A numeric set \(M\) containing 2003 distinct numbers is such that for every two distinct elements \(a, b\) in \(M\), the number \(a^2+ b\sqrt 2\) is rational. Prove that for any \(a\) in \(M\) the number \(q\sqrt 2\) is rational.
Solve the inequality: \(\lfloor x\rfloor \times \{x\} < x - 1\).