y $9$ .

To write numbers bigger than $9$ we repeat the same numbers but in a certain order.

For example, using $1$ and $3$ , $13$ and $31$ can be written, so that the position in which the symbols or digits are placed determines the value of the final number: $13$ is less than $31$ .

According to this method, a number consists of units, tens, hundreds, etc.

Example

$13$ has $1$ ten and $3$ units.

$31$ has $3$ tens and $1$ unit.

$131$ has $1$ hundred, $3$ tens and $1$ unit.

This classification also can be expressed this way:

Thirteen is ten times one plus three.

Thirty one is three times ten plus one.

Hundred thirty one is one hundred times one plus three times ten plus one.

Numerically, the previous classifications are written as follows:

$13 = 1 \cdot 10 + 3$

$31 = 3 \cdot 10 + 1$

$131 = 1 \cdot 100 + 3 \cdot 10 + 1$

An equivalent way of expressing the same is:

$13 = 1 \cdot 10^{1} + 3 \cdot 10^{0}$

$31 = 3 \cdot 10^{1} + 1 \cdot 10^{0}$

$131 = 1 \cdot 10^{2} + 3 \cdot 10^{1} + 1 \cdot 10^{0}$

The three numbers can be separated by powers of $10$ . To represent the units it is possible to use $10$ raised to $0$ , since $a^{0} = 1.$

The tens are represented by $10$ raised to $1$ and the hundreds with $10$ raised to $2$ . If needed, the thousands would be represented by $10$ raised to $3$ , the tens of thousands with $10$ raised to $4$ , and so on.

Example

$13.031 = 1 \cdot 10^{4} + 3 \cdot 10^{3} + 0 \cdot 10^{2} + 3 \cdot 10^{1} + 1 \cdot 10^{0}$

In this case, as there are no hundreds, it is represented by multiplying $0$ by the power that corresponds to the hundreds ( $10$ raised to $2$ ).

All these numbers are expressed in the system of decimal numeration and that's why they can be separated into powers of $10$ . This is the most known system to count and to group objects, but is not the only one.

Example

The following numbers are expressed in systems other than the decimal:

$(11011)_{2}$

$(1 B)_{16}$

Although both are equivalent to the same decimal number, $27$ .

In the first case, $(11011)_{2}$ , the subscript indicates that the base of the system is $2$ , also known as binary system. In this system only two symbols or digits are used, $0$ and $1$ , and the decomposition is done in powers of $2$ : $(11011)_{2} = 1 \cdot 2^{4} + 1 \cdot 2^{3} + 0 \cdot 2^{2} + 1 \cdot 2^{1} + 1 \cdot 2^{0}$

That is, this number decomposes in $1$ group of $16$ , $1$ de $8$ , none of $4, 1$ de $2$ and $1$ of $1$ .

If the operation is solved the equivalent number "translated" to the decimal system is obtained:

$16 + 8 + 0 + 2 + 1 = 27$

To decompose easily, it is just necessary to bear in mind that the first digit of the number represents the maximum power, and the exponent is decreasing as we move to the right.

Using the previous case, we can deduce that a number with $5$ digits has $5$ powers, which will be decreasing from $4$ to $0$ .

The second example is the same number, $27$ , but expressed in the numeration system in base $16$ or hexadecimal, as the subscript indicates: $(1 B)_{16}$

The base also indicates the number of symbols or numbers that are used in the system. The binary system, whose base is $2$ uses two numbers: $0$ and $1$ .

In the system hexadecimal, $16$ are used, from $0$ to $15$ , but to avoid confusion we use the letters from A to F to refer to the symbols from $10$ to $15$ . So now we can understand that, in the example, B represents the number $11$ .

In the system hexadecimal, the decomposition is carried out in powers of $16$ : $(1 B)_{16} = (1 (11))_{16} = 1 \cdot 16^{1} + 11 \cdot 16^{0} = 16 + 11 = 27$

So that $1 B$ in hexadecimal system involves having $1$ group of $16$ and $11$ of $1$ .

The following examples will give us a little bit of practice in looking at the decimal equivalent of numbers expressed in other systems.

Example

$(111)_{3}$

At first sight, we can see that it is a number in base $3$ (ternary system), so that it uses $3$ symbols or digits $(0, 1$ and $2)$ and the decomposition is done in powers of $3$ : $(111)_{3} = 1 \cdot 3^{2} + 1 \cdot 3^{1} + 1 \cdot 3^{0} = 9 + 3 + 1 = 13$

As the number has $3$ digits, the powers will decrease from $2$ to $0$ , so that the number is formed by $1$ group of $9, 1$ of $3$ and $1$ of $1$ .

Example

$(23)_{5}$

It is a number in base $5$ , therefore, it uses $5$ symbols: $0, 1, 2, 3$ and $4$ . The decomposition is done in powers of $5$ :

$(23)_{5} = 2 \cdot 5^{1} + 3 \cdot 5^{0} = 10 + 3 = 13$

Example

$(15)_{8}$

It is a number in base $8$ or octal, it uses $8$ symbols (from $0$ to $7$ ) and the decomposition is done in powers of $8$ :

$(15)_{8} = 1 \cdot 8^{1} + 5 \cdot 8^{0} = 8 + 5 = 13$

These last three examples are the same decimal number, $13$ .