Integer S

[karthikpandian19]

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.

[rijul007]

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.

(1) 1 is in S.

Acc to condition(i)
1 and -1 both are in the set S
Acc to condition (ii)
1*(-1) = -1 is in the set

Doest tell us anything about -4


(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


Option B

[pemdas]

where in st(2) it tells us a=b (or -a=b) so that you consider ab? If you were allowed to set such equations then it would be D not B, as you can consider well difference (d) and set arithmetic seq. or consider ratio to set a geometric seq. We need to keep by the conditions given within statements (1,2).

I go for c, Combined st(1&2) 1 and 2 are in S, either a=1(-1) and b=2(-2) or a=2(-2), b=1(-1)
possible set -> S {-1,1,2-2}

answer No, Sufficient

c
[spoiler]in st(1) a can be 1 or b can be 1, missing either b or a for cond. 2
in st(2) the same logic as per st(1).
Combined st-s: a,b are set, -4 is not in set S[/spoiler]

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.

(1) 1 is in S.

Acc to condition(i)
1 and -1 both are in the set S
Acc to condition (ii)
1*(-1) = -1 is in the set

Doest tell us anything about -4


(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


Option B

[rijul007]

where in st(2) it tells us a=b (or -a=b) so that you consider ab? If you were allowed to set such equations then it would be D not B, as you can consider well difference (d) and set arithmetic seq. or consider ratio to set a geometric seq. We need to keep by the conditions given within statements (1,2).

I go for c, Combined st(1&2) 1 and 2 are in S, either a=1(-1) and b=2(-2) or a=2(-2), b=1(-1)
possible set -> S {-1,1,2-2}

answer No, Sufficient

c
[spoiler]in st(1) a can be 1 or b can be 1, missing either b or a for cond. 2
in st(2) the same logic as per st(1).
Combined st-s: a,b are set, -4 is not in set S[/spoiler]

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.

(1) 1 is in S.

Acc to condition(i)
1 and -1 both are in the set S
Acc to condition (ii)
1*(-1) = -1 is in the set

Doest tell us anything about -4


(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


Option B

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

I hope you knw that the red part is the question
and blue part contains the two statements...

the two statements in red are always true for S...

[pemdas]

a=b is true, but is a=b or -a=b true? you make assumption than b=-a in st(2)
a=2, a=-a=-2
b=-2 implies b=-a

You are told 2 is in S.
2 can be a or b. You consider 2 as a (or b) and set a=-a. Then you make assumption b exists in set S too You set b=-2 and multiply ab=2*(-2)

I insist on C

[rijul007]

a=b is true, but is a=b or -a=b true? you make assumption than b=-a in st(2)
a=2, a=-a=-2
b=-2 implies b=-a

You are told 2 is in S.
2 can be a or b. You consider 2 as a (or b) and set a=-a. Then you make assumption b exists in set S too You set b=-2 and multiply ab=2*(-2)

I insist on C

I am really not getting what you're doing with a = -a or b = a thing here


S is a set that can have any number of elements.....


Statement 2 says
2 is in S

From condition 1, we can imply that
If 2 is in S, -2 is also in S
Right?

Now from condition 2 we can imply that
If 2 and -2 are in S, then 2*(-2) is in S

So the statement would be sufficient

[pemdas]

st(2) says 2 is in S. Question has two conditions
condition 1) if a is in S, then -a is in S NOT if b is in S, then -b is in S
>> applied to st(2) 2 is in S and it can be a or b. We don't know if a=2 or b=2, we only know that 2 is in S. You make an assumption here: a=2. There's one value, 2, and two possible elements in set S, a and b. You assign a=2 and let -a is in S with 2 and -2 put in set S. Then you assign b a new value from the derived value of -a, and you make the other assumption -a=b and conclude set S contains 2,-2 and 2*(-2) ... applying condition 2) below

condition 2) if each of a and b is in S, then ab is in S
>> you've already established the value of a (arbitrarily, you set a=2 and you are sure b=!2, but a=2 although 2 is in set S may imply a or b is 2 as both are in set S. Further you assign -a is in S and -a=b. Next, you enforce condition 2) a*b=2*(-2)

do you understand?

[neelgandham]

I insist on C

Ah! I see you what you say. Let us remove the variables a and b and replace them with An Integer and Another Integer respectively.

S is a set of integers such that
i) if An Integer is in S, then -(An Integer) is in S, and
ii) if each of An Integer and Another Integer is in S, then Product of these Integers is in S.

