practising C, C++ and JAVA programming with examples: Number system

What is number system?
Answer: The system concerned with the number and represented by a sequence of digits is called number system. It plays a vital role in computing and electronics. Number system also refers to the digits, its arrangement, positional value, and base of number system.
What is base or radix of number system?
Answer: The total number of digits used by the particular number system is called base or radix of that number system. For example, Base of Binary number system is 2 because it uses two digits 0 and 1 only.
List the types of number system with their bases.
Answer: There are four types of number system and their bases are given below:
Number System Base
(i) Binary Number System 2
(ii) Octal Number System 8
(iii) Decimal Number System 10
(iv) Hexadecimal Number System 16
Define Binary number system.
Answer: The number system having base two(2) and consists of digits: 0 and 1 is called Binary number system.
Define Octal number system.
Answer: The number system having base eight(8) and consists of digits: 0,1,2,3,4,5,6, and 7 is called Octal number system.
Define Decimal (Denary) number system.
Answer: The number system having base ten(10) and consists of digits: 0,1,2,3,4,5,6,7,8 and 9 is called Decimal(Denary) number system.
Define Hexadecimal number system.
Answer: The number system having base sixteen(16) and consists of digits: 0,1,2,3,4,5,6,7,8, 9 and alphabets A,B,C,D,E and F is called Hexadecimal number system.
Where,
A B C D E F
10 11 12 13 14 15
Why do digital computers use binary number system for its operations?
Answer: The number system having base two and consists of digits: 0 and 1 is called Binary number system. It has two bits 0 and 1. An electronic circuit has two states either ON & OFF state. The bit 1 represents high voltage (ON state) and the bit 0 represents the low voltage (OFF state) of an electronic circuit. That's why a digital computer uses binary number for its operations.
Fundamentals Rule for Number Conversion:
Number System Conversion:
We mainly focused on twelve(12) types of conversion for four types of number systems which can be grouped as follows based on the common method to be used for conversion.
(i) Decimal to other systems(Binary / Octal / Hexadecimal) conversion
(ii) Other systems to Decimal conversion
(iii) Binary to Octal and Hexadecimal Conversion
(iv) Octal and Hexadecimal to Binary Conversion
(v) Octal to Hexadecimal Conversion
(vi) Hexadecimal to Octal Conversion
Decimal to Binary, Octal, and Hexadecimal Conversion
A decimal number is converted into other number systems by using successive division by to be converted number base(e.g. if binary then by 2) where remainders noted during successive divisions are written in a bottom-up approach to get the required number system.

The fraction part of any decimal number is converted into other numbers by successive multiplication to the given number with a respective base of to be converted number (e.g. if binary then by 2). The process is terminated when we get zero(0) in the fraction part of the product. When we fail to get zero(0), then we may terminate the process after 5th round. The integer values that come from each successive multiplication are written in the top-down approach.
Conversion 1: (42)₁₀=(?)₂
2 42 Remainder
2 21 0
2 10 1
2 5 0
2 2 1
2 1 0
0 1
∴(42)₁₀=(101010)₂
Conversion 2: (42.54)₁₀=(?)₂
2 42 Remainder
2 21 0
2 10 1
2 5 0
2 2 1
2 1 0
0 1
∴(42)₁₀=(101010)₂
Also,
Fraction x 2 = Product Integer Part
0.54x2=1.08 1
0.08x2=0.16 0
0.16x2=0.32 0
0.32x2=0.64 0
0.64x2=1.28 1
∴(42.54)₁₀=(101010.10001)₂
Conversion 3: (123)₁₀=(?)₈
8 123 Remainder
8 15 3
8 1 7
0 1
∴(123)₁₀=(173)₈
Conversion 4: (123.54)₁₀=(?)₈
8 123 Remainder
8 15 3
8 1 7
0 1
∴(123)₁₀=(173)₈
Also,
Fraction x 8 = Product Integer Part
0.54x8=4.32 4
0.32x8=2.56 2
0.56x8=4.48 4
0.48x8=3.84 3
0.84x8=6.72 6
∴(123.54)₁₀=(173.42436)₈
Conversion 5: (123)₁₀=(?)₁₆
16 123 Remainder
16 7 11 (B)
0 7
∴(123)₁₀=(7B)₁₆
Conversion 6: (123.54)₁₀=(?)₁₆
16 123 Remainder
16 7 11 (B)
0 7
∴(123)₁₀=(7B)₁₆
Also,
Fraction x 16 = Product Integer Part
0.54x16=8.64 8
0.64x16=10.24 10(A)
0.24x16=3.84 3
0.84x16=13.44 13(D)
0.44x16=7.04 7
∴(123.54)₁₀=(7B.8A3D7)₁₆
Binary, Octal, and Hexadecimal to Decimal Conversion
Other systems(Binary, Octal & Hexadecimal) are converted into decimal number by calculating sum of product of each given digit and its corresponding place value ( in terms of power of its base which begins from 0 and increases for integer part and from -1 and decreases for a fractional part)
Conversion 7 : (101010)₂=(?)₁₀
Answer:
Face Value 1 0 1 0 1 0
Place Value 2⁵ 2⁴ 2³ 2² 2¹ 2⁰

