III says that ab < 0.ShruAj wrote:If a - b > a + b, where a and b are integers, which of the following must be true?
I. a < 0
II. b < 0
III. ab < 0
The answer is - [spoiler]only (II)[/spoiler]
But I could derive III as well by multiplying both the sides by a resulting in :
-ab > ab or 2ab <0 ; ab<0
Could anyone tell me where i went wrong?
This means that neither a nor b can equal 0.
It also means that either a > 0 and b < 0 or a < 0 and b > 0.
a > 0 and b < 0 is obviously possible, as we can see by plugging in a = 2 and b = -2
a - b = 4 > 0 = a + b
However, III does not have to be the case. a and b could both be less than 0, and a - b > a + b, would still work.
Let's use a = -2 and b = -4
a - b = 2 > -6 = a + b
Where you went wrong is in multiplying an inequality by a variable without knowing whether the variable represents a positive number or a negative number.
If a is negative, then when you multiply both sides by a, the inequality becomes reversed, as inequalities always do when multiplied by negative numbers.
So in multiplying by a you get two possibilities -ab > ab or -ab < ab.
