-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.

















