# Number System - Formulas, Examples, Shortcuts and Video

## 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.

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