CAT 2025 Lesson : Number Systems - Conversion to Base 10

2. Positional Number system

As discussed in the Number Theory lesson, we follow a positional number system with base $\bm{10}$. This simply means there are $10$ distinct digits in the system ⇒ $0, 1, 2, 3, 4, 5, 6, 7, 8, 9.$

In this positional number system, the position of a digit has a certain value called the Place Value. The actual value of the digit in that position is called Face Value.

In the number $\bm{34.25}$ of the base $10$ number system, the face values are the digits $3$, $4$, $2$ and $5$, while $10^{1}, 10^{-1}$ and $10^{-2}$ are their respective place values.

When we multiply the face values with their respective place values and add them up, we get the value of the number.

$34.25 = (3 \times 10^{1}) + (4 \times 10^{0}) + (2 \times 10^{-1}) + (5 \times 10^{-2})$

The base can be changed to any integer greater than $1$. The following rules, however, continue to apply
1) A base b number system uses b digits.
2) The place values to the left of the point are $b^{0}, b^{1}, b^{2}, ...$
3) The place values to the right of the point are $b^{-1}, b^{-2}, b^{-3}, ...$

∴ A system with base $6$ has $6$ distinct digits, generally represented using $0, 1, 2, 3, 4, 5.$
(Note that $10$ is not a digit in the base $10$ system. Likewise, $6$ is not a digit in the base $6$ system.)

When we write a number in a base other than $10$, we usually express it as $(number)_\text{base}$. Value of $34.25$ in the base $6$ number system is as follows

$(34.25)_{6} = (3 \times 6^{1}) + (4 \times 6^{0}) + (2 \times 6^{-1}) + (5 \times 6^{-2})$

For number systems with base greater than $10$, additional symbols need to be created for digits. We typically use the letters of the alphabet, $a, b, c$, etc. for the digits $10, 11, 12$, etc.

∴ In the base $12$ number system, the $12$ digits used will be $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a, b.$

3. Conversion to Decimal System

We have been trained to add, subtract, multiply or divide in the base $10$ number system. Therefore, while adding or multiplying numbers, we automatically add or multiply in the base $10$ number system.

∴ To convert a number from base b to base $\bm{10}$, we multiply the face values with the respective place values and add.

So, $(243)_{6} = (2 \times 6^{2}) + (4 \times 6^{1}) + (3 \times 6^{0})$
$= 72 + 24 + 3$
$= 99$

∴ $\bm{(243)_{6}} = \bm{(99)_{10}}$

A number in base $6$ can only have digits from $0, 1, 2, 3, 4$ and $5$. But, when we convert $(243)_{6}$ to base $10$, the expansion contains digits $6$, $7$ and $9$. These digits are of the Base $10$ system. Therefore, the conversion automatically happens when we expand and write.

Example 2

$(153)_{8} + (136)_{14} = (x)_{10}$ . Then, $x =$

Solution

$(153)_{8} = (1 \times 8^{2}) + (5 \times 8) + 3 = 107$

$(136)_{14} = (1 \times 14^{2}) + (3 \times 14) + 6 = 244$

In the decimal system,
$(153)_{8} + (136)_{14} = 107 + 244$
$= (351)_{10}$

Answer: $351$

3.1 Conversion to decimals after the point

Note that digits to the right of the point assume negative powers of the base. For example, if the base is $2$ and the number is $101.11$, the value in the decimal system is $(1 \times 2^{2}) + (0 \times 2^{1}) + (1 \times 2^{0}) + (1 \times 2^{-1}) + (1 \times 2^{-2})$ $= 4 + 0 + 1 + \dfrac{1}{2} + \dfrac{1}{4} = 5.75$

Example 3

What is the decimal equivalent of $(12.24)_{5}$?

Solution

$(12.24)_{5} = (1 \times 5^{1}) + (2 \times 5^{0}) + (2 \times 5^{-1} + (4 \times 5^{-2})$

$= 5 + 2 + \dfrac{2}{5} + \dfrac{4}{25}$

$= 5 + 2 + 0.4 + 0.16 = 7.56$

Answer: $7.56$