Question

In a Boolean algebra $B$ with respect to $\text{'}+\text{'}$ and $\text{'}·\text{'}$, $x\text{'}$ denotes the negation of $x\in B$. Then,

• A. $x-x\text{'}=1$ and $x·x\text{'}=1$.
• B. $x+x\text{'}=1$ and $x·x\text{'}=0$.
• C. $x+x\text{'}=0$ and $x·x\text{'}=0$.
• D. $x-x\text{'}=1$ and $x·x\text{'}=0$.

Answer 1

The correct option is B

$x+x\text{'}=1$ and $x·x\text{'}=0$.

Step 1: Explanation for the correct option

For option B : $x+x\text{'}=1$ and $x·x\text{'}=0$

The negotiation of a given set is defined as the universal set without the mentioned set.

Thus, the negotiation of set $x$ can be given by, $x\text{'}=1-x$.

$\Rightarrow x+x\text{'}=1$

By the complement law, $x·x\text{'}=0$.

Therefore, $x+x\text{'}=1$ and $x·x\text{'}=0$.

So option B is correct.

Step 2: Explanation for the incorrect options

For option A: $x-x\text{'}=1$ and $x·x\text{'}=1$.

It is given that, $x-x\text{'}=1$.

It is only valid when, $x=1$ and $x\text{'}=0$, as $x+x\text{'}=1$.

Therefore, option (A) is incorrect.

For option C: $x+x\text{'}=0$ and $x·x\text{'}=0$.

It is given that, $x+x\text{'}=0\ne 1$.

Therefore, option (C) is incorrect.

For option D: $x-x\text{'}=1$ and $x·x\text{'}=0$.

It is given that, $x-x\text{'}=1$.

It is only valid when, $x=1$ and $x\text{'}=0$, as $x+x\text{'}=1$.

Therefore, option (D) is incorrect.

Hence, (B) is the correct option.