Number System In Fundamental Of Computer ~ IBPS Exam Preparation Guide 2016-2017.......... Best Guidelines For Ibps Exam

Introduction of Number System-

Every Computer store numbers, letters, and other special characters in a coded form. Before going into the details of these codes, it is essential to have a basic understanding of number system. This topic also introduces some of commonly used number system by computer professionals and the relationship between them.

Binary Number System-

The Binary number system is exactly like the decimal system except that the base is instead of 10. We have only two symbols or digits (0 and 1) that can be used in this number system. Note that the largest single digit is 1 (one less then base). Again, each position in a binary number represents a power of the base (2). As such, in this system, the rightmost position is the unit’s $({2^0})$ position, the second position from the right Is the ${2^0}$ s ( ${2^1}$ ) position and proceeding in this way we have 4’s $({2^2})$ position, 8’s $({2^3})$ position, 16’s $({2^4})$ position, and so on. Thus, the decimal equivalent of the binary number 10101 written a $({10101_2})$ is

$(1*{2^4}) + (0*{2^3}) + (1*{2^2}) + (0*{2^1}) + (1*{2^0})$

Or 16+0+4+0+1

Or 21

In order to be specific about which system we are referring to, it is common practice to indicate the base as subscript. Thus we write

${10101_2} = {21_{10}}$

Binary digit is often referred to by the common abbreviation bit. A binary number consisting of n bits is called an n-bit number. Table 1-1 lists all the 3bit numbers along with their decimal equivalent.

3 bit numbers with their Decimal values.

Binary	Decimal Equivalent
000	0
001	1
010	2
011	3
100	4
101	5
110	6
111	7

Remember that we have only two digit, o and 1, in the binary system, and hence the binary equivalent of the decimal number 2 has to be stated as 10 (read as one, zero). It may be seen that a 3-bit number can have one of the 8 values. In fact. It can be shown that any decimal number in the range $0to{2^{n - 1}}$ can be represented in the binary form as an n-bit number.

Octal Number System –

In the Octal number system the base is 8. So in this system there are only eight symbol or digits: 0,1,2,3,4,5,6,7. Here also the largest single digit is 7 (one less than the base). Again each position in an octal number represents a power of the base (8). Thus the decimal equivalent of the octal number 2057 written as ${2057_8}$ is:

$(2*{8^3}) + (0*{8^2}) + (5*{8^{^1}}) + (7*{8^0})$

Or 1024+0+40+7

Or 1071

So we have ${2057_8} = {1071_{10}}$

Observe that since there are only 8 digits in the octal number system, so 3 bits are $({2^3} = 8)$ sufficient to represent any octal number in binary

Hexadecimal Number System –

The hexadecimal number system is one with a base of 16. The base 16 suggests choices of 16 single character digits or symbols. The first 10 digit are the digit of decimal system 0,1,2,3,4,5,6,7,8,9. The remaining six digits are denoted by A,B,C,D,E,F representing the decimal values 10,11,12,13,14,15 receptively. Therefore, the letters A through F are number digits in hexadecimal number system. Thus largest single digit is F or 15. Again each position in a hexadecimal system represents a power of the base 16. Thus the decimal equivalent of the hexadecimal number 1AF written as $1A{F_{16}}$ is :

$(1*{16^2}) + (A*{16^1}) + (F*{16^0})$

Or (1*256)+(10*16)+(15*1)

Or 256+160+15

Or 431

Thus $1A{F_{16}} = {431_{10}}$

Since there are only 16 digit in the hexadecimal number system, so 4 bit $({2^4} = 16)$ are sufficient to represent any hexadecimal number in binary.

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Converting from one number system to another

Number expressed in decimal are much meaningful to us than are values expressed in any other number system. However, any number value in one number system can be represented in any other number system. Because the input and the final output values are to be in decimal, so we are required to convert number system in other system to decimal and vice-versa. There are many methods or techniques that can be used to convert number from one base to another.

Converting to Decimal number system from another number system – the following three steps are used to convert to a base 10 value from any other number system:

Step 1- Determine the column (positional) value of each digit (this depends on the position of the digit and the base of the number system).

Step 2- Multiply the obtained column values in step 1 by the digits in the corresponding column.

Step 3- sum the products calculated in step 2. The total is the equivalent value in decimal.

Example 1 - ${(11001)_2} = {(?)_{10}}$

Solution Step 1- Determine column values.

Column number	Column value (from right)
1	${2^0} = 1$
2	${2^1} = 2$
3	${2^2} = 4$
4	${2^3} = 8$
5	${2^4} = 16$

Step 2- multiple column value by corresponding column digits.

