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conty911
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S is a set of integers such that I. If x is in S, then x is in S. Updated on: 20 Sep 2024, 00:34
Originally posted by conty911 on 25 Apr 2012, 21:58.
Last edited by Bunuel on 20 Sep 2024, 00:34, edited 4 times in total.
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35% (medium)

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68% (01:29) correct 32% (01:37) wrong based on 343 sessions

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S is a set of integers such that

I. If x is in S, then –x is in S.
II. If both x and y are in S, then so is x + y.

Is –2 in S?

(1) 1 is in S.
(2) 0 is in S.

Show Spoiler
The answer A, but i didnot find the explanation given in the book convincing.Please help. I thought the ans. to be E. :?:
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Official Answer
E
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Bunuel
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S is a set of integers such that I. If x is in S, then x is in S. 25 Apr 2012, 22:27
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conty911
S is a set of integers such that
I) If x is in S, then –x is in S.
II) If both x and y are in S, then so is x + y.
Is –2 in S?

(1) 1 is in S.
(2) 0 is in S.

The answer
Show Spoiler
A
, but i didnot find the explanation given in the book convincing.Please help. I thought the ans. to be E. :?:
Show more

You are right answer should be E, not A.

S is a set of integers such that
I) If x is in S, then –x is in S.
II) If both x and y are in S, then so is x + y.
Is –2 in S?

(1) 1 is in S --> according to (I) -1 is also in S. Next, according to (II) -1+1=0 also in S. We don't know whether -2 is in the set, for example S could be {-1, 0, 1}. Not sufficient.

(2) 0 is in S. Not sufficient.

(1)+(2) Nothing new. Not sufficient.

Answer: E.

Similar questions to practice:
s-is-a-set-of-integers-such-that-i-if-a-is-in-s-then-a-is-109986.html
k-is-a-set-of-numbers-such-that-i-if-x-is-in-k-then-x-96907.html
if-p-is-a-set-of-integers-and-3-is-in-p-is-every-positive-96630.html
k-is-a-set-of-integers-such-that-if-the-integer-r-is-in-k-103005.html
s-is-a-set-of-integers-such-that-i-if-a-is-in-s-then-a-130000.html

Hope it helps.
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S is a set of integers such that I. If x is in S, then x is in S. 25 Apr 2012, 23:19
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Thankyou, it really helped me in resolving my confusion :) .The explanation given in the book for statement 1 was -1+(-1)=-2 rather than 1-1=0;so they must have assumed x=y;
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S is a set of integers such that I. If x is in S, then x is in S. 28 Apr 2012, 15:48
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IMO very bad question .... they have not clearly stated the domain conditions. Kindly use the good materials only.
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S is a set of integers such that I. If x is in S, then x is in S. 09 Oct 2014, 10:14
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conty911
S is a set of integers such that
I) If x is in S, then –x is in S.
II) If both x and y are in S, then so is x + y.
Is –2 in S?

(1) 1 is in S.
(2) 0 is in S.

The answer
Show Spoiler
A
, but i didnot find the explanation given in the book convincing.Please help. I thought the ans. to be E. :?:
Show more

if such sets are considered infinite sets then A should be sufficient.

{1 -1 0 1 2 -2 }
Explanation for the above set: given 1 is in S hence -1 is also in S ( using property 1 )now -1 +1=0 is also in S by property 2 so we have 0 in S now in the set we have 1 -1 0, using property 2 if 1 and 0 are in set then 1+0 =1 is also in set hence we have 1 -1 0 1 in the set, again using property 2, if 1 and 1 are in set then 1+1=2 is also in set ,hence 2 is in set and using property 1 , if 2 is in S then -2 is also in S.

Hence if 1 is in S then -2 is also in S provided that the set is infinite

if the set can be stopped in between or if the set is a finite set then the set could be {1 -1 0 }

According to the OA the set is being considered infinite.
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S is a set of integers such that I. If x is in S, then x is in S. 09 Oct 2014, 20:56
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Why aren't the sets determined by I and II (independently) finite? That would make D the answer.

Bunuel
conty911
S is a set of integers such that
I) If x is in S, then –x is in S.
II) If both x and y are in S, then so is x + y.
Is –2 in S?

(1) 1 is in S.
(2) 0 is in S.

The answer
Show Spoiler
A
, but i didnot find the explanation given in the book convincing.Please help. I thought the ans. to be E. :?:

You are right answer should be E, not A.

S is a set of integers such that
I) If x is in S, then –x is in S.
II) If both x and y are in S, then so is x + y.
Is –2 in S?

(1) 1 is in S --> according to (I) -1 is also in S. Next, according to (II) -1+1=0 also in S. We don't know whether -2 is in the set, for example S could be {-1, 0, 1}. Not sufficient.

(2) 0 is in S. Not sufficient.

(1)+(2) Nothing new. Not sufficient.

Answer: E.