=1x2⁵+0x2⁴+1x2³+0x2²+1x2¹+0x2⁰
=1×32+0×16+1×8+0×4+1×2+0×1
=32+0+8+0+2+0
=(42)₁₀
∴(101010)₂=(42)₁₀
Conversion 8: (101.101)₂=(?)₁₀
Answer:
Face Value 1 0 1 .1 0 1
Place Value 2² 2¹ 2⁰ 2^-1 2^-2 2^-3

=1x2²+0x2¹+1x2⁰+1x2^-1+0x2^-2+1x2^-3
=1×4+0×2+1×1+1×0.5+1×0.25+0×0.125
=4+0+1+0.5+0+0.125
=(5.625)₁₀
∴(101.101)₂=(5.625)₁₀
Conversion 9: (345)₈=(?)₁₀
Answer:
Face Value 3 4 5
Place Value 8² 8¹ 8⁰

=3x8²+4x8¹+5x8⁰
=3×64+4×8+5×1
=192+32+5
=(229)₁₀
∴(345)₈=(229)₁₀
Conversion 10: (31.76)₈=(?)₁₀
Answer:
Face Value 3 1 .7 6
Place Value 8¹ 8⁰ 8^-1 8^-2

=3x8¹+1x8⁰+7x8^-1+6x8^-2
=3x8+1x1+7x0.125+6x0.015625
=24+1+0.875+0.09375
=(25.96875)₁₀
∴(31.76)₈=(25.96875)₁₀
Conversion 11: (ABC)₁₆=(?)₁₀
Answer:
We Know ,
A B C D E F
10 11 12 13 14 15

Now,
Face Value A B C
Place Value 16² 16¹ 16⁰

=Ax16²+Bx16¹+Cx16⁰
=10×256+11×176+12×1
=2560+176+12
=(2748)₁₀
∴(ABC)₁₆=(2748)₁₀
Conversion 12: (3F.7A)₁₆=(?)₁₀
Answer:
Face Value 3 F .7 A
Place Value 16¹ 16⁰ 16^-1 16^-2

=3x16¹+Fx16⁰+7x16^-1+Ax16^-2
=3x16+15x1+7x0.0625+10x0.00390625
=48+15+0.4375+0.0390625
=(63.4765625)₁₀
∴(3F.7A)₁₆=(63.4765625)₁₀
Binary to Octal Conversion

A binary number is converted into an octal number by making a group of 3 BITS and grouping should be done from right to left for integer part and left to right for fraction part. So, this method is called Grouping Method. If there are less than 3 BITS in the last group then required number of zeroes can be added in the left-hand side for integer part and in the right-hand side for the fraction part to make the group of 3 BITS. Then equivalent octal value of each group is written using Binary - Octal conversion table.

Table I: Binary-Octal Conversion Table
Binary (2) Octal (2³=8)
000 0
001 1
010 2
011 3
100 4
101 5
110 6
111 7
Conversion 13: (1011101)₂=(?)₈
Answer:
3 BITS Group 001 011 101
Octal Value 1 3 5
∴(1011101)₂=(135)₈
Conversion 14: (1011101.1011001)₂=(?)₈
Answer:
3 BITS Group 001 011 101 .101 100 100
Octal Value 1 3 5 .5 4 4
∴(1011101.1011001)₂=(135.544)₈

Binary to Hexadecimal Conversion
A binary number is converted into hexadecimal number by making a group of 4 BITS and grouping should be done from right to left for integer part and left to right for fraction part. So, this method is called Grouping Method. If there are less than 4 BITS in the last group then required number of zeroes can be added in the left-hand side for integer part and in the right-hand side for the fraction part to make the group of 4 BITS. Then equivalent octal value of each group is written using Binary - Hexadecimal conversion table.

