Number Systems

Part 1. Introduction

What is a Number System?

A number system is a system of representing numbers using a set of symbols. Each symbol, or digit, has a specific value, and the combination of digits determines the overall value of the number.

Part 2. Decimal Number System

We use decimal numbers in day to day life. This is the first number system we learn way back to our grade school. They are based on the number 10. They are represented by 0,1,2,3,4,5,6,7,8 and 9.

Decimal number representation.

Back in elementary school, we learned about place value. We understood that numbers are made up of digits, and each digit has a specific value based on its position.

Example No. 1

For Example, the Number 1995.

We represent this number as:

Position	Digit	Multiplier
Ones	5	x 10^0	5
Tens	9	x 10^1	90
Hundreds	9	x 10^2	900
Thousands	1	x 10^3	1000

After adding all the numbers, we get 1995.

Example No. 2

For Example, the Number 3.141.

We represent this number as:

Position	Digit	Multiplier
Thousandths	1	x 10^-3	.001
Hundredths	4	x 10^-2	.04
Tenths	1	x 10^-1	.1
Ones	3	x 10^0	3

After adding all the numbers, we get 3.141.

Part 3. Binary Number System

Binary numbers, the foundation of computer systems, use only two digits: 0 and 1. This simple system allows computers to process information efficiently.

Note: Binary numbers, denoted by the subscript 2.

Binary number representation.

Unlike decimal numbers that use 10 as base, binary numbers use 2 as base.

Example No. 3

For Example, the Number 11100011₂

We represent this binary number as:

Index	Digit	Multiplier
0	1	x 2^0	1
1	1	x 2^1	2
2	0	x 2^2	0
3	0	x 2^3	0
4	0	x 2^4	0
5	1	x 2^5	32
6	1	x 2^6	64
7	1	x 2^7	128

By adding the numbers, we get 99.

Example No. 4

For Example, the Number 11111111₂

We represent this binary number as:

Index	Digit	Multiplier
0	1	x 2^0	1
1	1	x 2^1	2
2	1	x 2^2	4
3	1	x 2^3	8
4	1	x 2^4	16
5	1	x 2^5	32
6	1	x 2^6	64
7	1	x 2^7	128

By adding the numbers, we get 255.

Example No. 5

For Example, the Number 1111.1111₂

We represent this binary number as:

Index	Digit	Multiplier
-4	1	x 2^-4	0.0625
-3	1	x 2^-3	0.125
-2	1	x 2^-2	0.25
-1	1	x 2^-1	.5
0	1	x 2^0	1
1	1	x 2^1	32
2	1	x 2^2	64
3	1	x 2^3	128

By adding the numbers, we get 15.9375.

Part 4. Octal Number System

Octal numbers are a number system based on eight digits: 0, 1, 2, 3, 4, 5, 6, and 7. They are often used in computer science and digital electronics as a more compact representation of binary numbers.

Note: Octal numbers, denoted by the subscript 8.

Octal number representation.

Octal numbers use the base 8.

Example No. 6

For Example, the number 1567₈

We represent this octal number as:

Index	Digit	Multiplier
0	7	x 8^0	7
1	6	x 8^1	48
2	5	x 8^2	320
3	1	x 8^3	512

By adding the numbers, we get 887.

Part 5. Hexadecimal Number System

This number is more compact if compared to binary and octal numbers. Its use 0,1,2,3,4,5,6,7,8,9, A, B, C, D, E and F.

Note: Hexadecimal numbers, denoted by the subscript 16.

Hexadecimal Number Representation.

Hexadecimal number used the base 16.

Example No. 7

For Example, the number 15AC₁₆

We represent this Hexadecimal number as:

Index	Digit	Multiplier
0	C = 12	x 16^0	12
1	A = 10	x 16^1	160
2	5	x 16^2	1280
3	1	x 16^3	4096

By adding the numbers, we get 5548.

Trivia!!!

The binary system was used in early computers like the ENIAC (Electronic Numerical Integrator and Computer) in the 1940s. These computers used vacuum tubes to represent 0s and 1s.

Part 6. Conversion from Decimal to Binary Numbers

There are two methods for converting DEC to BIN: The Direct Method and Successive Division.

Part 6.1. Direct Method

This is the step by step procedure using a direct method.

Step 1: Repeatedly subtract the largest power of 2 that is less than or equal to the decimal number.