16 8 4 2 1

*1 *1 *0 *0 *1

16 8 0 0 1

Step 3- Sum the product

16+8+0+0+1=25

$thus{(11001)_2} = {(25)_{10}}$

Example 2 - ${(4706)_8} = {(?)_{10}}$

Solution Step 1-

Column number	Column value (from right)
1	${8^0} = 1$
2	${8^1} = 8$
3	${8^2} = 64$
4	${8^3} = 512$

Step 2- multiple column value by corresponding column digits.

512 64 8 1

*4 *7 *0 *6

2048 448 0 6

Step 3- Sum the product

2048+448+0+6=2502

$Hence,{(4706)_8} = {(2502)_{10}}$

Example 3- ${(1AC)_{16}} = {(?)_{10}}$

Solution – ${(1AC)_{16}} = 1*{16^2} + A*{16^1} + C*{16^0}$

Example 4- ${(4052)_7} = {(?)_{10}}$

Solution – ${(4052)_7} = 4*{7^3} + 0*{7^2} + 5*{7^1} + 2*{7^0}$

$= {(1409)_{10}}$

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Converting from base 10 to a new base (Division – remainder technique)

following four step are used to convert a number from base 10 to a new base-

Step 1- Divide the decimal number to be converted by the value of the new base:

Step 2- Record the remainder from step 1 as the rightmost digit of the new base

number.

Step 3- Divide the quotient of the previous divide by the new base.

Step 4- Record the remainder from step 3 as the next digit of the new base

number.

Repeat step 3 and 4 recording remainder from right to left, until the quotient

Becomes zero in step 3. Note that the last remainder thus obtained will be the

most significant digit of the new base number.

Example 1. ${(25)_{10}} = {(?)_2}$

Solution :

Step 1 & 2: 25/2 =12 and remainder 1

Step 3 & 4: 12/2 =6 and remainder 0

Step 3 & 4: 6/2 = 3 and remainder 0

Step 3 & 4: 3/2 =1 and remainder 1

Step 3 & 4: 1/2 = 0 and remainder 1

$Hence{(25)_{10}} = {(11001)_2}$

Example 2. ${(42)_{10}} = {(?)_2}$

Solution: 2 42 remainder

2 21 0

2 10 1

2 5 0

2 2 1

2 1 0

0 1

$Hence{(42)_{10}} = {(101010)_2}$

Example 3. ${(952)_{10}} = {(?)_8}$

Solution: 8 952 remainder

8 119 0

8 14 7

8 1 6

0 1

$Hence{(952)_{10}} = {(1670)_8}$

Example 4. ${(428)_{10}} = {(?)_{16}}$

Solution: 16 428 12 = C remainder

16 26 10 = A

16 1 1

$Hence{(428)_{10}} = {(1AC)_{16}}$

Example 5. ${(100)_{10}} = {(?)_5}$

Solution: 5 100 0 remainder

5 20 0

5 4 4

$Hence{(100)_{10}} = {(400)_5}$

Converting from a base other than 10 to a base other then 10.

The following two step are used to convert a number from a base other than 10 to a base other than 10.

Step 1. Convert the original number to a decimal

Step 2. Convert the decimal number so obtained to the new base

Example 1. ${(545)_6} = {(?)_4}$

Solution- Step 1. Convert from base 6 to a base 10

${545 = 5*{6^2} + 4*{6^1} + 5*{6^0}}$

{ = 5*36 + 4*6 + 5*1}

{ = 180 + 24 + 5}

${{{(209)}_{10}}}$

$convert{(209)_{10}}tobase4$

4 209 1 remainder

4 52 0

4 13 1

4 3 3

${Hence{{(209)}_{10}} = {{(3101)}_4}}$

${so{{(545)}_6} = {{(209)}_{10}} = {{(3101)}_4}}$

${thus,{{(545)}_6}{{(3101)}_4}}$

Shortcut method for Binary to Octal Conversion.

Step- 1. Divide the binary digits into groups of three (staring from the right)

Step- 2. Convert each group of tree binary digit into one octal digit. Since there are only digits (0 to 7) in the octal number system, so 3 bits are sufficient to represent any octal number in binary.

Example 1- ${(101110)_2} = {(?)_8}$

Solution-

Step- 1. Divide the binary digits into groups of three (staring from the right)

Step 1. Divide the binary digits into group of 3 starting from right

101 110

Step 2. Convert each group into one digit of octal (use binary to decimal conversion)

${(101)_2} = 1*{2^2} + 0*{2^1} + 1*{2^0}$

= 4 + 0 + 1

$= {5_8}$

${(110)_2} = 1*{2^2} + 1*{2^1} + 0*{2^0}$

= 4 + 2 + 0

$= {6_8}$

$Hence{(101110)_2} = {(56)_8}$

Shortcut method for Octal to Binary Conversion.

