Let x and y be real numbers. Prove that if x≤y+ϵ for every positive real number ϵ, then x≤y. Why do we set ϵ = 1/2(x−y)?

Question

I understand in the solution that we are finding the contrapositive of $x ≤ y$ which is $x > y$ but why do we set set ϵ= $1\over2$(x−y) ? Where does the $1\over2$ come in and why do we subtract $y$ from $x$?

Answer 1

Well, you assume that $x>y$ and then need to find an $\epsilon$ such that $x\leq y+\epsilon$ does not hold. The given $\epsilon$ is such that $y+\epsilon$ is right in the middle between $y$ and $x$. (Any other $\epsilon = c(x-y)$ with $0<c<1$ would work, too.)

Answer 2

Suppose that $x>y$. Then $$x-y>\frac{x-y}{2}>0$$ If we let $\epsilon=\frac{x-y}{2}$ then we can rewrite this as $$x-y>\epsilon > 0$$ or equivalently, $$x>y+\epsilon>y>0$$ In particular, we found a positive $\epsilon$ such that $x>y+\epsilon$ i.e. $x\leq y+\epsilon$ doesn't hold.