Step 2: Continue this process until the remainder is zero. Note: A negative remainder is not allowed.

Step 3: For each power of 2 subtracted, write a '1'.

Step 4: List the '1's and '0's in the order in which the subtractions were performed, starting with the highest power of 2 and proceeding to the lowest."

Example No. 7

Convert 55 to BIN:

Hint: The largest power of 2 we can use is 2⁵ = 32.

Minuend	Subtrahend	Difference	Binary Equivalent
55	32	23	1
23	16	7	1
7	8	-1	0
7	4	3	1
3	2	1	1
1	1	0	1

Read it from top to bottom, the answer is 110111₂

Example No. 8

Convert 78 to BIN:

Hint: The largest power of 2 we can use is 2⁶ = 64.

Minuend	Subtrahend	Difference	Binary Equivalent
78	64	14	1
14	32	-18	0
14	16	-2	0
14	8	6	1
6	4	2	1
2	2	0	1
0	1	-1	0

Read it from top to bottom, the answer is 100110₂

Example No. 9

Convert 345 to BIN:

Hint: The largest power of 2 we can use is 2⁸ = 256.

Minuend	Subtrahend	Difference	Binary Equivalent
345	256	89	1
89	128	-39	0
89	64	25	1
25	32	-7	0
25	16	9	1
9	8	1	1
1	4	-3	0
1	2	-1	0
1	1	0	1

Read it from top to bottom, the answer is 1010110011₂

Part 6.2. Successive Division Method

 This is the step by step procedure using a Successive Division method.

Step 1: Successively divide the decimal number by 2.

Step 2: Place the resulting quotient directly below the dividend.

Step 3: Place the remainder opposite the quotient for clarity.

Step 4: Repeat steps 1-3 until the quotient is less than 2.

Step 5: The equivalent binary number is formed by reading the remainders from bottom to top. The first remainder is the least significant bit (LSB), while the last remainder is the most significant bit (MSB).

Example No. 10

Convert 55 to BIN:

Dividend	Divisor	Quotient	Remainder
55	2	27	1
27	2	13	1
13	2	6	1
6	2	3	0
3	2	1	1
1	2	0	1

Read it from bottom to top, the answer is 110111_2.

Example No. 11

Convert 78 to BIN:

Dividend	Divisor	Quotient	Remainder
78	2	39	0
39	2	19	1
19	2	9	1
9	2	4	1
4	2	2	0
2	2	1	0
1	2	0	1

Read it from bottom to top, the answer is 1001110_2.

Example No. 12

Convert 345 to BIN:

Dividend	Divisor	Quotient	Remainder
345	2	172	1
172	2	86	0
86	2	43	0
43	2	21	1
21	2	10	1
10	2	5	0
5	2	2	1
2	2	1	0
1	2	0	1

Read it from bottom to top, the answer is 101011001₂

Part 7. Conversion from Binary to Octal Numbers

To convert a binary number to octal, we must group the binary digits into groups of three, starting from the right. Then, convert each group into its equivalent decimal number.

Example No. 13

Convert 100011010101010₂ to OCT

Solution:

Group the number into group of three

100 011 010 101 010

	5^th 3 bit	4^th 3 bit	3^rd 3 bit	2^nd 3 bit	1^st 3 bit
	100	011	010	101	010
3 bit binary string to Decimal	4	3	2	5	2

Read it from left to right, the answer is: 43252₈

Example No. 14

Convert 11101010101111010101₂ to OCT

Solution:

Group the number into group of three

011 101 010 101 111 010 101

	7^th 3bit	6^th 3bit	5^th 3bit	4^th 3bit	3^rd 3bit	2^nd 3bit	1^st 3bit
	011	101	010	101	111	010	101
3 bit binary string to Decimal	3	5	2	5	7	2	5

Read it from left to right, the answer is: 3525725₈

Example No. 15

Convert 11111111111111111111111₂ to OCT

Solution:

Group the number into group of three

011 111 111 111 111 111 111 111

	8^th 3bit	7^th 3bit	6^th 3bit	5^th 3bit	4^th 3bit	3^rd 3bit	2^nd 3bit	1^st 3bit
	011	111	111	111	111	111	111	111
3 bit binary string to Decimal	3	7	7	7	7	7	7	7

Read it from left to right, the answer is: 3777777₈

Part 8. Conversion from Octal to Binary Numbers

Converting octal to binary is the reverse process of converting binary to octal. To convert octal to binary, first convert each octal digit to its equivalent 3-bit binary representation, then concatenate the resulting binary strings.

