Why does $a>b$ imply $(a-b)/2 > 0$ Sorry if this is an easy explanation but I'm having trouble understanding why $a > b$ implies that $(a-b)/2 > 0$. I'm trying to understand this theorem and its proof:
$a<b+\epsilon$ for all $ϵ>0$ implies $a\leq b$
I understand that the best way to go about it is to prove by contradiction. So you would assume $a > b$. From the other proofs I've seen it says that $a>b$ implies $(a-b)/2 > 0$.
I tried reorganizing it, but don't understand where the divide by $2$ comes from.
Sorry if this has been asked before or if I'm missing something simple.
 A: Let consider the following chain of implications
$$a>b \implies a-b>b-b \implies a-b>0 \implies \frac{a-b}2>\frac 02 \implies \frac{a-b}2>0$$
that is
$$a>b \implies \frac{a-b}2>0$$
Note also that in the same way we can prove that the reverse is also true $\frac{a-b}2>0 \implies a>b$ that is
$$a>b \iff \frac{a-b}2>0$$
Moreover, as noticed in the comments, for any $n>0$ we can show in the same way that
$$a>b \iff \frac{a-b}n>0$$

The proof for the particular theorem you are referring to:

*

*$a<b+\epsilon$ for all $ϵ>0$ implies $a\leq b$
should be as follows:

*

*let assume by contradiction $a>b$

*then for any $n>0$ we have $\frac{a-b}n>0$ and therefore exists $\varepsilon>0$ such that

$$0<\frac{\varepsilon}n\le\frac{a-b}n \implies a-b\ge \varepsilon \implies a\ge b+\varepsilon$$

As an alternative, we can proceed by a direct proof by exhaustion, that is:

*

*$a=b \implies a-b=0<\varepsilon \implies a-b<\varepsilon  $, $\forall \varepsilon>0$


*$a<b \implies a-b<0 <\varepsilon \implies a-b<\varepsilon  $, $\forall \varepsilon>0$


*$a>b$ let assume $\varepsilon =\frac{a-b}2>0$ then $a-b>\varepsilon$
therefore
$$a<b+\varepsilon, \:  \forall \varepsilon>0 \implies a=b \lor a<b$$
