3. Basics of Digital Systems - Mathematical Operations with the Binary Base

Operations with binary base are equal the decimal base. The difference is when we need to store numbers in a finite amount of bits. Let’s do some operations with 8 bits, where the MSB is the signal bit and the others 7 bits are the bits of numerical value.

3.1. Sum

104₁₀+ 2₁₀= 0110 1000₂+ 0000 0010₂= 0110 1010₂= 106₁₀

As the result isn’t greater than 0111 1111₂, for this operation didn’t occurred overflow.

104₁₀+ 24₁₀= 0110 1000₂+ 0001 1000₂= 1000 0000₂= 0₁₀

As the result is greater than 0111 1111₂the result is wrong and occurred overflow.

3.2. Subtraction

-104₁₀- 2₁₀= -(104₁₀+ 2₁₀)= -(0110 1000₂+ 0000 0010₂) = 1110 1010₂= - 106₁₀

As the result isn’t greater than 0111 1111₂, for this operation didn’t occurred overflow.

-104₁₀- 24₁₀= -(104₁₀+ 24₁₀) = -(0110 1000₂+ 0001 1000₂) = 1000 0000₂

As the result is greater than 0111 1111₂ the result is wrong and occurred overflow.

-104₁₀+ 24₁₀= -(104₁₀- 24₁₀) = -(0110 1000₂- 0001 1000₂) = 1101 0000₂

3.3. Multiplication

2₁₀* 60₁₀= 0000 0010₂* 0011 1100₂= 0111 1000₂= 120₁₀

-2₁₀* 60₁₀= -0000 0010₂* 0011 1100₂= -0111 1000₂= 1111 1000₂= - 120₁₀

In none of the cases occurred overflow.

3.4. Division

60₁₀/2₁₀= 0011 1100₂/ 0000 0010₂= 0001 1110₂= 30₁₀

If the division results in a decimal number, the decimal places are discarded.

3.5. Two Complement

The way we do arithmetic operations in a paper, adding transport numbers and deciding which number is greater to change the signal and perform a subtraction, is difficult to implement in a digital system. To overcome that there is another way to represent finite binary numbers and it’s called two complement. It is easier to perform calculations of sum and subtraction using two complement representation.

First we define with how many bits we are working. After that we perform an operation to find the two complement of a negative number. If you have to do the complement of a number x of n bits, the operation is this, y₂= ( \(2^n\) - x)₂. For example we are working with 3 bits and we want to know the two complement of -010₂, doing the operation,

y₂ = 1000₂ – 010₂ = 110₂, and this is -010₂ represented in two complement.

Adding and subtracting numbers is easy.

3₁₀ – 2₁₀ = 011₂ – 010₂ = 011₂ + (1000₂ - 010₂) =011₂ + 110₂ = 1001₂ = 010₂, in this case we have to discard the MSB, the result is correct as the operands had different sign.

3₁₀ +2₁₀ = 011₂ + 010₂ = 101₂ , the result is wrong as the sign of the operand are equal and the sign of the result is different, occurred overflow.

-3₁₀ – 2₁₀ = 101₂ + 110₂ = 1011₂ , the result is wrong as the sign of the operand are equal and the sign of the result is different, occurred overflow.

-3₁₀ – 1₁₀ = 101₂ + 111₂ = 1100₂ = 100₂, the result is correct as the sign of the operand are equal and the sign of the result is equal.

Binary number	3 bits complement two
011	011
010	010
001	001
000	000
100	111
101	110
110	101
111	100