How do I find an $\varepsilon > 0$ so that $x - \varepsilon > k$ implies $x > k$ ($k$ is a constant)?

Question

Say I have $x - \varepsilon > k$, and we know that $x > k$, and I want to find some positive term for $\varepsilon$ in terms of $x$ so that once I solve $x - \varepsilon > k$ for $x$, I get $x > k$. One such term is $\varepsilon = \frac{x - k}{2}$, so we have $x - (\frac{x - k}{2}) = \frac{x + k}{2} > k$ which implies that $x + k > 2k$ which means that $x > k$. How do I solve for $\varepsilon$ in order to obtain something like $\varepsilon = \frac{x - k}{2}$?

Answer 1

Add $\varepsilon-k$ to both sides of the inequality $x-\varepsilon>k$ to obtain $$x-k>\varepsilon\,,$$ i.e. any $\varepsilon$ will do it between $0$ and $x-k$.

Answer 2

Let $$ x - \varepsilon > k $$ where $\varepsilon > 0$.

Add $\varepsilon$ to each side: $$ x > k + \varepsilon. $$

But since $\varepsilon > 0$, we have $k + \varepsilon > k$. Therefore $$ x > k + \varepsilon > k, $$ or more simply, $$ x > k. $$

In short, any $\varepsilon$ such that $\varepsilon > 0$ will do.

Answer 3

You already solved for it.

$\epsilon =\frac{x-k}{2} $ will do fine.

What sort of problem do you have with this?