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This should follow from the fact that $x>y$ if and only if $x-y>0$.

Because we are saying there exists an $a \in S$ such that this holds true, my thought was to do a proof by contradiction (supposing it holds true for all $a$ and finding one example where it doesn't). However, I'm having a hard time negating this statement. Right now I have:

If $y>x$, then for all $a$ such that $a \le 0$, $x+a\ne y$.

Which seems to be true. I may be barking up the wrong tree here, help approaching this proof would be much appreciated.

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    $\begingroup$ Let $b=y-x$ (because you can subtract in a field) ... . What can you say about $b$ from (i) The algebraic axioms for a field (ii) the order properties of an ordered field ? $\endgroup$ – Mark Bennet Oct 12 '17 at 22:51
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    $\begingroup$ This is what I'm thinking. . . For $x, y \in S$, suppose $y>x$: that is, $y-x>0$. Let $b \in S$ be the element $b=y-x$. Then we know that $b>0$. Also, we can add $x$ to both sides to give us $b+x=y$, or $x+b=y$, which is what we were trying to show. I think that should hold up! $\endgroup$ – Ephraim Oct 13 '17 at 1:29
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For $x,y\in S$, suppose $y>x$: that is, $y−x>0$. Let $b \in S$ be the element $b=y−x$. Then we know that $b>0$. Also, we can add $x$ to both sides to give us $b+x=y$, or $x+b=y$, which is what we were trying to show.

(credit to Mark Bennet for pointing me in the right direction)

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