If we have that $A \leq B$ and that $C > B + \epsilon$, where $\epsilon>0$, does it immediately follow that $\mid A-C \mid > \epsilon$? If we have that $A \leq B$ and that $C > B + \epsilon$, where $\epsilon >0$, does it immediately follow that $\mid A-C \mid > \epsilon$?
I am a bit confused what is going on because it is quite obvious that if $A$ will always be $\epsilon$ away from $C$, that the magnitude will always be greater than $\epsilon$. However, since from $\mid A-C \mid > \epsilon$ we get that 
$$
A-C > \epsilon \ \ \ or \ \ \  A-C < -\epsilon
$$
then the second condition, $A-C < -\epsilon$, never holds. Am I perhaps misunderstanding something here? Thanks!
 A: You have $$A\le B<B+\epsilon<C,$$
hence $$A\le B<C-\epsilon<C.$$
Therefore, since $C>A$,
$$0<\epsilon<C-A=|C-A|.$$
A: Since you see that $A-C > \epsilon$ is clearly true, multiplying a (-1) on both sides of the inequality renders $C-A < -\epsilon$, so there is no problem. You can also see a derivation of this from TZakrevskiy's answer. Furthermore, the reason why $A-C$ is negative is because $C$ is greater than A, which is also clear with given information in the problem.
After convincing yourself that there is no problem with $A-C < -\epsilon < 0$, we could also write in conjunction with $|A-C|>\epsilon$ the following truth 
$$|A-C| = |(-1)(C-A)| = |(-1)| \cdot |C-A| = |C-A| > \epsilon$$
which just further insinuates my earlier point.
Now, in regards to logical statements consisting of the logical operator 'or', if we have a proposition $p=q \lor r$, and it turns out that one of the atoms $q$ or $r$ is false (not both), then $p$ is still true. So, even if it were false that $A-C < -\epsilon$, the fact that $A-C > \epsilon$ is true renders the proposition
$$A-C > \epsilon \, \, \, \mathrm{or} \, \, \, C-A < -\epsilon$$
true as well. Here is another example.
Suppose $x < 0$. Because this is true, we can also write $x \leq 0$ even though the case where $x=0$ is false. The fact that $x$ could be negative or equal to zero is true given the fact that x is negative.
