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GATE CSE 2015 | Set 3 | Question: 2

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Let $\#$ be the binary operator defined as

$X\#Y = X'+Y'$ where $X$ and $Y$ are Boolean variables.

Consider the following two statements.

  • $(S_1)$ $(P\#Q)\#R = P\#(Q\#R)$
  • $(S_2)$ $Q\#R = (R\#Q)$

Which are the following is/are true for the Boolean variables $P, Q$ and $R$?

  1. Only $S_1$ is true
  2. Only $S_2$ is true
  3. Both $S_1$ and $S_2$ are true
  4. Neither $S_1$ nor $S_2$ are true

5 Answers

Best answer
43 43 votes
Answer: B

$(P\#Q)\#R=(P'+Q')\#R$
$\quad \quad =P.Q+R'$

whereas,
$P\#(Q\#R)=P'+(Q\#R)'$
$\quad \quad=P'+(Q'+R')'$
$\quad \quad=P'+QR$
• edited by
20 20 votes
X#Y=X'+Y'

      =(XY)'     (DEMORGAN'S LAW)

so this # operatoe is nand operator

Nand is commutative but not associative. So B is the answer
6 6 votes

It is given that 

X#Y = X'+Y' where X and Y are Boolean variables.

Consider the following two statements.

(S1) is

(P#Q)#R = P#(Q#R)

L.H.S is  (P#Q)#R                                                                            

=(P'+Q')#R   (Using given relation)                                             

=(P'+Q')'#R=> P.Q+R' (Using Demorgans theorem )

    R.H.S is  P#(Q#R)

 =P#(Q'+R')

=P'+(Q'+R')' => P'+Q.R

L.H.S !=R.H.S

So S1 is not true.

(S2) is  Q#R = (R#Q)      

L.H.S Q#R=Q'+R'

and R.H.S is R#Q=R'+Q'

so L.H.S=R.H.S

So S2 is true.

                                                         

0 0 votes
ANSWER: B only S2 is true

S1 : (P#Q)#R = P#(Q#R)

       (P'+Q')#R = P#(Q'#R')

       (P'+Q')'+R' != P'+(Q'+R')'    
      S1 is not true (it is not associative)

S2 : Q#R = R#Q

       Q'+R' = R'+Q'
      S2 is true (it is commutative)

       

 
0 0 votes
S1 :

(P#Q)#R = (P'+Q')#R = (P'+Q')'+R'

P#(Q#R) = P#(Q'+R') = P'+(Q'+R')'

It clearly sees that it is not true.

S2 :

Q#R  = Q'+R'

R#Q = R'+Q'

clearly sees that Q'+R' == R'+Q' .

so only S2 TRUE.thank you
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