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In the following, $A$ stands for a set of apples, and $S(x, y)$ stands for "$x$ is sweeter than $y$. Let

$$\Psi \equiv \exists x : x \in A$$

$$\Phi \equiv \forall x \in A : \exists y \in A : S(x, y).$$

Which of the following statements implies that there are infinitely many apples (i.e.,, $A$ is an inifinite set)?

  1. $\Psi \wedge \Phi \wedge [\forall x \in A : \neg S(x, x)]$
  2. $\Psi \wedge \Phi \wedge [\forall x \in A : S(x, x)]$
  3. $\Psi \wedge \Phi \wedge [\forall x,y \in A : S(x, x) \wedge S(x, y) \rightarrow S(y,y)]$
  4. $\Psi \wedge \Phi \wedge [\forall x \in A : \neg S(x, x)] \wedge [\forall x, y, z \in A : S(x, y) \wedge S(y, z) \rightarrow s(y, x)]$
  5. $\Psi \wedge \Phi \wedge [\forall x \in A : \neg S(x, x)] \wedge [\forall x, y, z \in A : S(x, y) \wedge S(y, z) \rightarrow s(x, z)]$
asked in Mathematical Logic by Veteran (75.6k points)   | 174 views

2 Answers

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Here is my Approach ....i am also getting (E) as the answer 

let A be {1,2}  (say apple 1, apple 2)

there is atleast one element in A : satisfied

for every element x in A there is a y in A such that S(x,y) : Since nothing is told about the symmetry of the relation, we can have S(1,2) and S(2,1)  : so,satisfied

So as of now, A = {1,2} and S={(1,2),(2,1)}

now we can go throught the options :

(A) S(x,x) is not possible ....i dont have that in S ....so satisified .....and still finite A

(B)  S(x,x) should be there......well, i will make S={(1,2)(2,1),(1,1),(2,2)}....satisfied and still finite A

(C) take the same case as above ....satisfied and still finite A

(D) Now i cant have (1,1) and (2,2)

but even with S={(1,2).(2,1)} this condition is satisfied ....still finite A

(E) i cant do this with (1,2) and (2,1) ...because the trasitivity makes it (1,1) which should not be there .....and whatever elements i add the transitivity will lead me to (x,x) .....because any element added to A should occur atleast once in the left side of an ordered pair.......so only solution is infinite A ...

answered by (413 points)  
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$\Psi$ is the set containing at least one apple. $\Phi$ is the set containing all apples which are sweeter than at least one apple. 
Now,
(A) - Set of some apples such that they are sweeter than some apples and no Apple is sweeter than itself. Whether A is finite or infinite this statement always gives such apples therefore, this statement does not necessarily imply A is infinite. 

(B) - Set of some apples such that they are sweeter than some & itself. Whether A is finite or infinite this set is allways null as no Apple can be sweeter than itself. Therefore this statement doesn't imply A is infinite. 

(C)- Same as above with little variations.

(D)-  This statement also gives empty set whether A is finite or infinite due to given implication therefore this statement doesn't imply A is infinite. 

(E)- Set of some apples such that they are sweeter than some apples & no Apple among them is sweet than itself BUT every Apple among them is sweeter than some other apple. Now, if A is finite this statement is bound to give empty set since if A is finite we always can arrange Apples in order of their increasing degree of sweetnes & first apple can't have property "sweeter than some apple". But if A is infinite & that too unbounded set $(-\infty,+\infty)$(regarding degree of sweetnes) we can always have some Apples sweeter than others. Therefore they can have property "every apples among them can be sweeter than any other apple". Now considering this statement to be not null, statement will imply A is infinite. 

Therefore I think E is the answer. Lemme know if I'm wrong.

answered by Loyal (3.2k points)  
edited by
How can output of A be finite. Every apple should have an apple other than itself which is less sweet than it. So if A is finite as you are saying is possible, then how would that condition hold for least sweet apple?
Let A = { 1,2,3} , numbers denoting degree of sweetness. ∅ = { 2,3}

(A) can state about this set as well -

{1,2,3} ^ {2,3} ^ Universe = { 2,3} .
Okk...now what E is adding to that example?
Yes. You're right. I think somewhere is mistake. Let me think on this. Thank you for providing a loophole.
Sure.
Please let me know when you come up with an answer. :)
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