Help with algebraic manipulation of an inequality

[pzzldstudent]

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.

[sutupidmath]

ok , here it is what i think


since y>x=> y-x>0, now since 0<t<1, we have

0<t(y-x)<y-x add an x on both sides and we get

x<t(y-x)+x<y-x+x

x<ty-tx+x<y

x<ty+(1-t)x<y

what we actually need to show

[Dick]

If x<y then xt+x(1-t)<yt+x(1-t)<yt+y(1-t).

[pzzldstudent]

thanks for all the replies. i will try all these and reply back when i've gotten more work done on my own.

thank you very much!

[pzzldstudent]

ok , here it is what i think


since y>x=> y-x>0, now since 0<t<1, we have

0<t(y-x)<y-x add an x on both sides and we get

x<t(y-x)+x<y-x+x

x<ty-tx+x<y

x<ty+(1-t)x<y

what we actually need to show

Awesome! Thank you very much.:smile: That was very clear. I totally understood it and get it now! :cool: