Numbering systems are, in fact, pure mathematics. And because computers are computing devices, learning mathematics makes you more efficient to understand and use computers.

**Number systems are an organized and systematic way of representing numbers**. Each numbering system has its own set of symbols and a set of rules to determine how a quantity is represented and how operations can be performed. For example, Roman Number System has symbols like I, II, III, IV, V, …. and Hindu-Arabic System (Decimal) has symbols like १, २, ३, ४, ५…. or 1, 2, 3, 4, 5….

A numbering system is **positional** if the position of a number matters the value it represents. For example, in 002000 and 000020 the value ‘2’ represents is different because it is on different positions in those two numbers. In first number the 2 represents 2 * 10 ^ 3 (2 into 10 to the power 3) = 2000 whereas in second number 2 represents 2 * 10 ^ 1 (2 into 10 to the power 1) = 20.

A numbering system is **non-positional** if each symbol represents same value regardless of its position. The Roman numbering system is an example of the non-positional numbering system.

Among these numbering systems, we are more concerned with positional numbering systems.

There are different number systems in positional systems based on how many symbols it has. Such as:

**Binary** – 2 different symbols – 0 & 1

**Quinary** – 5 differnt symbols – 0, 1, 2, 3 and 4

**Octal** – 8 different symbols – 0, 1, 2, 3, 4, 5, 6 and 7

**Decimal** – 10 different symbols – 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9

**Hexadecimal** – 16 different symbols – 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F

The value a number in any system represents is **the product of the sum of its digits and its positional value**. for example:

1101 in binary represents

1 * 2 ^ 3 + 1 * 2 ^ 2 + 0 * 2 ^ 1 + 1 * 2 ^ 0 = 8 + 4 + 0 + 1 = 13

2031 in quinary represents

2 * 5 ^ 3 + 0 * 5 ^ 2 + 3 * 5 ^ 1 + 1 * 5 ^ 0 = 250 + 0 + 15 + 1 = 266

1360 in octal represents

1 * 8 ^ 3 + 3 * 8 ^ 2 + 6 * 8 ^ 1 + 0 * 8 ^ 0 = 512 + 192 + 48 + 0 = 752

8316 in decimal represents

8 * 10 ^ 3 + 3 * 10 ^ 2 + 1 * 10 ^ 1 + 6 * 10 ^ 0 = 8000 + 300 + 10 + 6 = 8316

3A2F in hexadecimal represents

3 * 16 ^ 3 + A * 16 ^ 2 + 2 * 16 ^ 1 + F * 16 ^ 0 = 12288 + 2560 + 32 + 15 = 14895

In above example, the numbers in red color are place value. * is used as multiplication sign and ^ as raised power.

For our computer operator and similar jobs examination, we need to learn how to convert a number expressed in one system into another (conversion) and perform binary addition, subtraction, multiplication, and division of binary numbers (operations). Among these, you already learned one method!

## Converting numbers from any system into decimal numbering systems

No mater in which numbering system the given number is, you can **get the decimal equivalent** by

1) multiply each digit by its face value

2) add all the products.

In above examples, we have converted binary 1101 into decimal 13; quinary 2031 into decimal 267; octal 1360 into decimal 752 and hex 3A2F into decimal 14895.

**Converting decimal numbers into any other systems**

To convert given decimal into any other required system, you will perform the following action

1) Divide the number by the base of new system succesively (integer division – no decimals)

2) Note down the remainders in reverse order

For example:

**a) Convert 74 into binary.**

Solution,

Successive division

74 / 2 = 37 remainder 0

37 / 2 = 18 remainder 1

18 / 2 = 9 remainder 0

9 / 2 = 4 remainder 1

4 / 2 = 2 remainder 0

2 / 2 = 1 remainder 0

1 / 1 = 0 remainder 1

Collect remainder in reverse order (bottom to top)

= binary 1001010

**b) Convert 379 into octal**

Solution:

Successive division

379 / 8 = 47 remainder 3

47 / 8 = 5 remainder 7

5 / 8 = 0 remainder 5

Collect remainder in reverse order

= octal 573

**c) Convert 984 into Hexadecimal**

Successive division

984 / 16 = 61 remainder 8

61 / 16 = 3 remainder 13 (i.e. D)

3 / 16 = 0 remainder 3

Collect remainder in reverse order

= octal 3D8

**Quick method to convert decimal into binary**

You can find the binary equivalent of a decimal number by

1) break the number into the sum of different powers of 2

Numbers -> 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, …. are the numbers that can be raised power of 2

Powers of 2 -> 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

2) note down the presence and absense of positions

Example:

**Convert 1568 into binary**

Break the number. The closest number to 1568 that can be raised power of 2 is 1024 (i.e. 2^10)

= 1024 + 544 [Again, the closesest number to 544 is 512 (i.e. 2^9)]

= 1024 + 512 + 32 [Again, the closest number to 31 is 16 (i.e. 2^4)]

= 1024 + 512 + 32

= 2 ^ 10 + 2^9 + 2^5

You don’t need to show all the procedure as above. Just write the closest number and then write closest of the remainder and so on. Finally, when you get the sequence of numbers like below, the job is almost done!

= 1024 + 512 + 32

Power of 2 of above numbers = 10 + 9 + 5

Now, write 1 for all the positions that are present and 0 for the positions that are absent. In above example, position 10,9 and 5 are present other positions are absent. So the binary number will be:

Therefore binary 11000100000 is equivalent to 1567.

If you practice for some time with your mind-work, you’ll be able to convert into binary very quickly. All you need to remember is the numbers that can be raised the power of 2.

## Binary number into Octal

Any number given in Binary can be very easily converted into Octal numbering system

1. Create groups from right most with 3 bits

2. Convert each group into binary

**Example**:

10110100110 into octal

*Solution:* Grouping 10 110 100 110

Convert each group 2 6 4 6

Therefore, 2646 in octal.

### Octal number into binary

Any given octal number can be converted into binary by:

1. Convert each digit into binary.

2. Add leading zeros if there are not 3 bits for every digit

**Example**:

Convert octal 6125 into binary

Solution: convert each digit 6 1 2 5

110 1 10 101

Make 3 bits for each digit 110 001 010 101

Therefore, 110001010101 is binary equivalent.

### Hex number into Binary

Converting the hexadecimal number into the binary is similar to that of octal to binary. Only the difference is the number of bits in each group. In the case of hexadecimal, there should be 4 bits in each group.

**Example**:

Convert 3A2F into binary

Solution: Convert each digit

3 A (10) 2 F (15) [A is 10, B is 11, C is 12, D is 13, E is 14 and F is 15]

11 1010 10 1111

Make 4 bits for each conversion by adding leading zeros

0011 1010 0010 1111

Therefore, 11 1010 0010 1111 is the binary equivalent.

### Binary into Hexadecimal

This is also similar to the binary into octal. Only remember that there should be 4 bits.

Example: 1011101001000110

Solution: Grouping 4 bits from right 1011 1010 0100 0110

Convert each group 11 (B) 10(A) 4 6

Therefore, BA46 is hex equivalent.

Another type of question included in PSC Exam is binary operation – add, subtract, multiply and divide binary numbers.

## Binary Addition

Binary addition is fairly simple and fun to operate. Following are the rules related to binary addition:

1) 1 + 0 = 1 [pretty obvious, isn’t it?]

2) 0 + 1 = 1 [Addition is associative. a+b = b+a, so same result]

3) 0+0 = 0 [What’s new? this is what we are always doing!]

4) 1+1=10 [Here is what binary differs. In the decimal system, 1+1 = 2 and that 2 is 10 in binary. So it is fairly understandable.] While you perform addition, you will write 0 on the sum and take 1 as the balance to add up to the sum of preceding digit.