Example No. 16

Convert 43252₈ to BIN

	5^th digit	4^th digit	3^rd digit	2^nd digit	1^st digit
	4	3	2	5	2
Decimal to 3bit Binary String	100	011	010	101	010

Read it from left to right, the answer is: 100011010101010₂

Example No. 17

Convert 3525725₈ to BIN

	7^th digit	6^th digit	5^th digit	4^th digit	3^rd digit	2^nd digit	1^st digit
	3	5	2	5	7	2	5
Decimal to 3bit Binary String	011	101	010	101	111	010	101

Read it from left to right, the answer is: 11101010101111010101₂

Example No. 18

Convert 3777777₈ to BIN

	7^th digit	6^th digit	5^th digit	4^th digit	3^rd digit	2^nd digit	1^st digit
	3	7	7	7	7	7	7
Decimal to 3bit Binary String	011	111	111	111	111	111	111

Read it from left to right, the answer is: 11111111111111111111111₂

Part 9. Conversion from Binary to Hexadecimal Numbers

To convert a binary number to octal, we must group the binary digits into groups of four, starting from the right. Then, convert each group into its equivalent decimal number.

Example No. 19

Convert 1110101000101010101010001₂ to Hexadecimal

Group the number into group of three

0001 1101 0100 0101 0101 0101 0001

	7^th 4 string bit	6^th 4 string bit	5^th 4 string bit	4^th 4 string bit	3^rd 4 string bit	2^nd 4 string bit	1^st 4 string bit
	0001	1101	0100	0101	0101	0101	0001
4 bit binary string to Decimal	1	D	4	5	5	5	1

Read it from left to right, the answer is: 1D45551₁₆

Example No. 20

Convert 1110111100011100011111000111100011111₂ to Hexadecimal

Group the number into group of three

1111 0001 1111 1000 1111 1000 0011 1110 1101 0001

	7^th 4 string bit	7^th 4 string bit	7^th 4 string bit	7^th 4 string bit	6^th 4 string bit	5^th 4 string bit	4^th 4 string bit	3^rd 4 string bit	2^nd 4 string bit	1^st 4 string bit
	1111	0001	1111	1000	1111	1000	0011	1110	1101	0001
4 bit binary string to Decimal	F	1	F	8	F	8	3	E	D	1

Read it from left to right, the answer is: 1DE38F8F1F₁₆

Part 9. Conversion from Hexadecimal to Binary Numbers

Converting hex to binary is the reverse process of converting binary to hex. To convert hex to binary, first convert each hex digit to its equivalent 4-bit binary representation, then concatenate the resulting binary strings.

Example No. 21

Convert 1D45551₁₆ to Binary

	7^th digit	6^th digit	5^th digit	4^th digit	3^rd digit	2^nd digit	1^st digit
	1	D	4	5	5	5	1
Decimal to 4bit Binary String	0001	1101	0100	0101	0101	0101	0001

Read it from left to right, the answer is: 0001110101000101010101010001₂

Example No. 22

Convert 1628DDF₁₆ to Binary

	7^th digit	6^th digit	5^th digit	4^th digit	3^rd digit	2^nd digit	1^st digit
	1	6	2	8	D	D	F
Decimal to 4bit Binary String	0001	0110	0010	1000	1101	1101	1111

Read it from left to right, the answer is: 0001011000101000110111011111₂

Part 10. Conversion from Decimal to Octal Numbers

Step 1: Successively divide the decimal number by 8.

Step 2: Place the resulting quotient directly below the dividend.

Step 3: Place the remainder opposite the quotient for clarity.

Step 4: Repeat steps 1-3 until the quotient is less than 2.

Step 5: The equivalent octal number is formed by reading the remainders from bottom to top. The first remainder is the least significant bit (LSB), while the last remainder is the most significant bit (MSB).