Is -4 in S?

(1) 1 is in S.

If 1 is one of the Integers in the set, then -1 is also in the set(from i)
If 1 and -1 are in the set, then -1*1 = -1 is also in the set(from ii)
Insufficient

(2) 2 is in S.

If 2 is one of the Integers in the set, then -2 is also in the set(from i)
If 2 and -2 are in the set, then -2*2 = -4 is also in the set(from ii)
sufficient to answer the question

Hence B as Riju mentioned above. Let me know if you aren't convinced.

[pemdas]

Nil, with what you say we can construct arithmetic seq. a,-a,
a=1,-a=-1, difference=2
... -1,1,3,5,7 ...

or geometric sequence
a=1, -a=-1, -a/a=-1
...-1,1,-1,1,-1...

Basically what you are saying applied with condition 2) can extend our arithmetic and/or geometric sequences. What you are saying is a can be equal to b to any value. With such loose restriction we won't be able to decide if -4 is in S or not. This can be both cases.

It's very precise if a is in S then -a is in S (not b is in S and -b is in S)
At the same time it says if each a,b in set S - this implies set S may contain two elements and we cannot know if 2=a or 2=b. May be 2 is -a and a=-2?

do you see how aberrant your sol?

[rijul007]

@pemdas

You have wrongly interpreted the question..

In the conditions, a and b are general elements in the Set... They do not represent any particular element.....

The conditions should be interpreted as
i) If a number is in S, then negative of that number is in S

ii) I any two nos are in S, then their product is in S.

[neelgandham]

Nil, with what you say we can construct arithmetic seq. a,-a,
a=1,-a=-1, difference=2
... -1,1,3,5,7 ...
or geometric sequence
a=1, -a=-1, -a/a=-1
...-1,1,-1,1,-1...

We can construct BUT we shouldn't. I can understand your overemphasis on the variables (a,b) but they just denote two different integers and nothing else.

It's very precise if a is in S then -a is in S (not b is in S and -b is in S)

No, again you are taking the variables seriously. The question says, if a is in S then -a is in S, i.e if Any Integer A is in the set S then -A is also in the set S. As mentioned earlier forget about the values A and B and treat them as two unique integers. If you are still not convinced, never mind, blame it on my written skills:-).I shall ask one of the experts or Quantwhizs to explain it to you

[lunarpower]

i received a message regarding this thread.

the following response is perfectly correct and succinct:

@pemdas

You have wrongly interpreted the question..

In the conditions, a and b are general elements in the Set... They do not represent any particular element.....

The conditions should be interpreted as
i) If a number is in S, then negative of that number is in S

ii) I any two nos are in S, then their product is in S.

yep. in these kinds of things, the letters/variables are generalized elements.

you'll see the same thing when function symbols are defined. for instance, "x # y is defined to mean x - 2y" means that
any number # any other number
is the first number minus twice the second one.
i.e., "x" and "y" stand for any two numbers at all; there is no issue of figuring out particular values for them to represent.

same thing for any formula at all. think about formulas such as Area = (pi)(r^2) for a circle; again, "r" can stand for absolutely any radius at all.
etc.

[pemdas]

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

[shankar.ashwin]

Would like to add a few points too;

Using B : we know the set contains 2 and '-2'

Now consider a = 2 and b = -2,

we get a third element in the set ab = -4.

Now at this point we could also have a = 2 , b = -4 (OR) a = -2, b = -4 (including the third element as well)

now we have ab = -8 (or) 8

To *strictly* satisfy both statements in the question stem, the SET should contain infinite elements such that ANY number (a,b) in the set also contains (ab). But we aren't really bothered if the set contains finite elements or not.

If the set were to satisfy both statements, we can conclude it contains -4 as well.


The Set S would look like (2,-2,4,-4,8,-8,16 ------- )

Now extending the same logic to Statement I which says '1' is in Set S.

We know -1 also is in set S

Again a=1 and b = -1

Now the third element (ab) = 1*(-1) = -1.

Now the SET S contains (1,-1,-1) now for any two numbers in Set S (a and b),

should satisfy a=-a and if 'a' and 'b' are present 'ab' should also be present. But notice for any number in the Set S, these 2 operations would again result in either a '1' or '-1'. So the set contain only 2 distinct elements FOR SURE (1 and -1) It maybe possible that there are other elements but we cannot conclude it from the question stem and statement 1 alone.

[rijul007]

@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.