sachindia wrote:K is a set of numbers such that
a) if x is in K, then -x is in K and
b) if each of x and y is in K, then xy is in K.
Is 12 in K?
1)2 is in K.
2)3 is in K.
If x is in K, then -x is in K.
In other words:
If a particular value is in K, then -(THAT VALUE) also is in K.
If x and y are in K, then xy is in K.
In other words:
If any two particular values are in K, then THEIR PRODUCT also is in K.
The conditions above apply to EVERY value in K.
Thus, each condition will yield an INFINITE number of values in K, as we will see when we evaluate the two statements.
Statement 1: 2 is in K
Thus, -2 is in K.
Thus, 2 * -2 = -4 is in K.
Thus, -(-4) = 4 is in K.
Thus, 2*4 = 8 is in K.
Thus, -2*4 = -8 is in K.
Thus, 4 * - 4 = -16 is in K.
Thus, -(-16) = 16 is in K.
Thus, K = {...-16, -8, -4, -2, 2, 4, 8, 16...}.
But we don't know what other values might be in K, so 12 might be in K or 12 might not be in K.
Insufficient.
Statement 2: 3 is in K
Thus, -3 is in K.
Thus, 3 * -3 = -9 is in K.
Thus, -(-9) = 9 is in K.
Thus, 3*9 = 27 is in K.
Thus, -3*9 = -27 is in K.
Thus, K = {...-27, -9, -3, 3, 9, 27...}.
But we don't know what other values might be in K, so 12 might be in K or 12 might not be in K.
Insufficient.
Statements 1 and 2 combined:
Since both 4 and 3 are in K, 4*3 = 12 is in K.
Sufficient.
The correct answer is
C.
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