Let x and y be real numbers such that x<y. There exists z that is a real number such that x<z<y.
The Attempt at a Solution
I wrote the following: We are given that x<y. We know from previous proof that (1/2)<1.
Consider y such that y=x+1
x+1<x and we know (1/2)<1. Therefore x+(1/2)<x+1
Also, x<x+(1/2). Therefore x<x+(1/2)<x+1 which shows the existence of a z such that x<z<y.
For this problem my professor marked me wrong at the part where I said let y=x+1. He just draws a red line and leaves it up to us to figure what we did wrong. I do not see how this argument is invalid though. Can someone explain?