Integer S

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by pemdas » Sat Dec 17, 2011 2:33 am
@rijul007 by following your logic, we can state since Question stem provides a set S contains integers, and asks whether -4 is in the set S, we don't need any statements AT ALL!

-4 is an integer (like jack is anyone of us) so it's surely in the set S.

I am not convinced and would like to know the source and compare this to any similar OG question, if the original question could not be found in OGs. I doubt such an ambiguous question would appear on G-day.
rijul007 wrote:
pemdas wrote:@Ron, I am still under impression that there is at least one assumption involved with this question if we select answer B (not C). The question states: S is a set of (ALL-implied) integers such that to contain a and -a (two different numbers, if a=!0 and one number if a=0), b (the third different number) and ab (the fourth different number). We know that at least four different numbers can be in the set S -> a,-a,b,ab.

Now, Ron, don't we make an assumption by selecting answer B, that there are two numbers which are the same -a=b. Otherwise how we effect the condition 2) and allow for ab=-4 such as 2*(-2)=-4?

We've agreed from beginning that there can be four different numbers: a,-a,b,ab.

You are doing the same mistake again. You are considering a and b as particular elements of the set.

Kindly re-read the prev posts


Heres another ex to make it clearer
You must have heard of the saying
All work and no play makes jack a dull boy.
Here jack is not a particular guy.
Jack represents all those who work all day and dont play.

Similarly in this ques., a and b are no particular elements. They can be any elements of the set.

Another eg.,
Lets say we have an arithematic progression
2,4,6,8,10,....

If a(n) is the nth term of the AP
And we are given a condition
a(n) = a(n-1) +2

Here, does a(n) represent a particular element of the AP?
No, its a general term that represents any term.

In this ques lets say we do have two particular nos a and b
Then the set would turn out to be
S = {a,b,-a,-b,ab,-ab,ab^2,ba^2,....}
It would have infinite no of elements
except for a case {1,-1}, which has just two elements

I hope am able to clarify your doubt.
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by rijul007 » Sat Dec 17, 2011 3:35 am
pemdas wrote:@rijul007 by following your logic, we can state since Question stem provides a set S contains integers, and asks whether -4 is in the set S, we don't need any statements AT ALL!

Yes, set S contains integer, but does that mean it contains ALL the integers?

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by pemdas » Sat Dec 17, 2011 3:59 am
you mean that in st(2) and opting ans. B
rijul007 wrote:
pemdas wrote:@rijul007 by following your logic, we can state since Question stem provides a set S contains integers, and asks whether -4 is in the set S, we don't need any statements AT ALL!

Yes, set S contains integer, but does that mean it contains ALL the integers?
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by rijul007 » Sat Dec 17, 2011 4:10 am
pemdas wrote:you mean that in st(2) and opting ans. B
rijul007 wrote:
pemdas wrote:@rijul007 by following your logic, we can state since Question stem provides a set S contains integers, and asks whether -4 is in the set S, we don't need any statements AT ALL!

Yes, set S contains integer, but does that mean it contains ALL the integers?
(2) 2 is in S.

Condition 1:
2 and -2 both are in set S

Condition 2:
2*(-2) = -4 is also in S

Sufficient

Can you point out where I ASSUMED that set S contains ALL integers?

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by Ian Stewart » Sat Dec 17, 2011 4:47 pm
I don't really have anything to add to the comments Ron made above, but I don't find the question at all ambiguous. I do, however, think these questions can seem abstract if you haven't seen a similar question before. As for this:
pemdas wrote:
I am not convinced and would like to know the source and compare this to any similar OG question, if the original question could not be found in OGs. I doubt such an ambiguous question would appear on G-day.
there are two very similar questions to this one in the Official Quant Review book (the green book) - Q158 in the PS section, and Q70 in the DS section.
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by karthikpandian19 » Sun Dec 18, 2011 9:23 pm
This question is from GROCKIT MBA
I had previously subscribed for this.
pemdas wrote:it would be interesting to know the source of original question.
@karthikpandian19, please update us per source of this question.

@Ron, I am still under impression that there is at least one assumption involved with this question if we select answer B (not C). The question states: S is a set of (ALL-implied) integers such that to contain a and -a (two different numbers, if a=!0 and one number if a=0), b (the third different number) and ab (the fourth different number). We know that at least four different numbers can be in the set S -> a,-a,b,ab.

Now, Ron, don't we make an assumption by selecting answer B, that there are two numbers which are the same -a=b. Otherwise how we effect the condition 2) and allow for ab=-4 such as 2*(-2)=-4?

We've agreed from beginning that there can be four different numbers: a,-a,b,ab. How it comes that we consider two of no-s -a=b set in certain relation (value) if they can be in any relation (value)?

again, in your formula example, Pi*r^2, I agree r can be any radius BUT it must be always within the range of lengths made by connecting points equidistant from a center with the center.

