Algebra is a wide topic of mathematics that deals with algebraic expressions and terms. In computer and other branches of science, algebra is used for various purposes. Boolean algebra is a subtype of algebra used in machine language.
It is a widely used topic in computer systems as it takes the binary variables 0 & 1 for solving the problems and describing the expressions. In this article, we are going to explain Boolean algebra along with examples and solutions.
Boolean algebra also known as binary algebra is the division of mathematics that deals with various operations on logical values and includes binary numbers. The binary numbers are the base 2 numbers such as 0 & 1.
It is a widely used concept as it denotes the binary variables 0 & 1 as false & true respectively. For example, if the statement is true it is denoted by 1 and if the statement is false, it is denoted by 0. In set theory, Boolean algebra plays a vital role.
Here are the gates of binary algebra.
The gates of the Boolean algebra denote the arithmetic operations such as addition & multiplications on the given values but it is rather different from the actual plus and multiplication.
The gates of the Boolean algebra can be described by using the truth table. Here is a brief introduction to the gates of Boolean algebra.
In this gate of binary algebra, when both the terms are true then the result is true otherwise it is false. Such as, the result must be 1 if both the terms are 1 and the if the terms are (0 & 1), (1 & 0), and (0 & 0) gives zero.
The conjunction gate is also known as the And gate and it gives the output by multiplying the numbers. It is generally denoted by a cap sign “^” and written as P ^ Q. Let X & Y are two numbers with binary variables then the conjunction gate is applied such as:
| X | Y | X.Y |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
In this gate of binary algebra, when both the terms are false then the result is false otherwise it is true. Such as, the result must be 0 if both the terms are 0 and if the terms are (0 & 1), (1 & 0), and (1 & 1) give one.
The disjunction gate is also known as the OR gate and it gives the output by adding the numbers. It is generally denoted by a sign “v” and written as P v Q. Let X & Y are two numbers with binary variables then the disjunction gate is applied such as:
| X | Y | X + Y |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
In the negation gate, the values are inverted such as if the value is true the negation inverts it as a false, and if the value of false it is inverted to true. In general, negation is used to reverse the process.
Such as if the value is 1 it gives 0 and if the value is 0 it gives 1. The negation gate is also known as the NOT gate.
| Input | Output– |
| 0 | 1 |
| 1 | 0 |
In the NAND gate, the AND & NOT gates are involved in finding the values. In general, the NAND gate is the opposite of AND gate as it reverses the process that AND gate does. In this gate, when both the terms are 1 it gives 0 otherwise it gives 1.
Let X & Y are two numbers with binary variables then this gate is applied such as:
| P | Q | (P.Q)– |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
In the XOR gate, the OR & NOT gates are involved in finding the values. In general, the XOR gate is the opposite of the OR gate as it reverses the process that the OR gate does. In this gate, when both the terms are 0 it gives 1 otherwise it gives 0.
Let X & Y are two numbers with binary variables then this gate is applied such as:
| P | Q | (P + Q) |
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
There are two ways of solving the problems of Boolean algebra, one is by using a calculator and the other is the manual method.
By using calculator
There are hundreds of online tools present online for helping students to solve their problems with steps. A Boolean calculator is a helpful tool that gives step-by-step solutions to problems.
Here are the steps to use this calculator.
Manually
Example 1
Find the truth table of the given expression by using the gates of Boolean algebra.
(X + Z) * (X + Y) + Z
Solution
Step 1: Write the algebraic expression.
(X + Z) * (X + Y) + Z
Step 2: By taking the total number of terms identify the rows of the truth table.
Term in the given expression = n = 3
According to formula
2m = 23 = 2 x 2 x 2 = 8
Hence, there will be 8 rows in the truth table.
Step 3: According to the given terms make the truth table for three variables.
| X | Y | Z | X + Z | X + Y | (X + Y) + Z | (X + Z) * (X + Y) + Z |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Example 2
Find the truth table of the given expression by using the gates of Boolean algebra.
((P * Q) + R) + ((P + Q) * R)
Solution
Step 1: Write the algebraic expression.
((P * Q) + R) + ((P + Q) * R)
Step 2: By taking the total number of terms identify the rows of the truth table.
Term in the given expression = n = 3
According to formula
2n = 23 = 2 x 2 x 2 = 8
Hence, there will be 8 rows in the truth table.
Step 3: Make a truth table.
| P | Q | R | P * Q | P + Q | ((P * Q) + R) | ((P + Q) * R) | ((P * Q) + R) + ((P + Q) * R) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
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Final Words
Now you are witnessing that this sub-type of algebra is not a difficult topic. You can easily find the solution to the problems of binary algebra with the help of logic gates. In this post, all the basics of binary algebra are covered with examples.
