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

[skate_nerd]

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]

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:

$$\mathbb{N}$$

 

[Evgeny.Makarov]

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.

 

[skate_nerd]

So you're thinking that when it says "for some integers \(x\) and \(y\)" it means all different possible values?

 

[Evgeny.Makarov]

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 $x$.

 

[skate_nerd]

Ahhh okay thank you. That was very helpful.