This question is from Kaplan
If x is a positive number, is x less than 1 ?
1) x > sqrt(x)
2) -sqrt(x) > -x
My thought process was :
considering statement 2, multiply both sides by negative 1. It gives the same statement as 1.
Now with statement 1, if x = 4, x > 2 which is x > sqrt(4) => x>1
if x = 1/4, then x > -1/2 which is x >sqrt(1\4) => x<1.
So my answer was both of the statements put together are insufficient.
's reasoning was there can't be any number x which is less than 1 and satisfies the condition x > sqrt(x). Obviously Kaplan
is assuming that x is a positive integer. But the question says "x is a positive number" which means it can be positive fractional number.
Could some one please clarify whether it is a valid assumption by Kaplan
? Am I missing missing something here ?