Given x < y:
(1-t)x < (1-t)y
tx < ty
x = tx + (1-t)x < ty + (1-t)x
y = ty + (1-t)y > ty + (1-t)x
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.
Thanks to everyone's help, I think I got it:
My final proof:
Let x, y be in the reals with x , y; and t in the reals with 0 < t < 1.
Since y > x, that implies y - x > 0. And since 0 < t < 1, we have
0 < t(y-x) < y - x. Add an x to both sides to get
x < t(y-x) + x < y - x + x
x < ty - tx + x < y
x < ty + (1 - t)x < y.
QED.