Question

re state the following statement with appropriate conditions to make it true statement. " For every real number x, x2 > _ x ". (Where 2 is square of x and _is under the symbol >. )

Answer 1

Given statement      x² ≥ x
       x² - x ≥ 0
       (x - 0) (x - 1) ≥ 0

  If x is between 0 and 1 then LHS is negative.
   But for x ≤ 0, and x ≥ 1 the LHS is ≥ 0.

Hence,   x² ≥ x   for  x ≤ 0   and  for  x≥ 1