Question

If \[A = B = 1\] , then in terms of Boolean algebra the value of \[A.B + A\;\] is not equal to
A) \[B.A + B\]
B) \[B + A\]
C) \[B\]
D) $\overline A .B$

Answer 1

Hint AND operator is a Boolean operator that returns a value of TRUE $(1)$ if both its operands are TRUE$(1)$ and FALSE $(0)$ otherwise. The boolean expression for AND gate, $Q=A.B$ , where Q is the output.
OR operator is a Boolean operator that returns a value of TRUE$(1)$ if any of its operands are TRUE$(1)$ and FALSE$(0)$ if both operands are FALSE $(0)$. The boolean expression for OR gate, $Q=A+B$ , where Q is the output.
NOT operator is a unary Boolean operator that returns a value of TRUE $(1)$ if the operand is FALSE $(0)$ and vice versa. The Boolean expression for NOT gate
$Q =\overline A$, where Q is the output.

Complete step by step solution:
For\[A = B = 1\],
\[ \Rightarrow A.B + B\] Substituting values of \[A\] and \[B\]
\[ = 1.1 + 1\]
(The \[1.1\] will give answer as \[1\] , as AND operator \[ \to A.B\] is used)
\[ = 1 + 1\]
(\[ = 1 + 1\] will give answer as \[1\], as OR operator \[ \to A + B\] is used)
$ = 1$
Therefore,
(A) \[B.A + B\]
\[ = 1.1 + 1\]
(The \[1.1\] will give answer as \[1\], as AND operator \[ \to A.B\] is used)
\[ = 1 + 1\]
(\[1 + 1\] will give answer as \[1\], as OR operator \[ \to A + B\] is used)
$ = 1$

(B) \[B + A\]
Substituting values of \[A\] and \[B\]
\[ = 1 + 1\]
(\[1 + 1\]Will give answer as $1$ ,as OR operator \[ \to A + B\]is used)
$ = 1$

(C) \[B\]
Substituting value of \[B\]
$ = 1$

(D) $\overline {A.} B$
Substituting values of \[A\] and \[B\]
$ = \overline 1 .1$
( $1\overline {} $ can also be written as $0$ )
\[ = 0.1\]
(The \[0.1\] will give answer as $0$ , as AND operator \[ \to A.B\] is used)
\[ = 0\]

Hence, option D is the correct answer.

Note: Logic gates take input in form of 0 and 1 and the algebra dealing with these numbers is known as Boolean algebra. In logic gates boolean algebra rules must be followed:
$1 + 1 = 1\left( { \ne 2} \right)$
$1.1 = 1$
$\overline 1 = 0$
$1.0 = 0$ $0.1 = 0$