Table II: Binary-Hexadecimal Conversion Table
Binary (2) Hexadecimal (2⁴=16)
0000 0
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 8
1001 9
1010 A
1011 B
1100 C
1101 D
1110 E
1111 F
Conversion 15: (1011101)₂=(?)₁₆
Answer:
4 BITS Group 0101 1101
Hexadecimal 5 13(D)
∴(1011101)₂=(5D)₁₆
Conversion 16: (1011101.101111)₂=(?)₁₆
Answer:
4 BITS Group 0101 1101 .1011 1100
Hexadecimal 5 13(D) .11(B) 12(C)
∴(1011101.101111)₂=(5D.BC)₁₆

Octal to Binary Conversion
Octal number is converted into binary number by breaking each digits of the given octal number into a group of 3 BITS using Table I: Binary-Octal Conversion Table. So, this method is called Breaking Method.
Conversion 17: (567)₈=(?)₂
Answer:
Octal Digits 5 6 7
Equivalent BITS 101 110 111
∴(567)₈=(101110111) ₂
Conversion 18: (567.123)₈=(?)₂
Answer:
Octal Digits 5 6 7 .1 2 3
Equivalent BITS 101 110 111 .001 010 011
∴(567.123)₈=(101110111.001010011)₂

Hexadecimal to Binary Conversion
Hexadecimal number is converted into binary number by breaking each digits of the given hexadecimal number into a group of 4 BITS using Table II: Binary-Hexadecimal Conversion Table. So, this method is called Breaking Method.
Conversion 19: (A43)₁₆=(?)₂
Answer:
Hexadecimal Digits A 4 3
Equivalent BITS 1010 0100 0011
∴(A43)₁₆=(101001000011) ₂
Conversion 20: (4A3.EF)₁₆=(?)₂
Answer:
Hexadecimal Digits 4 A 3 .E F
Equivalent BITS 0100 1010 0011 .1110 1111
∴(4A3.EF)₁₆=(010010100011.11101111)₂

Octal to Hexadecimal Conversion

Octal number is converted into hexadecimal number by using any one of the following two double conversion methods.
Method I:
Octal → Binary
Binary → Hexadecimal
Method II:
Octal → Decimal
Decimal → Hexadecimal
Conversion 21: (567)₈=(?)₁₆
Answer:Method - I
Here, given octal number is first converted into equivalent binary number as:
Octal Digits 5 6 7
Equivalent BITS 101 110 111
∴(567)₈=(101110111)₂
Now, obtained binary number is converted into hexadecimal number as:
4 BITS Group 0001 0111 0111
Hexadecimal Digits 1 7 7

So, (101110111)₂=(177)₁₆
∴(567)₈=(177)₁₆Method - II
Here, given octal number is first converted into equivalent decimal number.
Face Value 5 6 7
Place Value 8² 8¹ 8⁰

=5x8²+6x8¹+7x8⁰
=5×64+6×8+7×1
=320+48+7
=(375)₁₀
Now, the obtained decimal number is converted into hexadecimal number as:
16 375 Remainder
16 23 7
16 1 7
0 1
So, (375)₁₀=(177)₁₆
∴(567)₈=(177)₁₆
Conversion 22: (123.456)₈=(?)₁₆
Answer: Here, given octal number is first converted into equivalent binary number as:
Octal Digits 1 2 3 .4 5 6
Equivalent BITS 001 010 011 .100 101 110
∴(123.456)₈=(1010011.100101110)₂
Now, obtained binary number is converted into hexadecimal number as:
4 BITS Group 0101 0011 .1001 0111 0000
Hexadecimal Digits 5 3 .9 7 0

So, (1010011.100101110)₂=(53.970)₁₆
∴(123.456)₈=(53.97)₁₆
Hexadecimal to Octal Conversion
Hexadecimal number is converted into octal number by using any one of the following two double conversion methods.
Method I:
Hexadecimal → Binary
Binary → Octal
Method II:
Hexadecimal → Decimal
Decimal → Octal
Conversion 23: (5E7)₁₆=(?)₈
Answer:Method - I
Here, given octal number is first converted into equivalent binary number as:
Hexadecimal Digits 5 E 7
Equivalent BITS 0101 1110 0111
∴(5E7)₁₆=(10111100111)₂
Now, obtained binary number is converted into octal number as:
3 BITS Group 010 111 100 111
Octal Digits 2 7 4 7

So, (10111100111)₂=(2747)₈
∴(5E7)₁₆=(2747)₈Method - II
Here, given hexadecimal number is first converted into equivalent decimal number.
Face Value 5 E 7
Place Value 16² 16¹ 16⁰