Example No. 23

Convert 456 to Octal number:

Dividend	Divisor	Quotient	Remainder
456	8	57	0
57	8	7	1
7	8	0	7

Let’s read it from bottom to top, the answer is: 710₈

Example No. 24

Convert 1453 to Octal number:

Dividend	Divisor	Quotient	Remainder
1453	8	181	5
181	8	22	5
22	8	2	6
2	8	0	2

Let’s read it from bottom to top, the answer is: 2655₈

Example No. 25

Convert 8595 to Octal number:

Dividend	Divisor	Quotient	Remainder
8595	8	1074	3
1074	8	134	2
134	8	16	6
16	8	2	0
2	8	0	2

Let’s read it from bottom to top, the answer is: 20623₈

Part 11. Conversion from Decimal to Hexadecimal Numbers

Step 1: Successively divide the decimal number by 16.

Step 2: Place the resulting quotient directly below the dividend.

Step 3: Place the remainder opposite the quotient for clarity.

Step 4: Repeat steps 1-3 until the quotient is less than 2.

Step 5: The equivalent oct number is formed by reading the remainders from bottom to top. The first remainder is the least significant bit (LSB), while the last remainder is the most significant bit (MSB).

Example No. 26

Convert 456 to Hexadecimal number:

Dividend	Divisor	Quotient	Remainder
456	16	28	8
28	16	1	C
1	16	0	1

Let’s read it from bottom to top, the answer is: 1C8₁₆

Example No. 27

Convert 75839 to Hexadecimal number:

Dividend	Divisor	Quotient	Remainder
75839	16	4739	F
4739	16	296	3
296	16	18	8
18	16	1	2
1	16	0	1

Let’s read it from bottom to top, the answer is: 1283F₁₆

Fun Fact!!! Octal was once more common in computing due to the early use of 6-bit character codes, which could be easily grouped into 3-bit octal digits. However, hexadecimal has become the preferred choice in modern computing due to its wider use in various data formats and programming languages.

Click Here for Activity No 1!!!

Part 12. Addition and Subtraction of Binary Numbers

Part 12.1 Addition of Binary Numbers

To add binary numbers, we must understand how to add single binary bits.

Basic Rules of Addition

0 + 0 = 1

0 + 1 = 1

1 + 0 = 1

1 + 1 = 0 plus a carry-over 1

Carry-over operations in binary addition are analogous to those in decimal arithmetic. Given that 1 is the maximum digit in the binary system, any sum exceeding 1 necessitates a carry-over.

Example No. 28

Add 11110101 to 11101011

Solution:

	9^th Bit	8^th Bit	7^th Bit	6^th Bit	5^th Bit	4^th Bit	3^rd Bit	2^nd Bit	1^st Bit
Carry Over		1	1	1	1	1	1	1
Augend		1	1	1	1	0	1	0	1
Addend		1	1	1	0	1	0	1	1
Sum	1	1	1	1	0	0	0	0	0

Let’s read it from left to right, the answer is: 11110000₂

Example No. 29

Add 11101010101110 to 1110011001011

Solution:

	15^th Bit	14^th Bit	13^th Bit	12^th Bit	11^th Bit	10^th Bit	9^th Bit	8^th Bit	7^th Bit	6^th Bit	5^th Bit	4^th Bit	3^rd Bit	2^nd Bit	1^st Bit
Carry Over	1	1	1	0	0	0	1	0	0	0	1	1	1	0
Augend		1	1	1	0	1	0	1	0	1	0	1	1	1	0
Addend			1	1	1	0	0	1	1	0	0	1	0	1	1
Sum	1	0	1	0	1	1	1	0	1	1	1	1	0	0	1

Let’s read it from left to right, the answer is: 101011101111001₂

Part 12.2 Subtraction of Binary Numbers

To subtract binary numbers, we must understand how to subtract single binary bits.

Basic Rules of Subtraction

0 - 0 = 0

0 - 1 = 1 with a borrow of 1

1 - 0 = 1

1 - 1 = 0

Example No. 30

Subtract 1000 from 1011

Solution:

	5^th Bit	4^th Bit	3^rd Bit	2^nd Bit	1^st Bit
Borrow
Minuend		1	0	1	1
Subtrahend		1	0	0	0
Difference		0	0	1	1

Let’s read it from left to right, the answer is: 0011₂

Example No. 31

Subtract 101110 from 11100101

Solution:

	8^th Bit	7^th Bit	6^th Bit	5^th Bit	4^th Bit	3^rd Bit	2^nd Bit	1^st Bit
Borrow		0	0	1	1	0
Minuend	1	1	1	0	0	1	0	1
Subtrahend	0	0	1	0	1	1	1	0
Difference	1	0	1	1	0	1	1	1

Let’s read it from left to right, the answer is: 10110111₂

Part 13. Multiplication and Division of Binary Numbers

Part 13.1 Multiplication of Binary Numbers

To multiply binary numbers, we must understand how to multiply single binary bits.

