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

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The discussion revolves around proving that the set \(S = \{k \in \mathbb{N} : k = mx + ny \text{ for some integers } x \text{ and } y\}\) is non-empty, given non-zero integers \(m\) and \(n\). A common misconception is that if \(x = y = 0\), then \(k = 0\), which is not a natural number, leading to the assumption that \(S\) could be empty. However, it is clarified that \(S\) includes all combinations of \(mx + ny\) for various integer values of \(x\) and \(y\), not just zero. Thus, there exist positive integers that can be formed by choosing appropriate values for \(x\) and \(y\), ensuring that \(S\) is indeed non-empty. The conversation also touches on the correct LaTeX formatting for the set of natural numbers, which is \(\mathbb{N}\).
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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.
 
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Re: not quite convinced of this proposition i need to prove...

skatenerd said:
...(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:

$$\mathbb{N}$$
 
Re: not quite convinced of this proposition i need to prove...

skatenerd said:
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.
 
So you're thinking that when it says "for some integers \(x\) and \(y\)" it means all different possible values?
 
skatenerd said:
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 $x$.
 
Ahhh okay thank you. That was very helpful.
 
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