Question

What will be the input of A and B for the Boolean expression.
\[\left( {\overline {A + B} } \right)\left( {\overline {A.B} } \right) = 1\]
A) 0,0
B) 0,1
C) 1,0
D) 1,1

Answer 1

Hint: Input is defined as the amount of energy that is put in a device and output is defined as the amount of energy that comes out. The type of energy can change but not the amount in a device.

Complete step by step solution:
Given data:
The equation is \[\left( {\overline {A + B} } \right)\left( {\overline {A.B} } \right) = 1\]
This can be solved by trial and error method.
(A) For 0,0
\[\Rightarrow \left( {\overline {A + B} } \right)\left( {\overline {A.B} } \right) = \left( {\overline {0 + 0} } \right)\left( {\overline {0.0} } \right)\]
\[\Rightarrow \left( {\overline {A + B} } \right)\left( {\overline {A.B} } \right) = \left( 1 \right)\left( 1 \right) = 1\]
(B) For 0,1
\[\Rightarrow \left( {\overline {A + B} } \right)\left( {\overline {A.B} } \right) = \left( {\overline {0 + 1} } \right)\left( {\overline {0.1} } \right)\]
\[\Rightarrow \left( {\overline {A + B} } \right)\left( {\overline {A.B} } \right) = \left( 1 \right)\left( 0 \right) = 0\]
(C) For 1,0
\[\Rightarrow \left( {\overline {A + B} } \right)\left( {\overline {A.B} } \right) = \left( {\overline {1 + 0} } \right)\left( {\overline {1.0} } \right)\]
\[\Rightarrow \left( {\overline {A + B} } \right)\left( {\overline {A.B} } \right) = \left( 1 \right)\left( 0 \right) = 0\]
(D) For 1,1
\[\Rightarrow \left( {\overline {A + B} } \right)\left( {\overline {A.B} } \right) = \left( {\overline {1 + 1} } \right)\left( {\overline {1.1} } \right)\]
\[\Rightarrow \left( {\overline {A + B} } \right)\left( {\overline {A.B} } \right) = \left( {\overline 0 } \right)\left( {\overline 0 } \right) = 0\]

Hence the correct option is A.

Note: 1. The logical statement which is either True or False is called a Boolean expression. It will compare the data of any type when both parts have the same basic data. It is generally used in programming languages which gives a Boolean value when calculated.
2. The branch of the algebra will have the values of variables are the truth values and generally denoted by 1 and 0. Boolean algebra is somewhat hard. All the numbers except 0 are considered as true while 0 is considered false.
3. Boolean Algebra is used to analyze the digital gates and the circuits. George Boole in 1854 invented Boolean Algebra. The other names of Boolean Algebra are Logical Algebra, binary Algebra. These are also used to simplify logic circuits.
4. The variables used will have only two values. Binary 0 for low value and Binary 1 for high value. Overbar is used to represent the complement of a variable. Boolean laws are of six types namely Associative law, Commutative law, Distributive law, AND law, OR law, Inversion law.