Basic Rules of Multiplication

0 x 0 = 0

0 x 1 = 0

1 x 0 = 0

1 x 1 = 1

Example No. 32

Multiply 11001 by 101

Solution:

	8^th Bit	7^th Bit	6^th Bit	5^th Bit	4^th Bit	3^rd Bit	2^nd Bit	1^st Bit
Multiplicand				1	1	0	0	1
Multiplicator						1	0	1
				1	1	0	0	1
			0	0	0	0	0
		1	1	0	0	1
Summation		1	1	1	1	1	0	1

Let’s read it from left to right, the answer is: 1111101₂

Part 13.1 Division of Binary Numbers

To divide binary numbers, we must understand how to divide single binary bits.

0 / 1 = 0

1 / 1 = 1

Example No. 33

Divide 11011001 by 1011

Solution:

Click Here for Activity No 2!!!

Part 14. Representation of Negative Numbers

There are two common methods for representing negative numbers using binary digits: signed-magnitude and complement number systems. These methods are widely used in computer systems.

Part 14.1 Signed-magnitude System

In binary numbers, an extra bit is used to indicate the sign, often referred to as the sign bit. This sign bit is typically located at the most significant bit (MSB) position, where 1 represents a positive value and 0 represents a negative value. The remaining bits hold the magnitude of the number.

For Examples

01010101₂ = -85

11010101₂ = 85

01111111₂ = -127

11111111₂ = 127

Part 14.2 Complement Number System

There are two types of complements used in number systems: radix complement and radix minus one complement.

Part 14.2.1 Diminished Radix Complement or Radix-Minus-One Complement or (r-1)’s Complement System

Radix minus one complement, also known as (r − 1)'s complement, is a method of representing negative numbers in a positional numeral system. In this system, to find the negative representation of a number, each digit is subtracted from the radix (base) minus one and then the entire result is complemented (all 0s changed to 1s and vice versa).

For example, in the decimal system (radix 10), the 9's complement of 25 is 74.

Solution:

9 – 2 = 7

9 – 5 = 4

Let’s read it from top to bottom, the answer is 74.

In binary (radix 2), the 1's complement of 1011 is 0100.

1 – 1 = 0

1 – 0 = 1

1 – 1 = 0

1 – 1 = 0

Let’s read it from top to bottom, the answer is 0100.

Radix minus one complements are used in some older computer architectures and are still relevant in certain applications like checksum calculations and error detection. However, they have largely been replaced by two's complement representation, which is more efficient and widely used in modern computers.

Part 14.2.2 Radix Complement or True Complement System

Radix complement, also known as r's complement, is a method of representing negative numbers in a positional numeral system. In this system, to find the negative representation of a number, the entire number is subtracted from the radix (base) raised to the power of the number of digits.

For example, in the decimal system (radix 10), the 10's complement of 25 is 75.

Solution:

The radix-minus-one complement of 25 is 74. Add one in 74 = 75.

In binary (radix 2), the 2's complement of 1011 is 0101.

Solution:

The radix-minus-one of 1011 is 0100. Then add one to the 0100, the answer is 0101.

Part 14.3 Subtraction using Complement Number System

Part 14.3.1 Using Radix-minus-one Complement

Procedure:

Step 1. Copy the minuend.

Step 2. Obtain the radix-minus-one complement of the subtrahend.

Step 3. Add the radix-minus-one complemented subtrahend to the minuend.

Step 4. Handle carry-over: a. If the outmost carry-over from step 3 is '1', add this to the least significant digit obtained in step 3. b. If the outmost carry-over from step 3 is '0', take the radix-minus-one complement of the result in step 3 and prefix the negative sign.

Example No. 34

Subtract 17 from 92

Solution.

Step 1. Copy the minuend.

92

Step 2. Obtain the radix-minus-one complement of the subtrahend.

