Let R denote the set of all real numbers and Q the set of all rational numbers. Statement to prove: If x and y are in R with x < y, show that x < ty + (1-t)x < y for all t, 0 < t < 1. My work on the proof so far: Given x and y are real numbers with x < y. By theorem we know there exists an r in Q such that x < r < y. Take r = t. So x < t < y. That's all I have so far. My professor said this proof was more of algebraic manipulation. I am stuck as to how I can algebraically manipulate the inequality to get to x < ty + (1-t)x < y.