Step 1. Convert each octal digit to a 3 digit binary number

Step 2. Combine all the resulting binary group into a single binary number.

Example 1- ${(562)_8} = {(?)_2}$

Solution –

Step 1. Convert each octal digit to 3 binary digits.

${5_8} = {101_2}$

${6_8} = {110_2}$

${2_8} = {010_2}$

Step 2. Combine the binary group

${(562)_8} = \frac{{101}}{5}$ $\frac{{110}}{6}$ $\frac{{010}}{2}$

$Hence,{(562)_8} = {(101110010)_2}$

Shortcut method for Binary to Hexadecimal Conversion.

Step 1. Divide the binary digits into group of four (staring from the right)

Step 2. Convert each group of four binary digits to one Hexadecimal digit. Remember that Hexadecimal digits 0 to 9 are equal to decimal digit 0 to 9, and hexadecimal digits A to F are equal to decimal digits 10 to 15.

Example 1- ${(11010011)_2} = {(?)_{16}}$

Solution –

Step 1. Divide the binary digit into groups of 4.

1101 0011

Step 2. Convert each group of 4 binary digits to 1 hexadecimal digit.

${(1101)_2} = 1*{2^3} + 1*{2^2} + 0*{2^1} + 1*{2^0}$

= 8 + 4 + 0 + 1

$= {13_{10}}$

= D16

${(0011)_2} = 0*{2^3} + 0*{2^2} + 1*{2^1} + 1*{2^0}$

= 0 + 0 + 2 + 1

$= {3_{16}}$

$Hence{(11010011)_2} = {(D3)_{16}}$

Shortcut method for Hexadecimal to Binary Conversion.

Step 1. Convert the decimal equivalent to each Hexadecimal digit to 4 binary digit.

Step 2. Combine all the resulting binary group (of 4 digit each into a single binary number)

Example 1 ${(2AB)_{16}} = {(?)_2}$

Solution –

Step 1. Convert the decimal equivalent to each Hexadecimal digit to 4 binary digit.

${2_{16}} = {2_{10}} = {0010_2}$

${A_{16}} = {10_{10}} = {1010_2}$

${B_{16}} = {11_{10}} = {1011_2}$

Step 2. Combine all the resulting binary group

$2A{B_{{{16}^ - }}} = \frac{{0010}}{2}$ $\frac{{1010}}{A}$ $\frac{{1011}}{B}$

$= {(101010111100)_2}$

$Hence{(ABC)_{16}} = {(101010111100)_2}$

Fractional number.

In binary number system, fractional numbers are formed in the same general way as in the decimal system. Just as in the decimal system.

Example 1

$0.235 = (2*{10^{ - 1}}) + (3*{10^{ - 2}}) + (5*{10^{ - 3}})$

And $68.53 = (6*{10^1}) + (8*{10^0}) + (5*{10^{ - 1}}) + (3*{10^{ - 2}})$

Similarly in binary number system.

$0.101 = (1*{2^{ - 1}}) + (0*{2^{ - 2}}) + (1*{2^{ - 3}})$

And $10.01 = (1*{2^1}) + (0*{2^0}) + (0*{2^{ - 1}}) + (1*{2^{ - 2}})$

Thus the binary point serves the same purpose as the decimal point.

Example 1 ${(110.101)_2} = {(?)_{10}}$

Solution –

${(110.101)_2} = (1*{2^2}) + (1*{2^1}) + (0*{2^0})$

$= (1*{2^{ - 1}}) + (0*{2^{ - 2}}) + (1*{2^{ - 3}})$

$= 4 + 2 + 8 + \frac{1}{2} + 0 + \frac{1}{8}$

$= {6.625_{10}}$

$Hence,{(110.101)_2} = {(6.625)_{10}}$

Example 2 ${(127.54)_8} = {(?)_{10}}$

Solution –

${(127.54)_8} = (1*{8^2}) + (2*{8^1}) + (7*{8^0}) + (5*{8^{ - 1}}) + (4*{8^{ - 2}})$

$= 64 + 16 + 7 + \frac{5}{\begin{array}{l} \\ 8 \end{array}} + \frac{4}{\begin{array}{l} \\ 64 \end{array}}$

= ${87.6875_{10}}$

$Hence,{(127.54)_8} = {(87.6875)_{10}}$

Example 3 ${(2B.C4)_{16}} = {(?)_{10}}$

Solution –

${(2B.C4)_{16}} = 2*{16^1} + B*{16^0} + C*{16^{ - 1}} + 4*{16^{ - 2}}$

$= 32 + 11 + \frac{{12}}{{16}} + \frac{4}{{256}}$

$= {43.765625_{10}}$

IBPS Exam Preparation Guide 2016-2017.......... Best Guidelines For Ibps Exam

Number System In Fundamental Of Computer

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