If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...
(1) ab = 2. First of all this means that a and b are of the same sign.
If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.
If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.
We have that the answer to the question is NO. Sufficient.
(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.
Answer: A.
THIS QUESTION IS DISCUSSED HERE:
new-set-number-properties-149775-60.html#p1205389 I think Bunuel you missed something for statement 2 what if a=1 and b=2 in this case ab =2 and [a] + [b] =3