9 – 1 = 8

9 – 7 = 2

Let’s read it from top to bottom: 82

Step 3. Add the radix-minus-one complemented subtrahend to the minuend.

92 + 82 = 174

Step 4. Handle carry-over: a. If the outmost carry-over from step 3 is '1', add this to the least significant digit obtained in step 3. b. If the outmost carry-over from step 3 is '0', take the radix-minus-one complement of the result in step 3 and prefix the negative sign.

74 + 1 = 75 This is the Answer.

Example No. 35

Subtract 111 from 110010

Solution.

Step 1. Copy the minuend.

110010

Step 2. Obtain the radix-minus-one complement of the subtrahend.

1	1	1	1	1	1
0	0	0	1	1	1
1	1	1	0	0	0

The Radix-minus-one of 000111 is 111000

Step 3. Add the radix-minus-one complemented subtrahend to the minuend.

110010 + 111000 = 1101010

Step 4. Handle carry-over: a. If the outmost carry-over from step 3 is '1', add this to the least significant digit obtained in step 3. b. If the outmost carry-over from step 3 is '0', take the radix-minus-one complement of the result in step 3 and prefix the negative sign.

Add 1 to 101010 = 101011₂ this is our answer.

Example No. 36

Subtract 110010 from 000111

Solution:

Step 1. Copy the minuend.

000111

Step 2. Obtain the radix-minus-one complement of the subtrahend.

The radix-minus-one complement of subtrahend is 001101

Step 3. Add the radix-minus-one complemented subtrahend to the minuend.

Add 000111 and 001101, the answer is 010100.

Step 4. Handle carry-over: a. If the outmost carry-over from step 3 is '1', add this to the least significant digit obtained in step 3. b. If the outmost carry-over from step 3 is '0', take the radix-minus-one complement of the result in step 3 and prefix the negative sign.

The radix-minus-one complement of 010100 is 101011 and add negative sign.

Our answer is -101011.

Part 14.3.2 Using Radix Complement

Procedure:

Step 1. Copy the minuend.

Step 2. Obtain the true complement of the subtrahend.

Step 3. Add the true complemented subtrahend to the minuend.

Step 4. Handle carry-over: a. If the outmost carry-over from step 3 is '1', discard it. b. If the outmost carry-over from step 3 is '0', take the true complement of the result in step 3 and prefix the negative sign.

Step 5. The obtained result in step 4 is the difference between the two numbers.

Example No. 37

Subtract 17 from 92

Solution:

Step 1. Copy the minuend.

92

Step 2. Obtain the true complement of the subtrahend.

Let’s get first the radix-minus-one complement of 17

9 – 1 = 8

9 – 7 = 2

The radix-minus-one complement of 17 is 82, then add one.

Therefore, the true compliment of 17 is 83.

Step 3. Add the true complemented subtrahend to the minuend.

Add 83 and 92, the sum is 175.

Step 4. Handle carry-over: a. If the outmost carry-over from step 3 is '1', discard it. b. If the outmost carry-over from step 3 is '0', take the true complement of the result in step 3 and prefix the negative sign

Our answer is 75.

Example No. 38

Subtract 111 from 110010

Solution:

Step 1. Copy the minuend.

110010

Step 2. Obtain the true complement of the subtrahend.

The radix-minus-one complement of 000111 is 111000.

Therefore, the true compliment of 000111 is 111000 + 1 = 111001.

Step 3. Add the true complemented subtrahend to the minuend.

Add 110010 and 111001, the sum is 1101011.

Step 4. Handle carry-over: a. If the outmost carry-over from step 3 is '1', discard it. b. If the outmost carry-over from step 3 is '0', take the true complement of the result in step 3 and prefix the negative sign

Our answer is 101011₂.

Example No. 39

Subtract 110010 from 000111

Solution:

Step 1. Copy the minuend.

000111

Step 2. Obtain the true complement of the subtrahend.

The radix-minus-one complement of 110010 is 001101.

Therefore, the true compliment of 110010 is 001101 + 1 = 001110.

Step 3. Add the true complemented subtrahend to the minuend.

Add 000111 and 001110, the sum is 010101.

Step 4. Handle carry-over: a. If the outmost carry-over from step 3 is '1', discard it. b. If the outmost carry-over from step 3 is '0', take the true complement of the result in step 3 and prefix the negative sign.

