Those are 3 questions I need to answer:
a. Can the dual of a Boolean expression be equal to the expression itself? If yes, give an example.
b. Can the complement of a Boolean expression be equal to the expression itself? If yes, give an example.
c. Can the complement of a Boolean expression be equal to its dual? If yes, give an example.
Those are my answers:
a. Yes, the dual of a Boolean expression can be equal to the expression itself.
Example 1:
$F = X + X = X$
$FD = XX = X + X = X = F$
Example 2:
$F = XY + YZ + XZ$
$FD = (X + Y)(Y + Z)(X + Z) =$
$(Y + XZ)(X+Z) =$
$YX + XZ + YZ = F$
b. No, because for any Boolean expression $F$, it will evaluate to either one or zero, and since if it evaluates to zero its complement will evaluate to one and vice versa, an expression whose complement evaluates to the same value as the expression itself does not exist.
c. ??
Now I want to firstly know if my reasoning for b is correct, and secondly, if a complement of a Boolean expression can be equal to its dual. As you have probably noticed, I'm very new to this, so bear with me if these seem to be very basic questions, all help will be appreciated and thanks in advance.
Your answers to (a) and (b) are correct. For (c), you can refer to this answer, which I partially reproduce here for convenience.
The dual function $f^*$ of a Boolean function $f$ of $n$ variables $x_1, \dots, x_n$ is defined by $$ f^*(x_1, \dots, x_n)= \overline{f}(\overline{x}_1,\dots, \overline{x}_n) $$ Thus, for instance if $f = x \vee x$, then $f^* = \overline{\overline{x}\vee \overline{x}} = x \wedge x$. If $f = 0$, then $f^* = \overline{0} = 1$. This is an example where the dual of $f$ is equal to its complement. If you prefer a Boolean function with a variable, you can take $f = x \vee \overline{x} = 1$. Then $$f^* = \overline{\overline{x} \vee \overline{\overline{x}}} = \overline{\overline{x} \vee x} = x \wedge \overline{x} = 0.$$ Again, the dual of $f$ is equal to its complement.