GATE CSE 2014 Set 3 | Mathematical Logic Question 11 | Discrete Mathematics

GATE CSE 2014 Set 3

MCQ (Single Correct Answer)

-0.3

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 of the following about L, M, and N is Correct?

Only L is TRUE.

Only M is TRUE.

Only N is TRUE.

L, M and N are TRUE.

GATE CSE 2014 Set 1

MCQ (Single Correct Answer)

-0.3

Consider the statement
"Not all that glitters is gold"
Predicate glitters$$(x)$$ is true if $$x$$ glitters and
predicate gold$$(x)$$ is true if $$x$$ is gold.

Which one of the following logical formulae represents the above statement?

$$\forall x:\,glitters\,\left( x \right) \Rightarrow \neg gold\left( x \right)$$

$$\forall x:\,gold\left( x \right) \Rightarrow glitters\left( x \right)$$ v

$$\exists x:\,gold\left( x \right) \wedge \neg glitters\left( x \right)$$

$$\exists x:\,glitters\,\left( x \right) \wedge \neg gold\left( x \right)$$

GATE CSE 2012

MCQ (Single Correct Answer)

-0.3

Consider the following logical inferences.
$${{\rm I}_1}:$$ If it rains then the cricket match will not be played. The cricket match was played.
Inference: there was no rain.

$${{\rm I}_2}:$$ If it rains then the cricket match will not be played. It did not rain
Inference:the cricket match was played. which of the following is TRUE?

Both $${{\rm I}_1}$$ and $${{\rm I}_2}$$ are correct inferences

$${{\rm I}_1}$$ is correct but $${{\rm I}_2}:$$ is not a correct inference

$${{\rm I}_1}$$ is not correct but $${{\rm I}_2}$$ is a correct inference

Both $${{\rm I}_1}$$ and $${{\rm I}_2}$$ are not correct inferences

GATE CSE 2012

MCQ (Single Correct Answer)

-0.3

What is the correct translation of the following statement into mathematical logic? "Some real numbers are rational"

$$\exists x\left( {real\left( x \right) \vee rational\left( x \right)} \right)$$

$$\forall x\left( {real\left( x \right) \to rational\left( x \right)} \right)$$

$$\exists x\left( {real\left( x \right) \wedge rational\left( x \right)} \right)$$

$$\exists x\left( {rational\left( x \right) \to real\left( x \right)} \right)$$