S is a set of integers such that i) if a is in S, then a is

We are told that: if some number a is in S, then –a is also in S, and if each of a and b is in S, then their product, ab is also in S.

(2) says that 2 is in S, then -2 also must be in S and since both 2 and -2 are in S then their product 2*(-2)=-4 must also be in S.

Hope it's clear.

Bunuel - Im also confused, could you kindly help?
(2) says that 2 is in S so this can be a or b right?
So if 2 is a then -2 is -a
if 2 is b then -2 is -b
Aren't these mutually exclusive i.e how can we take 2 to be a and -2 to be b?

Also another way I (mistakenly) interpret the statements as if 2 (a or b) is in set then b (or a) can be any nos in the set? for example 5 then we cant get -4


please help!

cheers

(2) says that 2 is in S so this can be a or b right?
So if 2 is a then -2 is -a
if 2 is b then -2 is -b
Aren't these mutually exclusive i.e how can we take 2 to be a and -2 to be b?

Also another way I (mistakenly) interpret the statements as if 2 (a or b) is in set then b (or a) can be any nos in the set? for example 5 then we cant get -4


please help!

cheers

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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.

It doesn't matter what you call the number.
i) States that if a number is in S, then its opposite is also in S. Meaning, if 1 is in S, then -1 is also in S. If -3 is in S, then -(-3) = 3 is also in S.
ii) States that if two numbers are in S, then their product is also in S. Doesn't matter even the two numbers are equal. For example, if 2 is in S, then 2*2 = 4 is also in S. If 2 and -2 are in S, then 2*(-2) = 4 is also in S, 2*4 = 8, -2*4 = -8,...

Hi Bunnel ,

I am a little confused with this one .
In stmt A ) i get the elements in the set as 1 and -1 ... i did not find 12 so i thought the Set S has only two elements .. hence We could answer the question that 12 is not present in the set ?

However later on i saw that this is not sufficient to answer the question how can 12 be present in the set with Stmt 1 yields on two values 1 , -1 ?

Thanks and Regards ,
Sheldon Rodrigues

1 and -1 may NOT be the only numbers in S. S could contain a whole bunch of other numbers. For example, the set could be:
{..., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ...}
{..., -5, -3, -1, 1, 3, 5, ...}

Hope it's clear.

Sorry, I am lost here, if we know that a =-2 then how do we now that b= 2 ? to reach the conclusion that -4 is in S.

Hi,
you don't require to know what is a or what is b..
a and b are just two integers in the SET...
similarly when we know 2 is there, -2 will be there..
if 2 and -2 are there, 2*-2=-4 is there..

further if -4 is there, 4 is there..
so 4*-4=-16 is also there..
also 2*4=8 is there.. and so on

so a and b stand for any two integers in the set..

Given :
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 --> according to (i) -1 is in S. Is -4 in S? -- Not sufficient.

(2) 2 is in S --> according to (i) -2 is in S --> according to (ii) -2*2=-4 is in S. Sufficient.

Answer: B.

(A) If \(1\) is in set, then \(-1\) is also in set. We don't know about other elements in the set and we cannot say whether \(-4\) will be there in the set. Insufficient.

(B) If \(2\) is in set, then \(-2\) is also in set, and \(2*-2=-4\) is also in the set. Sufficient.

Hence, B.

Hi All

Trying my hand to explain the Qs since it took me 15 mins to crack the discrepancy between a and b-

Lets forget about a and b for a minute and lets discuss the conditions in general-

1- states that if any integer is present in the set, then the integer's negative counter part is also present.
2- states that if 2 integers (any 2 integers) are present then their product is also present.

Option B-
2 is present

condition 1: -2 also present
condition 2: states since 2 is present (given) and from condition 1 (above) : (-2) is also present

then 2*(-2) = -4 will be present


Here, a, -a, and b are given just to show the relationship in individual conditions.


Hope it helps.