-
willbeatthegmat
- Senior | Next Rank: 100 Posts
- Posts: 71
- Joined: Sat Sep 20, 2008 5:48 am
- Thanked: 5 times
Statement 2 suggests the possibility that ab = 0, which is helpful if we decide to choose numbers. If a = -2, b = 0, then |a| > |b|, a < 0, and ab >= 0, --these numbers agree with the facts given -- and we find the answer to the question is 'no'. If, on the other hand, a = -2, and b = -1, we find the answer is 'yes'. So E.
More abstractly, using both statements, we know that a < 0, and from statement 2, that b <=0. This means that:
|a| = -a
|b| = -b
The question asks whether a|b| < a-b. If a and b are negative (or zero), this is equivalent to asking whether -ab < a - b, or rearranging:
-ab < a - b
0 < a +ab - b
b < a(b+1)
So, we want to know if this is always true when a < b <= 0. Notice that if b < -1, the left side of this inequality is negative, while the right side is positive (it is the product of two negatives), so the inequality will certainly be true. If b = -1, the inequality is also true, so if b <= -1, the answer is 'yes'. However, b can be zero, and if we plug in b=0, the inequality is false, since a is negative. So the answer to the question can be 'yes', but can also be 'no' *only in the case where b is equal to zero*. Hence E.