I am still questioning answer B

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by lunarpower » Mon Dec 19, 2011 6:35 am
pemdas wrote:@Ron, I am still under impression that there is at least one assumption involved with this question if we select answer B (not C). The question states: S is a set of (ALL-implied) integers such that to contain a and -a (two different numbers, if a=!0 and one number if a=0), b (the third different number) and ab (the fourth different number). We know that at least four different numbers can be in the set S -> a,-a,b,ab.
so i think this is the problem here -- stated in two different ways:
1/ you are assuming without cause that a and b stand for DIFFERENT integers;
or, alternatively,
2/ you are assuming without cause that the set contains at least two different numbers (or four, or whatever you're saying).

neither of these is true; in mathematics, when we say "if a and b belong to a set...", weirdly enough, the default is to allow the situation in which a and b both stand for the same element of the set.
for instance, if s contains absolutely no numbers except 0, it actually satisfies both of the postulates required here: (i) we know that 0 is in S, and -0 = 0 is also in S; (ii) the only possible a and b in this statement are 0 and 0, and indeed ab = 0 is also in the set.

i can totally understand your confusion here, because this is so weird compared to the way things are normally done in the real world.
for instance, if someone says "two people x and y are standing in a park...", then this person is obviously assuming that x and y stand for different individuals. however, that's just not the way that mathematics is done, by convention -- the default assumption is that different letters may, in fact, stand for the same number.
as a side note, this is why you see so many problems that explicitly mention "different integers" (because this is NOT assumed unless you actually say it) -- but why you will never ever see a problem that says "two integers which could be the same" (because this is already assumed).

hope this helps.
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by lunarpower » Mon Dec 19, 2011 6:39 am
by the way, here's another GMAT PREP problem:
If there are more than two numbers in a certain list, is each of the numbers in the list equal to 0?
(1) The product of any two numbers in the list is equal to 0.
(2) The sum of any two numbers in the list is equal to 0.
this problem can absolutely be re-written as follows:

In the sequence a(sub 1), a(sub 2), a(sub 3), ... [these are subscripts], ..., aN, is each element of the sequence equal to 0?
(1) For any integers u and v, where 1 < u < N and 1 < v < N, the product of a(sub u) and a(sub v) is 0.
(2) For any integers u and v, where 1 < u < N and 1 < v < N, the sum of a(sub u) and a(sub v) is 0.

do you see the connection?
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by pemdas » Mon Dec 19, 2011 11:32 am
yea, clear enough. Many thanks Ron!
lunarpower wrote:by the way, here's another GMAT PREP problem:
If there are more than two numbers in a certain list, is each of the numbers in the list equal to 0?
(1) The product of any two numbers in the list is equal to 0.
(2) The sum of any two numbers in the list is equal to 0.
this problem can absolutely be re-written as follows:

In the sequence a(sub 1), a(sub 2), a(sub 3), ... [these are subscripts], ..., aN, is each element of the sequence equal to 0?
(1) For any integers u and v, where 1 < u < N and 1 < v < N, the product of a(sub u) and a(sub v) is 0.
(2) For any integers u and v, where 1 < u < N and 1 < v < N, the sum of a(sub u) and a(sub v) is 0.

do you see the connection?
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by [email protected] » Tue Mar 13, 2012 5:32 pm
Everything said by you is correct, but even after that I have a doubt...

Assume both the possibilities given by you i.e -a = b or -a not equal to b,

then this is a data sufficiency question and in DS questions,

if even one possibility negates the statements, then those statements are not taken into consideration.

what you are saying in this solution actually means that if a = 2 and -a = b by any chance then the solution makes sense. But this is again a possibility.

How can you say that -4 has to be there in the set

So then there are two cases and under 1 case -4 is there and in another case -4 isn't.

Hence as per DS GMAT rules, the answer should not be taken as B.

I think because of the extra conditions, the whole idea of DS question has changed.

In this question, we actually took one possibility, proved it right and claimed that the statement 2 is correct.

We do not do this. We try to negate and see if the thing is possible from all sides. Thats what is meant to make the argument air tight.

This is really ambigious question for a DS gmat I feel...
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by lunarpower » Wed Mar 14, 2012 5:34 am
amit, the question is not ambiguous -- the problem lies in your understanding of the notation and mathematical language, not in the question statement. i know this may be somewhat of a blow to your ego, but you are not going to learn anything if your modus operandi is to continue denying the reality of how mathematical notation is used. are you on this forum to learn, or to argue?

i'm going to try to explain this in one more way, and then i will let this thread rest.
if the notation is confusing you, here's what these statements mean without the notation:
if a is in S, then -a is in S,
--> this means, "for any number that is in S, the opposite of that number is also in S."
this statement does NOT reference a particular value of "a"; it is a general truth about all values in the set.
if each of a and b is in S, then ab is in S.
--> this means, "for any two numbers that are in S (even if that's the same number twice), the product of those numbers is also in S."
again, this statement does NOT reference particular values of "a" and "b"; it's another general truth about all values in the set.

this is how this sort of notation works.
full disclosure: i am a mathematician by formal training; i have studied with books and professors from all around the world, and this is how this kind of notation is used absolutely everywhere. so, you should take a moment to
(a) learn the correct meaning of these statements, and
(b) write your own examples of situations and statements that work in the same way -- this is the best way to learn notation/language with which you are unfamiliar.

good luck.
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