# Thread: [SOLVED] Help with algebraic manipulation of an inequality

1. ## [SOLVED] Help with algebraic manipulation of an inequality

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.

2. 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

3. ## Clarification

Given x < y:

(1-t)x < (1-t)y -- I see here you just multiplied both sides by (1 - t)

tx < ty -- Multiplying through I get x - tx < y - ty. I don't 'see' how you got to this part tx < ty.

4. Originally Posted by ilikedmath
Given x < y:

(1-t)x < (1-t)y -- I see here you just multiplied both sides by (1 - t)

tx < ty -- Multiplying through I get x - tx < y - ty. I don't 'see' how you got to this part tx < ty.
He's used the property that $\displaystyle x<y$ and multiplied through by $\displaystyle t$.

5. ## I think I got it! :)

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.

6. I like that proof. Very elegant, given the scope of the problem.