Number System Study Notes for GATE & Computer Science Engineering Exams

• The base or radix of a number system is the number of digits present. The decimal numeral system has a base or radix of 10, where the set of 10 symbols (digits) is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. The weights are in powers of ten.
• In general, a bases-b number (a_n a_n-1 … a₀ . f₁ f₂ … f_m)_b has the value
  • (a_n × bⁿ) + (a_n-1 × b^n-1) +…+ (a₀ × b⁰) + (f₁ × b^-1) + (f₂ × b^-2) +…+ (f_m × b^-m)

Weighted Binary System:  These binary codes are those which follow the positional weighting principle, In which each position of a number represents a specific weight.

e.g., 8421, 2421, 5421

Sequential Code: A code is said to be sequential when two subsequent codes, seen as numbers in binary representation, differ by one. The 8421 and excess-3 codes are sequential, whereas 2421 and 5421 codes are not.

Non-weighted Codes: Non-weighted codes are codes which are not positionally weighted. That is each position within the binary number isn't assigned a hard and fast value, Example: Gray Code

Reflective Code: A code is said to be reflective when code for 9 may be its a complement for the code for 0 and so is for 8 and 1 codes, 7 and 2, 6 and 3, 5 and 4. Codes 2421, 5421 and excess-3 are reflective, whereas the 8421 code is not.

BCD (Binary Coded Decimal): It is a direct assignment of the binary equivalent. Thus to encode a decimal number using the common BCD encoding. Each digit is stored during a 4-bit number

BCD encoding for number 127 would be

1 2 7

(0001 0010 0111) → BCD equivalent of 127

whereas the pure binary number would be (01111111)₂

BCD

Add (148 + 157) = ?

When sum of two digits is bigger than or adequate to 9, then we'd like to feature then we'd like to feature 6 i.e., 0110

2421 Code

This is often weighted code, its weights are 2, 4, 2 and 1. A decimal number is represented in 4 bit form and therefore the total 4 bits weight is 2 + 4 + 2 + 1 = 9.

Hence, 2421 code represents to the decimal numbers which are from 0 to 9.

Excess-3 Code

Excess-3 code is a non-weighted code usually used to represent decimal numbers. The code receives its name from the fact that each binary code is the correspondence of 8421 code plus 0011 (3).

e.g.,

Gray Code

This is variable ,weighted code and is cyclic in nature. They means that it is arranged in order so that every transition from one value to the next value involves only one-bit Change.

Binary to Gray Code Conversion

1. Write down the number in binary codes.
2. The Most Significant Bit (MSB) of the grey code will be the same as the MSB of binary code.
3. Perform XOR operation on MSB and next bit to the MSB in the binary number.
4. Repeat step 3 till all bits of binary number have been XO Red, the resultant code is the grey code equivalent to binary code.

Gray Code to Binary Conversion

1. Start with the MSB of grey coded number.
2. Copy this bit as the MSB of the binary number.
3. Now, perform Ex-OR operation of this bit with the next bit of the binary number.
4. Repeat step 3 till all bits of grey coded number have been used in XOR operation. The resultant number is the binary equivalent of the grey number.

Complements

Complements are used in the digital computer system for simplifying the subtraction operation and for logical manipulation.

>>types of complements for each base r system as,

1. The r′ s complement.
2. The (r′ – 1)′s complement.

The r's Complement

Here given no is a positive number N with base r with an integer part of n digits. The r's complement of N is given as rⁿ – N for N ≠ 0 and 0 for N = 0.

e.g., 10's complement of (25.639)₁₀ is (10² - 25.639)

100 - 25.639 = 74.361, here the number of digits in integer part is 2

means n = 2

The (r – 1)′s Complement

Given, a positive number N with base r with an integer part of n digits and a fraction part of m digits, the (r – 1)'s complement of N is derived as rⁿ – r^–m – N.

• 9’s complement of (52520)₁₀ is (10⁵ – 1 – 52520)
  = 99999 – 52520 = 47479
  Because number of integer part is 5, so rⁿ = 10⁵ and no fractional part is
  Present' so r^–m = 10^–0 = 1
• 9’s complement of (0.3267)₁₀ is (10⁰ – 10^–4 – 0.3267)
  = 1 – 0.0001 – 0.3267
  = 0.9999 – 0.3267 = 0.6732
  No integer part, so 10ⁿ = 10⁰ = 1
• 1’s complement of (101100)₂ is (2⁶ – 2⁰)₁₀ – (101100)₂
  = (64 – 1)₁₀ – (101100)₂
  = (63)₁₀ = (101100)₂
  = 111111 – 101100 = 010011

Key Points

• 520 → Here, n = 3, but (052) → here n = 2.
• In the later example 0 is of no significance.

Representation of Integers

The are three possible ways to represent a number

1. Signed magnitude method
2. 1's complement method
3. 2's complement method

Signed Magnitude Method

Number is divided into two parts, one is sign bit and other parts for magnitude, In the example we are using 5-bit register to represent –6 and +6.

Range of Number For n bit register, MSB will be a sign bit and (n – 1) bits will be the magnitude.

Key Points

• The drawback of signed magnitude method is that 0 will be having 2 different representation one will be 10000 i.e., –0 and the other one will be 00000 + 0.

1's Complement Method

Positive numbers are represented in the same way as in sign-magnitude method. If the number is negative, then it is represented by the help of 1's complement method. For this, we need to represent the number with a positive sign and then take its 1's complement of this number.

e.g., Suppose we are using the 5-bit register. The representation of –6 will be as below.

Key Points

• The only drawback of 1's complement method is that there are two different representation for zero, one is –0 and other is +0.

2's Complement Method

Positive numbers are represented in the same way as in sign-magnitude. For representing a negative number, we take 2's complement of the corresponding positive number.

Properties of 2's Complement

This representation usually uses binary arithmetic operations on signed integers, resulting in the current 2's complement result.

• Positive Numbers Positive 2's complement numbers are represented as the simple binary.
• Negative Numbers Negative 2's complement numbers are represented as the binary number that when added to a positive number of the same magnitude equals zero.

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