In computers, the logical gates play a vital role in running different kinds of objects. In this post, we are going to explain Boolean algebra along with the definition, theorem, gates, and solved examples.
A division of mathematics that deals with the operations on logical values and includes binary variables is known as Boolean algebra. It deals with the binary variable i.e., the base two numbers 0 and 1.
The binary numbers represent the true and false values i.e., 0 for the false and 1 for the true that is why it is also known as binary algebra. It is also widely used in the set theory. Boolean gates or operations are involved in this kind of algebra.
The gates of the binary algebra are:
The arithmetic symbols (plus, minus, multiply, and divide) are used in the gates of Boolean algebra. The arithmetic operations perform in Boolean algebra are different in calculation from the usual arithmetic symbols.
The truth table is used in binary algebra to perform the calculations of logic gates with true or false values according to the theorem and condition. Let’s learn the working and calculations of the logical gates.
In conjunction, the values are true only when both the values are true otherwise it is false. For example, if both the terms are 1 then it gives 1 otherwise it gives 0. It is also known as the AND gate and deals with the product of the terms.
Conjunction or AND gate is denoted by a “^” such as A ^ B. It is generally used for the product of numbers.
Let P & Q are two numbers with binary variables then the AND gate is applied such as:
P | Q | P.Q |
0 | 0 | 0 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
In conjunction, the values are false only when both the values are false otherwise it is true. For example, if both the terms are 0 then it gives 0 otherwise it gives 1. It is also known as the OR gate and deals with the product of the terms.
Disjunction or OR gate is denoted by a “v” such as A v B. It is generally used for the sum of the numbers.
Let P & Q are two numbers with binary variables then the AND gate is applied such as:
P | Q | P + Q |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 1 |
In negation, the values are true when the other value is true and false when the other value is false. It reverts the process. For example, if both the term is 1 then it gives 1, and if 0 it gives 0. It is also known as the NOT gate and deals with the reverse of the terms.
Input | Output– |
0 | 1 |
1 | 0 |
The NAND gate is the combination of the AND and NOT gates. The values are false only when both the values are true otherwise it is true. For example, if both the terms are 1 then it gives 0 otherwise it gives 1.
Let P & Q are two numbers with binary variables then the NAND gate is applied such as:
P | Q | (P.Q)– |
0 | 0 | 1 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
The XOR gate is the combination of the OR and NOT gates. The values are true only when both the values are false otherwise it is false. For example, if both the terms are 0 then it gives 1 otherwise it gives 0.
Let P & Q are two numbers with binary variables then the XOR 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.
There are hundreds of online tools present online for helping the students to solve their problems with steps. A Boolean algebra calculator is a helpful tool that gives step-by-step solution to problems.
Here are the steps to use this calculator.
Example 1
Find the truth table of the given expression by using the gates of Boolean algebra.
(P * Q) + (Q * R) * Q
Solution
Step 1: Write the algebraic expression.
(P * Q) + (Q * R) * Q
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.
P | Q | R | P * Q | Q * R | (Q * R) * Q | (P * Q) + (Q * R) * Q |
0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 0 | 1 |
1 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 |
Use a Boolean expression calculator to verify the above result.
Example 2
Find the truth table of the given expression by using the gates of Boolean algebra.
(A + B) * ((A + C) * B)
Solution
Step 1: Write the algebraic expression.
(A + B) * ((A + C) * B)
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.
A | B | C | A + B | A + C | ((A + C) * B) | (A + B) * ((A + C) * B) |
0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 |
Now you can solve grab all the basics of Boolean algebra from this post. In this post, we have covered all the basics of Boolean algebra along with gates and examples. You can solve any problems by using truth tables.
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