A. | (a) only |
B. | (b) only |
C. | (a) and (b) |
D. | (a), (b) and (c) |
Option: D Explanation : Click on Discuss to view users comments. |
A. | S=1 , C0= 0 and S=0 , C0= 1 |
B. | S=0 , C0= 0 and S=1 , C0= 1 |
C. | S=1 , C0= 1 and S=0 , C0= 0 |
D. | S=0 , C0= 1 and S=1 , C0= 0 |
Option: A Explanation : 1 0 0=01->carry 0,sum=1;0 1 1=10->carry 1,sum=0; Click on Discuss to view users comments. D.Krishnakumar said: (8:15pm on Monday 29th June 2015)
(A) is correctExplanation1 0 0=01->carry 0,sum=1;0 1 1=10->carry 1,sum=0;
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A. | if (c ∧ b) → r and ¬b → ¬p, then (¬r∧p)→¬c |
B. | if (c ∨ b) → r and ¬b → ¬p, then (r∧p)→c |
C. | if (c ∧ b) → r and ¬p → ¬b, then (¬r ∨ p)→¬c |
D. | if (c ∨ b) → r and ¬b → ¬p, then (¬r∧p)→ ¬c |
Option: A Explanation : Click on Discuss to view users comments. Jayanthi Ganapathi said: (5:27am on Tuesday 28th July 2015)
option D is correct
Jayanthi Ganapathy said: (5:28am on Tuesday 28th July 2015)
Option A is correct not option D
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Match the following:
List - I |
List - II |
(a) (p →q) ⇔ (¬q → ¬p) |
(i) Contrapositive |
(b) [(p ∧q) →r ]⇔ [p→ (q →r)] |
(ii) Exportation law |
(c) (p →q)⇔ [(p ∧ ¬q) →o] |
(iii) Reductio ad absurdum |
(d) (p⇔q)⇔ [(p →q) ∧(q→p)] |
(iv) Equivalence |
A. | (a) (b) (c) (d) (i) (ii) (iii) (iv) |
B. | (a) (b) (c) (d) (ii) (iii) (i) (iv) |
C. | (a) (b) (c) (d) (iii) (ii) (iv) (i) |
D. | (a) (b) (c) (d) (iv) (ii) (iii) (i) |
Option: D Explanation : Click on Discuss to view users comments. Jayanthi Ganapathy said: (5:34am on Tuesday 28th July 2015)
(A) is contrapositive(B) is Exportation Law(C) is Reductio ad adburdum(D) is Equivalence
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Consider a proposition given as :
" x ≥ 6, if x2 ≥ 25 and its proof as:
If x ≥ 6, then x2 = x· x= 6· 6=36 ≥25
A. | (a) only |
B. | (c) only |
C. | (a) and (b) |
D. | (b) only |
Option: C Explanation : Click on Discuss to view users comments. Jayanthi Ganapathy said: (5:39am on Tuesday 28th July 2015)
The proof shown is converse and it assumes x>=6 to prove the statement.
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