Similar questions to practice:
s-is-a-set-of-integers-such-that-i-if-a-is-in-s-then-a-is-109986.html
k-is-a-set-of-numbers-such-that-i-if-x-is-in-k-then-x-96907.html
if-p-is-a-set-of-integers-and-3-is-in-p-is-every-positive-96630.html
k-is-a-set-of-integers-such-that-if-the-integer-r-is-in-k-103005.html
s-is-a-set-of-integers-such-that-i-if-a-is-in-s-then-a-130000.html

Hope it helps.
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S is a set of integers such that I. If x is in S, then x is in S. 10 Oct 2014, 01:53
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kamranjkhan
Why aren't the sets determined by I and II (independently) finite? That would make D the answer.

Bunuel
conty911
S is a set of integers such that
I) If x is in S, then –x is in S.
II) If both x and y are in S, then so is x + y.
Is –2 in S?

(1) 1 is in S.
(2) 0 is in S.

The answer
Show Spoiler
A
, but i didnot find the explanation given in the book convincing.Please help. I thought the ans. to be E. :?:

You are right answer should be E, not A.

S is a set of integers such that
I) If x is in S, then –x is in S.
II) If both x and y are in S, then so is x + y.
Is –2 in S?

(1) 1 is in S --> according to (I) -1 is also in S. Next, according to (II) -1+1=0 also in S. We don't know whether -2 is in the set, for example S could be {-1, 0, 1}. Not sufficient.

(2) 0 is in S. Not sufficient.

(1)+(2) Nothing new. Not sufficient.

Answer: E.

Similar questions to practice:
s-is-a-set-of-integers-such-that-i-if-a-is-in-s-then-a-is-109986.html
k-is-a-set-of-numbers-such-that-i-if-x-is-in-k-then-x-96907.html
if-p-is-a-set-of-integers-and-3-is-in-p-is-every-positive-96630.html
k-is-a-set-of-integers-such-that-if-the-integer-r-is-in-k-103005.html
s-is-a-set-of-integers-such-that-i-if-a-is-in-s-then-a-130000.html

Hope it helps.
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The set might be finite but that's not the point. We don't know whether -1, 0, and 1 are the only integers in the set. Please follow the links provide to understand the logic of these kind of questions better.
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S is a set of integers such that I. If x is in S, then x is in S. 10 Oct 2014, 04:40
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qlx
conty911
S is a set of integers such that
I) If x is in S, then –x is in S.
II) If both x and y are in S, then so is x + y.
Is –2 in S?

(1) 1 is in S.
(2) 0 is in S.

The answer
Show Spoiler
A
, but i didnot find the explanation given in the book convincing.Please help. I thought the ans. to be E. :?:

if such sets are considered infinite sets then A should be sufficient.

{1 -1 0 1 2 -2 }
Explanation for the above set: given 1 is in S hence -1 is also in S ( using property 1 )now -1 +1=0 is also in S by property 2 so we have 0 in S now in the set we have 1 -1 0, using property 2 if 1 and 0 are in set then 1+0 =1 is also in set hence we have 1 -1 0 1 in the set, again using property 2, if 1 and 1 are in set then 1+1=2 is also in set ,hence 2 is in set and using property 1 , if 2 is in S then -2 is also in S.

Hence if 1 is in S then -2 is also in S provided that the set is infinite

if the set can be stopped in between or if the set is a finite set then the set could be {1 -1 0 }

According to the OA the set is being considered infinite.
Show more


Responding to a pm:

I am not really sure what your query is but I think it is this: why isn't the set necessarily infinite?

You get that -1, 1 and 0 must be in the set - great. But this is not correct: "using property 2 if 1 and 0 are in set then 1+0 =1 is also in set hence we have 1 -1 0 1 in the set"

The set till now is {-1, 0, 1}
Both properties are already satisfied for each element. 1+0 = 1 is already present in the set. Why do you need to add another 1? How do you differentiate one 1 from another 1? The set could be just {-1, 0, 1} or it could have some other elements too.

The other question you quoted has a different rule. If k is in the set, k+1 must be in the set too. So the moment you put any one number in the set, all numbers after it will automatically be a part of this set.
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S is a set of integers such that I. If x is in S, then x is in S. 20 Sep 2024, 00:30
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2+1=3 which is not in your set.
qlx
conty911
S is a set of integers such that
I) If x is in S, then –x is in S.
II) If both x and y are in S, then so is x + y.
Is –2 in S?

(1) 1 is in S.
(2) 0 is in S.

The answer
Show Spoiler
A
, but i didnot find the explanation given in the book convincing.Please help. I thought the ans. to be E. :?:

if such sets are considered infinite sets then A should be sufficient.

{1 -1 0 1 2 -2 }
Explanation for the above set: given 1 is in S hence -1 is also in S ( using property 1 )now -1 +1=0 is also in S by property 2 so we have 0 in S now in the set we have 1 -1 0, using property 2 if 1 and 0 are in set then 1+0 =1 is also in set hence we have 1 -1 0 1 in the set, again using property 2, if 1 and 1 are in set then 1+1=2 is also in set ,hence 2 is in set and using property 1 , if 2 is in S then -2 is also in S.

Hence if 1 is in S then -2 is also in S provided that the set is infinite

if the set can be stopped in between or if the set is a finite set then the set could be {1 -1 0 }

According to the OA the set is being considered infinite.
Show more
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