Show that for all $a,b \in \mathbb{Z}$ with $a \neq 0$ exists $x \in \mathbb{Z}$ such that $ax>b$. Where we define "$>$" as $x>y$ iff $x-y \in \mathbb{N}$.

This is in the context of ordered rings so we know there exists a set which we call $\mathbb{N}$ such that is closed under adition and product and for all $n \in \mathbb{Z}$ one and only one of the cases holds: $n=0,n\in \mathbb{N}, -n\in \mathbb{N}$

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    $\begingroup$ What have you tried? $\endgroup$ – TokenToucan Apr 7 at 3:11
  • $\begingroup$ @TokenToucan I haven´t tried anything meaningfull I tried let $x=b+1$ or $b$ or creating a function $f_a(x)=ax$ and work with that but that didn't work. I'm having problem in the assumptions for this problem, that is, what I can assume to true vs what I have to show to be true. $\endgroup$ – Tomás Pacheco Apr 7 at 3:13
  • $\begingroup$ Try letting $x=b$ or $x=-b$ depending on whether $a$ is positive or not. Given any $a$ and $b$ where $a\neq 0$, you are looking for $x$ that satisfies $ax > b$. $\endgroup$ – TokenToucan Apr 7 at 3:15
  • $\begingroup$ @TokenToucan Thats my problem because now I have to show $ab>b$ assuming $a,b$ to be positive $\endgroup$ – Tomás Pacheco Apr 7 at 3:17

For $a>0$:

Suppose that there is no $x \in \mathbb{Z}$ such that $ax>b$, that is, for all integer $x$ we have $b - xa \geq 0$. Let $S = \{b - xa: x \in \mathbb{Z}, x >0\}$ is formed by non negative integers. So there is $m = \min S$ (by well-ordering principle). Since $m \in S$, m has the form $m = b-na$, for some $n \in \mathbb{Z}$.

Now, take the element $m' = b - (n+1)a$, which is in $S$. But

$m' = b - (n+1)a = b - na - a < b - na = m = \min S$. It is a contradiction.

So, there is a number $x \in \mathbb{Z}$ such that $ax >b$.

Try the same way for $a<0.$

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  • $\begingroup$ Thank you very much! In the book, the well ordering principle actually comes later so to avoid using that can we make an argument such as: If we call $L$ the set of lower-bounds of $S$ note that $0 \in L$ so $L$ is non-empty and therefore there exists inf$S$. But we know that in the integers inf$S=$min$S$ (it was a previous exercise). Is this correct? $\endgroup$ – Tomás Pacheco Apr 7 at 14:10

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