**Example**

Add 1011010 and 10110

Solution 1 0 1 1 0 1 0 1 0 1 1 0 ------------------ 1 1 1 0 0 0 0

- Addition is done from the last digit. Because both digits are zero, so write 0 on the answer line.
- On next digit (second last digit of both numbers), because both digits are 1, due to the rule #4 the sum is 10. So write 0 on the answer line and take 1 as balance.
- For the third last digit, because given numbers are 0 and 1, their sum is 1 (rule #2). This sum 1 and our previous balance 1 have to be added makes 10. So write 0 on the answer line and take 1 as balance for next digit.
- Now on fourth last digit we have 1 & 0. The sum is 1 and with the previous balance, updated sum is 10. Write 0 on the answer line and take 1 as balance for next digit.
- On fifth last digit, we have 1 & 1 to add. Sum became 10. because we had previous balance 1 the updated sum is 11 (10+1 = 11). Write 1 on the answer line and take 1 as balance.
- Because the second number does not have sixth last digit, 0 from the first number and previous balance 1 makes up the sum of 1. Write that 1 in answer line. Remember there is no balance on this step.
- For the seventh last or the first digit of first number, we don’t have digit from second number and also we don’t have previous balance. Therefore, copy that digit 1 into the answer line.

Thus, finally, we have 1110000 as the sum of given two numbers.

## Binary Subtraction

Binary subtraction is as simple as that addition was. In addition, you just need to remember to adjust the balance, in case of binary subtraction, you’ll remember to adjust the carry over.

The rules for subtraction is:

1) 0-0=0 [obvious!]

2) 1-0=1 [nothing different]

3) 1-1=0 [this is what we are always doing]

4) 0-1=1 with carry over 1 [Here is what binary operations come]

Example

Subtract 10011 from 10110101

Solution: 1 0 1 1 0 1 0 1 1 0 0 1 1 ------------------- 1 0 1 0 0 0 1 0

- Subtraction is done from last digit of the numbers. Because both the numbers has 1 on its last digit, the difference is 0 (1-1=0). Write that 0 on answer line.
- On second last digit, you need to subtract 0-1. Because the difference is 1 with carry over 1, write difference 1 on answer line. Remember there is carry over.
- On third last digit, because carry over 1 is there, subtract it from the 1 of first number. The adjusted difference is 0. From this 0 you need to subtract 0 of second number, because 0 – 0 = 0, write 0 on answer line. No carry over.
- On fourth last digit, both digits are 0, so write 0 on answer line. No carry over.
- On fifth last digit, both digits are 1. So, write 0 on answer line. No carry over.
- Because there are no digits left on second number copy other digits in answer line.

Thus, we got 10100010 as the answer.

### Binary Multiplication

Binary multiplication is very similar to the decimal multiplication except when you are adding. The rules of binary multiplication is as follows:

1) 1 x 1 = 1

2) 1 x 0 = 0

3) 0 x 1 = 0

4) 0 x 0 = 0

Everything perfectly same, isn’t it? Let’s see an example.

Example:

Multiply 1011 by 110.

Solution:

1011 x 110 -------------- 0 0 0 0 1 0 1 1 1 0 1 1 -------------- 1 0 0 0 0 1 0

### Binary division

Binary division is easier than decimal division as there are only two possibilities, either it will won’t go or if it goes, it goes for 1 time.

Example

Divide 110110 by 110

1001 ____________ 110 ) 110110 -110 ---------- 110 -110 ---------- x Therefore, 1001 is the quotient of division and nothing is the remainder. I hope I put forward the method of binary operations in a way that you can start practicing the operation. If you are still confused about numbering systems, conversion and operations, you can drop your question at User Questions Forum

## Further Reading

Number System: http://www.robotroom.com/NumberSystems.html

Binary Number Conversion: http://www.robotroom.com/NumberSystems2.html

Decimal Number Conversion: http://www.robotroom.com/NumberSystems3.html

Octal Number Conversion: http://www.robotroom.com/NumberSystems4.html

Hexadecimal Number Conversion: http://www.robotroom.com/NumberSystems5.html

Joshua Munaweza says

Wow nice information, helped me in my course.