0

Consider the bit strings $A=001100$ and $B=010101$ then $A\vee B`= ? $ ($B`=$bitwise NOT)

First I wrote $B`$ and get,

$$B`= 101010$$

then I took first "$0$" of $A$ and "$1$" of $B`$ then get (1),

$$A\vee B`= (001100)\vee(101010) = (101110)$$

But this is accepted wrong. What am I missing? Can't we negate $B$ in bit operation like this?

This was a multiple choice question and answers were like these

A-)001000 B-)011101 C-)101110 D-)010001

wave
  • 11
  • It depends on what ` is supposed to mean. If it means bitwise NOT, then you're correct. If it means something else, you may not be. – saulspatz Jun 05 '20 at 06:08
  • Thanks for checking. Yes it means bitwise NOT. – wave Jun 05 '20 at 06:11
  • Seems correct to me, at least assuming $A\vee B' = A\vee (B')$ and not $(A\vee B)'$ – Peter Franek Jun 05 '20 at 06:17
  • 1
    Yes, they are written as seperate. Otherwise we must use De Morgan law. I ask this because this was an exam question and this answer did not accepted. Thanks for checking. – wave Jun 05 '20 at 06:24
  • maybe it has something to do with priority of the operators. –  Jun 05 '20 at 06:42
  • Thanks for checking, I've also added choices of the question if you want to look. – wave Jun 05 '20 at 06:56

0 Answers0