NUMBER SYSTEM FACTS AND FORMULAS

A number system is nothing more than a code. For each distinct quantity there is an assigned symbol. The most familiar number system is the decimal system which uses 10 digits, that is, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

The main advantage of its simplicity and long use. Most of the ancient societies used this system. Even in our everyday life we use this system and is sometimes being taken as the natural way to count. Since this system uses 10 digits it is called a system to base 10. A binary number system is a code that uses only two basic symbols, that is, 0 and 1. This system is very useful in computers. Since, in this system, only two symbols are there, it can be used in electronic industry using ‘on’ and ‘off’ positions of a switch denoted by the two digits 0 and 1.

Decimal Number System :
Decimal number system used 10 digits, 0 through 9, that is, the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Binary Number System :
Binary means two. The binary number system uses only two digits, 0 and 1.

Basic or Radix :
The basic or radix of a number system is equal to the number of digits or symbols used in that number system. For example, decimal system uses 10 digits, so that base of decimal system (this is, decimal numbers) is 10.

Binary numbers have base 2.
A subscript attached to a number indicates that base of the number. For example, $100_{2}$ means binary 100. $100_{10}$ stands for decimal 100.

Weights: In any number to a given base, each digit, depending on its position in the number has a weight in powers of the base.

Note: Decimal to Binary Conversion
Step 1 : Divide the number by 2.
Step 2 : Divide Quotient of Step 1 by 2.
Continue the process till we get quotient = 0 and remainder as 1.
Then, the remainder from down upwards written form left to right give the binary number.
Illustration 4 Convert decimal 23 to binary.
Solution Images
Reading the remainders upwards and writing from left to right we get the binary equivalent of decimal 23 as 10111.
That is, Binary 10111 is equivalent to decimal 23 or we can write $10111_{2}=23_{10}$.

Note : Binary to Decimal Conversion
Following steps are involved to convert a binary number to its decimal equivalent.
Step 1 : Write the binary number.
Step 2 : Write the weights $2^{0},2^{1},2^{2},2^{3}$,… under the binary digits starting from extreme right.
Step 3 : Cross out any weight under a zero, that is, weight under zeros in the binary number should be deleted.
Step 4 : Add the remaining weights.

Binary Addition : In binary number system there are only 2 digits, that is, 0 and 1. In decimal system we carry 1 for every 10 whereas in binary system we carry 1 for every 2. Hence, rules of addition are as under.

0+0=0
0+1=1
1+0=1
1+1=10

Number System Important Examples

Question-1 : In the number $(5342)_{x}$.

solution : The weight of 2 is $x^{0}$
The weight of 4 is $x^{1}$
The weight of 3 is $x^{2}$
The weight of 5 is $x^{3}$
The sum of all the digits multiplied by their respective weights is equal to the decimal equivalent of that number and gives the total amount represented by that number.
$(5342)_{x}=(5x^{3}+3x^{2}+4x+2x^{0})_{10}$

Question-2 : 5 7 0 3 4 Number to the base 10,
solution :$10^{4} 10^{3} 10^{2} 10^{1} 10^{0}$ that is, decimal number weights
$\therefore 5\times 10^{4}+7\times 10^{3}+0\times 10^{2}+3\times 10+4\times 10^{0}$
= value represented or decimal equivalent 1 1 0 0 1 Number to the base 2
$2^{4} 2^{3} 2^{2}2^{1} 2^{0}$ that is, binary number weights
$\therefore 1\times 2^{4}+1\times 2^{3}+0\times 2^{2}+0\times 2^{1}+1\times 2^{0}$
= 16 + 8 + 1 = 25
= Decimal equivalent or value represented by
$11001_{2}$

Question 3 : Convert binary 1101 to its decimal equivalent.
Solution : 1 1 0 1 Binary number
$2^{3}2^{2} 2^{1}2^{0}$ weights
The weight $2^{1}$ is under 0 so it can be deleted. Sum of the remaining weights.
= $2^{3}+2^{2}+2^{0}$ =8 + 4+1 =13
$\therefore$ Decimal equivalent of binary 1101 =13
That is, $1101_{2}=13_{10}$.

Question- 4: Add 1010 to 10100
1o100+ 1010= 11110
Binary Subtraction
1.0 -0 =0
2. 1- 0 = 1
3. 1-1= 0
4. 10 -1 = 1
5. 0-1 = -1
[Complement of a binary number is the exact reverse of the given number]
Complement of 0 = 1
Complement of 1 = 0 , For subtraction of binary number the following method known as one’s complement method is used.

Note :
Subtraction of a lower number from a higher number. To determine which binary number is lower and which is higher, it is advisable to find their decimal equivalent.
Step I: Make the number of digits equal in both the numbers.
Step II: Take the complement of the second number, that is, take the complement of the number to be subtracted.
Step III: Add the complement obtained in Step II to the first number. The carry over obtained from this addition indicates that the answer shall be positive.
Step IV: This carry over is taken out and added to the first digit on the right, that is, extreme right digit.
Step V : The digits so obtained is the final answer.

Question -5 : Subtract 11 from 101.
Solution : Now, $101_{2}=4+1=5_{10},11_{2}=2+1=3_{10}$.
Clearly, 11 is smaller then101. Making the number of digits equal, we write 11 as 011.
Complement of 011=100
Adding 100 to 101, we get
101
$\frac{100}{(1)001}$ {Carry over is 1}
Taking out the carry over and adding to extreme right digit, we get
001
1/010
$\therefore$ The answer is 010 or 10.

Note :
Subtraction of a higher number from a lower number.
Step I : Take the complement of the second number.
Step II : Add the complement obtained in Step I to the first number. In this case there is no carry over indicating that the answer is negative.
Step III : Recomplement the digits obtained after adding the complement of the second number to the first number.
Step IV : Put a negative sign before the result obtained in Step IV.
Question- 5 Subtract 1110 from 1001.
Solution : Now, $1110_{2}=8+4+2=14_{10}$
1001=8+2 = $10_{10}$
Clearly, $1110_{2}> 1001_{2}$
Complement of 1110 = 0001.
Adding 0001 to 1001, we get
1001
0001
-------
1010 [ There is no carry over ]
Complement of 1010= 0101.
$\therefore$ The answer is -0101 or -101.

Number System Important Questions from Previous Year Exam

Number System Important Video