1. The problem statement, all variables and given/known data Prove that the following statement is either true or false: For all real numbers y, there is a real number x such that y = x + ceiling[x]. Note: ceiling[x] = n such that n-1 < x ≤ n. For example, if x = 3.2, then ceiling[x] = 4. 2. Relevant equations 3. The attempt at a solution I think the statement is false. So I am going to prove its negation is true. Negation: There exists a real number y such that for all real numbers x, y ≠ x + ceiling[x]. Proof: Suppose x is a real number. Suppose y = 2x + 1. Then y is a real number because x is a real number. (I think I can do this because I just have to show that my statement holds for at least one y.) Now suppose that y = x + ceiling[x]. Define x1 = ceiling[x] - x. Then 0 ≤ x1 < 1. So ceiling[x] = x1 + x. Plug this into y = x + ceiling[x]. So y = x + x1 + x = 2x + x1 But by one of my assumptions, y= 2x + 1. Therefore, x1 = 1. However, x1 < 1. This is a contradiction. So y ≠ x + ceiling[x]. Negation of the statement is proved, so the original statement is true. -- Is this correct? I showed it to a friend who told me this doesn't prove the right thing. Apparently this only shows that y ≠ x + ceiling[x], and doesn't actually prove my original statement. I'm confused because I think it works. Help!