Can you represent a ceiling function with a floor function?
ceiling(x) = -floor(-x) Proof that this works: If x is an integer, it obviously works. If x isn't an integer, we can still split it into an integer n and a number d in the range (0,1), so x = n + d. (If d is 0, the claim is obviously true, so we can ignore that case.) Then floor(x) = n and ceiling(x) = n + 1. Now, -x = -n - d. Since -n - d is between -n - 1 and -n, floor(-x) = -n - 1. So -floor(-x) = n + 1 = ceiling(x).
floor(x) = x^2/20 - 5 20*floor(x) = x^2 - a hundred 20*floor(x) + a hundred = x^2 The left side is an integer. for this reason, x^2 must be an integer. hence, x is an integer besides, so floor(x) = x. the unique equation now's such as x = x^2/20 - 5. fixing for x yields x = 10.