# Not quite convinced of this proposition i need to prove...

#### skatenerd

##### Active member
I am given that $$m$$ and $$n$$ are non zero integers and that $$S=\{k\in N:\,k=mx+ny$$ for some integers $$x$$ and $$y\}$$. (Side note: How do you type the double struck capital N for the set of natural numbers in LaTeX?)
I need to prove that $$S$$ is non-empty.

Well I am kind of stuck here thinking that if you had $$x=y=0$$ then $$k=0$$ making $$k$$ not defined in the natural numbers, therefore leaving $$S$$ to be none other than the empty set.

#### MarkFL

Staff member
Re: not quite convinced of this proposition i need to prove...

...(Side note: How do you type the double struck capital N for the set of natural numbers in LaTeX?)...
Use the code:

\mathbb{N}

to get:

$$\displaystyle \mathbb{N}$$

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Re: not quite convinced of this proposition i need to prove...

I am given that $$m$$ and $$n$$ are non zero integers and that $$S=\{k\in N:\,k=mx+ny$$ for some integers $$x$$ and $$y\}$$. I need to prove that $$S$$ is non-empty.

Well I am kind of stuck here thinking that if you had $$x=y=0$$ then $$k=0$$ making $$k$$ not defined in the natural numbers, therefore leaving $$S$$ to be none other than the empty set.
For given $m$ and $n$, the set $S$ is formed by numbers of the form $mx+ny$ for all possible integer $x$ and $y$, not just $x=y=0$. Even if $0\notin\mathbb{N}$ (which differs according to different conventions). there are other $x$ and $y$ that produce positive integers.

#### skatenerd

##### Active member
So you're thinking that when it says "for some integers $$x$$ and $$y$$" it means all different possible values?

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
So you're thinking that when it says "for some integers $$x$$ and $$y$$" it means all different possible values?
Yes, it may seem a little counterintuitive. The phrase
$k\in S\iff k=mx+ny\text{ for some integers }x\text{ and }y$
means
$k\in S\iff(\exists x,y,\,k=mx+ny)$
In particular,
$(\exists x,y,\,k=mx+ny) \implies k\in S$
which is logically equivalent to
$\forall x,y,\,(k=mx+ny\implies k\in S)$
or
$\forall x,y,\,mx+ny\in S.$
So, if we say that some object $x$ is interesting iff $P(x,y)$ holds for some $y$, it means that every pair $(x,y)$ satisfying $P$ gives rise to an interesting $x$.

#### skatenerd

##### Active member
Ahhh okay thank you. That was very helpful.