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Set of integers

Expert replies
by ayankm » Sat May 08, 2010 7:31 am
S is a set of integers such that
i) if a is in S, then -a is in S, and
ii) if each of a and b is in S, then ab is in S.

Is -4 in S?

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

Answer is -B-.
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Source: — Data Sufficiency |

by akhpad » Sat May 08, 2010 7:45 am
ayankm wrote:S is a set of integers such that
i) if a is in S, then -a is in S, and
ii) if each of a and b is in S, then ab is in S.

Is -4 in S?

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

Answer is -B-.
Statement 1:

1 is in S then -1 is in S
if each of 1 and -1 is in S, then -1 [-1*1] is in S

No information of -4; -4 may or may not be in S

Not Sufficient

Statement 2

2 is in S then -2 is in S
if each of 2 and -2 is in S, then -4 [-2*2] is in S

Yes, -4 is in S

Sufficient

Answer B
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by ayankm » Sat May 08, 2010 7:51 am
ayankm wrote: No information of -4; -4 may or may not be in S

Not Sufficient
My question now is, isnt the Condition 1 enough to say that -4 is not in S?
On what basis are we concluding that it MAY or MAY NOT be in S.
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by kevincanspain » Sat May 08, 2010 7:56 am
ayankm wrote:
ayankm wrote: No information of -4; -4 may or may not be in S

Not Sufficient
My question now is, isnt the Condition 1 enough to say that -4 is not in S?
On what basis are we concluding that it MAY or MAY NOT be in S.
On the basis that we do not know what other elements there may be in S
Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid
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by rockeyb » Sat May 08, 2010 8:46 am
ayankm wrote:S is a set of integers such that
i) if a is in S, then -a is in S, and
ii) if each of a and b is in S, then ab is in S.

Is -4 in S?

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

Answer is -B-.
The issue I have here is why are we assuming '-a' = 'b'

I dont see any were in the question saying '-a' = 'b' .

if 'a' and 'b' are two different integers then can we apply the 2 nd condition that is if each of a and b is in S, then ab is in S.

If that was the case then why dose problem statement not say 'a'* '-a' is in S .

Also if at all we consider 'b' = -2 then also we are not certain that -4 is in S .

Since when 'b' = -2 then -4 is in S and when 'b' = 1 then -4 may not be in S . So we are not certain if -4 is in S or not .

Can any one tell me what I am doing wrong here ?
"Know thyself" and "Nothing in excess"
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by harshavardhanc » Sat May 08, 2010 1:01 pm
rockeyb wrote:
ayankm wrote:S is a set of integers such that
i) if a is in S, then -a is in S, and
ii) if each of a and b is in S, then ab is in S.

Is -4 in S?

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

Answer is -B-.
The issue I have here is why are we assuming '-a' = 'b'

I dont see any were in the question saying '-a' = 'b' .

if 'a' and 'b' are two different integers then can we apply the 2 nd condition that is if each of a and b is in S, then ab is in S.

If that was the case then why dose problem statement not say 'a'* '-a' is in S .

Also if at all we consider 'b' = -2 then also we are not certain that -4 is in S .

Since when 'b' = -2 then -4 is in S and when 'b' = 1 then -4 may not be in S . So we are not certain if -4 is in S or not .

Can any one tell me what I am doing wrong here ?
rocky,

the question is saying that :

1) S will contain the additive inverse of every element.
2) for every pair of integers in S, you'll also find their product in S.

for e.g if 1, 2, and 3 are present in S.

We are sure that -1,-2, and -3 are also present in S ( per condition 1 )

and that 1*2 , 1*3, and 2*3 are also present in S ( per condition 2)

HTH.
Regards,
Harsha
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