# Real Analysis Real Number Inequality Proof

Prove:

For all $x,y$ in the real numbers [$(x>y)$ implies (there exists an e in the positive real numbers $x \ge y+e$)]

I can't see which of the real number axioms of addition, multiplication and order will help me prove this theorem.

note: this statement was deduced using the logical equivalence of the contrapositive so if there are any problems with the statement itself let me know.

If $x>y$ then $x-y>y-y=0$, so setting $e=x-y$ we get $$x=y+(x-y)=y+e$$ with $e>0$.