# Let $a,b\in\Bbb{N}.$ Show that there is a minimum natural number of the form $a-b m,$ where $m \in \Bbb{Z}$.

Let $$a,b\in\Bbb{N}.$$ Show that there is a minimum natural number of the form $$a-b m,$$ where $$m \in \Bbb{Z}$$.

If I define the set $$T=\{a-bm \in \Bbb N \mid a,b \in \mathbb{N}, m \in\mathbb{Z}\}$$, it's clear that $$T\neq\emptyset$$ because $$1\in T$$ and $$T\subset\mathbb{N}$$. Then by Well-ordering theorem $$T$$ has a minimum.

Is this right or I have to do something more?

• This is not right. Why do you think $1$ must be in $T$? What if $a=-3$ and $b=6$? I suggest you write out what $T$ looks like for several examples. Then you can see how to use the well ordering theorem. Commented Dec 17, 2020 at 22:32
• @EthanBolker $a\in\mathbb{N}$ and we can write $1= 1-b*0=$ in particular if we take $m=0$ T is containing all natural numbers Commented Dec 17, 2020 at 22:37
• If $a=15$ and $b=6$ then whatever the value of $m$, $a-mb$ will be divisible by $3$ so will never be $1$. Commented Dec 17, 2020 at 22:40
• @EthanBolker and if we define $T=\{a-bm \in \Bbb N \mid a,b \in \mathbb{N}, m \in\mathbb{Z}\}$ but only taking the natural numbers in the form a-bm Commented Dec 17, 2020 at 23:03
• See the answer from @ACheca Commented Dec 17, 2020 at 23:32

Reading your answer and your comment I think you're misunderstanding an important part of the problem: when you say let $$a,b\in \mathbb{N}$$ you're fixing these values. For the rest of the problem, $$a,b$$ are fixed. So the definition of $$T$$, which is dependent on $$a,b$$ should actually be:
$$T_{a,b}= \{ a-bm\in \mathbb{N} | m\in \mathbb{Z}\}$$
Note that in your definition of $$T$$, it seems that $$a,b$$ could be variable, and I think that was what made you think $$1\in T$$, when this is false in general. In the expression $$a-bm$$ everything is fixed except $$m$$.
With this in mind, try again the same argument you were trying before, having in mind that in general $$1$$ doesn't have to be in $$T$$.