Prove that for all real numbers x there is a y n which case x<y

[Andrax]

Homework Statement

prove that
($\forall$x$\in$R) ($\exists$y$\in$R) : x < y

Homework Equations

The Attempt at a Solution

x < y
y= x+1
then x<x+1
which is correct but I'm kinda not sure of this answer...

 

[hedipaldi]

y=x+1 satisfies x<y so what wrong with it?

 

[Andrax]

y=x+1 satisfies x<y so what wrong with it?

just wanted to make sure someone actually confused me saying it doesn't