# Proof that $ax>b$ has solution in the integers [closed]

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: $$n=0,n\in \mathbb{N}, -n\in \mathbb{N}$$

• What have you tried? – TokenToucan Apr 7 at 3:11
• @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. – Tomás Pacheco Apr 7 at 3:13
• 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 $ax > b$. – TokenToucan Apr 7 at 3:15
• @TokenToucan Thats my problem because now I have to show $ab>b$ assuming $a,b$ to be positive – 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 $$a<0.$$

• 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? – Tomás Pacheco Apr 7 at 14:10