GATE CSE 2014 | Set 3 | Question: 1

Question

Consider the following statements:

• P: Good mobile phones are not cheap
• Q: Cheap mobile phones are not good

L: P implies Q
M: Q implies P
N: P is equivalent to Q

Which one of the following about L, M, and N is CORRECT?

• A. Only L is TRUE.
• B. Only M is TRUE.
• C. Only N is TRUE.
• D. L, M and N are TRUE.

Comments

If N is true then it make L and M both true.

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let

A = good mobile phones

B = cheap mobiles phones

P = A imples not B

Q = B imples not A

So, Option D is correct

Answer 1

Correct Answer (D)

Lets break the given compound statements into atomic statements.

• A : Good mobile phones.
• B : Cheap mobile phones.

$P :(A\to \neg B) \iff (\neg A\vee \neg B)$
$Q :(B\to \neg A) \iff \big((\neg B\vee \neg A) \iff \neg A\vee \neg B)\big)$ (Disjunction is commutative),
Hence, $(P\iff Q)$ which means $(P\to Q)$ and $(Q \to P).$

Comments

Nice Explanation

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@Arjun sir

Please explain

How to break this statement  into conditional statement ??

• P: Good mobile phones are not cheap
• Q: Cheap mobile phones are not good

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Crystal clear....thqqq

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P and Q are contrapositives of each other.

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How do we know that implication will be used for P and Q as the statements don't have any one of the terminologies::

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great explanation

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How we know that implication comes in to P and Q....here statement is not used term like...if, only if...etc

Plz help

@Bikram sir

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@KUSHAGRA गुप्ता , @Hradesh patel Here the implication is not used directly but used with quantifiers. The given statements P and Q are examples of an implicit universal quantification where the keywords like for all, for every, are  not used directly but their meaning is implied automatically

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It is first order logic question not propositional logic question . For simplicity propositional logic is used in solution .

p: for all x if x is a good mobile phone then x is not cheap

q: for all x if x is cheap mobile phone then x is not good

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What does “TRUE” mean here, does it mean “Tautology”

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@Deepak Poonia sir can we do this like this way:

Let’s first define the predicate:

G(x): Good  M(x):Mobile Phone  C(x): cheap

P: $\forall x((G(x) \Lambda M(x)) \rightarrow \sim C(x) )$ $\equiv$ $\forall x(\sim G(x) V \sim C(x) V \sim M(x))$

Q: $\forall x((C(x) \Lambda M(x)) \rightarrow \sim G(x) )$ $\equiv$ $\forall x(\sim G(x) V \sim C(x) V \sim M(x))$

Hence, (P⟺Q) which means (P→Q) and (Q→P).

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@samarpita

Yes, we can do this way.

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@KUSHAGRA गुप्ता and @Hradesh patel along with the reading provided by @subbus, you can also practice these two imp. examples (Examples 26 & 27) in Section-1.4.12 titled, “Examples from Lewis Carroll”, Rosen 8^th Ed. These will be helpful for understanding the use implications in such statements.

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very nice explanation

Answer 2

P and Q are contra-positives to each other. A proposition and its contra-positive are always equivalent.

Equivalent and <--> are same, so p <--> q holds.

So option D)

Comments

can you please explain me the line "P is equivalent to Q, which means P implies Q ,and Q implies P" .

P implies Q means P->Q  and Q implies P means Q->P then how can we conclude these two from the fact that P is equivalent to Q.

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P is equivalent to Q means (P->Q)^(Q->P). So both (P->Q) and (Q->P) must be true to ensure this.

Answer 3

Here given statements P, Q are implicitly universally quantified statements

we can express P,Q statements using quantifiers as shown below

$\therefore $ Answer is option D

Answer 4

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Answer 5

P: Good mobile phones are not cheap (False: some good mobile phones are cheap)

Q: Cheap mobiles are not good ( False: some cheap phones are also good)

L: P→Q : False → False : True

M: Q→P: False → False: True

N: P=Q: True

Hence L,M,N are True ans is (D)

 

Please correct me if my approach is wrong.