The Radix-minus-one complement of 010101 is 101010. Then add 1 to 101010 to get the true complement.

Our answer is – 101011_2.

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Part 15. Binary Codes

Digital systems operate using binary code, a system of 1s and 0s. However, humans interact with information using decimal numbers. To bridge this gap, digital devices must convert data between binary and a more human-readable format. This involves representing each character (number, letter, or symbol) as a unique combination of 1s and 0s. This representation is called a code.

Part 15.1 Binary Coded Decimals (BCD) Codes

Arithmetic operations are typically performed using decimal numbers. However, computers work with binary numbers. Binary Coded Decimal (BCD) was developed to bridge this gap. BCD represents each decimal digit as a four-bit binary equivalent. Numbers 0 to 9 correspond to binary codes 0000 to 1001. Six combinations (1010 to 1111) remain unused. BCD codes are categorized into weighted and unweighted types.

Example No. 40

Convert 936 to BCD Code

Solution:

	3^rd Digit	2^nd Digit	1^st Digit
	9	3	6
Answer	1001	0011	0110

Example No. 41

Convert 7121 to BCD Code

Solution:

	4^th Digit	3^rd Digit	2^nd Digit	1^st Digit
	7	1	2	1
Answer	0111	0001	0010	0001

Part 15.1.1 Weighted BCD Code

The most common weighted BCD code is the 8-4-2-1 code. In this code, the digits in a binary number have weights of 8, 4, 2, and 1. The table below shows several weighted BCD codes.

Decimal Digit	Natural Binary 8-4-2-1	4-2-2-1	2-4-2-1		Biquianary
					5-0		4-3-2-1-0
0	0000	0000		0000		01		00001
1	0001	0001		0001		01		00010
2	0010	0010		0010		01		00100
3	0011	0011		0110		01		01000
4	0100	1000		0100		01		10000
5	0101	0111		1011		01		00001
6	0110	1100		1100		10		00010
7	0111	1101		1101		10		00100
8	1000	1110		1110		10		01000
9	1001	1111		1111		10		10000

Part 15.1.2 Unweighted BCD Code

Unweighted BCD codes are used in data processing, transmission, and measurement, but not in arithmetic operations due to their lack of positional weight. Examples include Excess-3 code, Gray code, and 2 out of 5 code. The table below shows these codes.

Decimal Digit	Excess-3 Code	Gray Code	2 out of 5
0	0011	0000	00011
1	0100	0001	00101
2	0101	0011	00110
3	0110	0010	01001
4	0111	0110	01010
5	1000	0111	01100
6	1001	0101	10001
7	1010	0100	10010
8	1011	1100	10100
9	1100	1101	1100

Note: Excess-3 code is always three greater than the same 8-4-2-1 representation.

Part 15.2 Character Codes

In addition to numbers, digital systems must handle text, symbols, and other non-numeric information. To do this, they use character codes that represent letters, punctuation marks, and special characters.

EBCDIC: An 8-bit code used in IBM systems. It represents characters using two 4-bit groups.

ASCII: The most common character code. It's a 7-bit code divided into data link control, graphic control, and alphanumeric characters. Used for data transfer between computers and devices.

Baudot Code: A 5-level code developed for teletype machines. Represents letters, numbers, punctuation, and special symbols using a "lowercase" and "uppercase" set.

Hollerith Code: Used in punched cards. Each column represents a digit, letter, or symbol. A 12-level code is used to represent data.

Part 15.3 Codes for Actions, Conditions, and States

These codes use simple binary numbers to control actions, indicate conditions, or represent the current state of hardware in a digital system.

Part 15.4 Codes for Detecting and Correcting Errors

When transmitting binary data, errors can occur, leading to misinterpretation. To address this, digital systems use error detection and correction methods. Parity bits, checksum codes, CRC codes, and Hamming codes are examples of these techniques.

Part 15.5 Codes for Data Transmission

Data transmission is a common operation in digital systems. Information is transmitted in binary form as voltage levels between sending and receiving circuits. The format of the signal on the line during transmission is determined by the line code. Manchester code and AMI are examples of line codes used for data transmission.

Fun Fact!!! Emojis originated in Japan in the 1990s as a way to add expressive elements to text messages. They were initially limited to a small set of basic smiley faces and other symbols, but have since expanded to include a wide variety of characters representing emotions, objects, and people.

END of the LESSON