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

skatenerd

Active member
Oct 3, 2012
114
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 30, 2012
2,492
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
Oct 3, 2012
114
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
Jan 30, 2012
2,492
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
Oct 3, 2012
114
Ahhh okay thank you. That was very helpful.