=5x16²+Ex16¹+7x16⁰
=5×256+14×16+7×1
=1280+224+7
=(1511)₁₀
Now, the obtained decimal number is converted into hexadecimal number as:
8 1511 Remainder
8 188 7
8 23 4
8 2 7
0 2
So, (1511)₁₀=(2747)₈
∴(5E7)₁₆=(2747)₈
Conversion 24: (5E7.3D)₁₆=(?)₈
Answer: Here, given hexadecimal number is first converted into equivalent binary number as:
Hexadecimal Digits 5 E 7 .3 D
Equivalent BITS 0101 1110 0111 .0011 1101
∴(5E7.3D)₁₆=(10111100111.00111101)₂
Now, obtained binary number is converted into octal number as:
3 BITS Group 010 111 100 111 .001 111 101
Octal Digits 2 7 4 7 .1 7 5
So, (10111100111.00111101)₂=(2747.175)₁₆
∴(5E7.3D)₁₆=(2747.175)₁₆

Number System	Base
(i) Binary Number System	2
(ii) Octal Number System	8
(iii) Decimal Number System	10
(iv) Hexadecimal Number System	16

Fraction x 2 = Product	Integer Part
0.54x2=1.08	1
0.08x2=0.16	0
0.16x2=0.32	0
0.32x2=0.64	0
0.64x2=1.28	1

Fraction x 8 = Product	Integer Part
0.54x8=4.32	4
0.32x8=2.56	2
0.56x8=4.48	4
0.48x8=3.84	3
0.84x8=6.72	6

Fraction x 16 = Product	Integer Part
0.54x16=8.64	8
0.64x16=10.24	10(A)
0.24x16=3.84	3
0.84x16=13.44	13(D)
0.44x16=7.04	7

Face Value	1	0	1	0	1	0
Place Value	2⁵	2⁴	2³	2²	2¹	2⁰

Face Value	1	0	1	.1	0	1
Place Value	2²	2¹	2⁰	2^-1	2^-2	2^-3

Face Value	3	4	5
Place Value	8²	8¹	8⁰

Face Value	3	1	.7	6
Place Value	8¹	8⁰	8^-1	8^-2

Face Value	A	B	C
Place Value	16²	16¹	16⁰

Face Value	3	F	.7	A
Place Value	16¹	16⁰	16^-1	16^-2

Table I: Binary-Octal Conversion Table
Binary (2)	Octal (2³=8)
000	0
001	1
010	2
011	3
100	4
101	5
110	6
111	7

3 BITS Group	001	011	101
Octal Value	1	3	5

3 BITS Group	001	011	101	.101	100	100
Octal Value	1	3	5	.5	4	4

Table II: Binary-Hexadecimal Conversion Table
Binary (2)	Hexadecimal (2⁴=16)
0000	0
0001	1
0010	2
0011	3
0100	4
0101	5
0110	6
0111	7
1000	8
1001	9
1010	A
1011	B
1100	C
1101	D
1110	E
1111	F

4 BITS Group	0101	1101
Hexadecimal	5	13(D)

4 BITS Group	0101	1101	.1011	1100
Hexadecimal	5	13(D)	.11(B)	12(C)

Octal Digits	5	6	7
Equivalent BITS	101	110	111

Octal Digits	5	6	7	.1	2	3
Equivalent BITS	101	110	111	.001	010	011

Hexadecimal Digits	A	4	3
Equivalent BITS	1010	0100	0011

Hexadecimal Digits	4	A	3	.E	F
Equivalent BITS	0100	1010	0011	.1110	1111

Octal Digits	5	6	7
Equivalent BITS	101	110	111

4 BITS Group	0001	0111	0111
Hexadecimal Digits	1	7	7

Face Value	5	6	7
Place Value	8²	8¹	8⁰

Octal Digits	1	2	3	.4	5	6
Equivalent BITS	001	010	011	.100	101	110

4 BITS Group	0101	0011	.1001	0111	0000
Hexadecimal Digits	5	3	.9	7	0

Hexadecimal Digits	5	E	7
Equivalent BITS	0101	1110	0111

3 BITS Group	010	111	100	111
Octal Digits	2	7	4	7

Face Value	5	E	7
Place Value	16²	16¹	16⁰

Hexadecimal Digits	5	E	7	.3	D
Equivalent BITS	0101	1110	0111	.0011	1101

3 BITS Group	010	111	100	111	.001	111	101
Octal Digits	2	7	4	7	.1	7	5

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Number system credit goes to PLK sir(with thanks😃)

We can follow him on:

https://plkcomputersir.blogspot.com/p/computer-11.html

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2	42	Remainder
2	21	0
2	10	1
2	5	0
2	2	1
2	1	0
	0	1

2	42	Remainder
2	21	0
2	10	1
2	5	0
2	2	1
2	1	0
	0	1

8	123	Remainder
8	15	3
8	1	7
	0	1

8	123	Remainder
8	15	3
8	1	7
	0	1