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I am working on homework right now, and I am not sure of how to solve this problem. I am not sure of how to come up with the translation. Any help is greatly appreciated. This is the problem:

Write this in English: ∀k ∈ 3Z ,∃S ⊆ N ,|S | = k . (Is it true?) What is the negation of this statement? (Is the negation true?)

I have made an attempt and think that it says, "For every integer, there exists 3 integers such that S is a subset of N, and |S| is equal to k."

I am not sure if this is correct or not.

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Your translation doesn't make any sense: What do the three integers have to do with $S$ at all? For starters, $3\mathbb{Z}$ is generally defined to be $\{3n : n \in \mathbb{Z}\}$, the set of all multiples of $3$. – user61527 Jan 31 '14 at 21:29
I know it doesn't make sense. I am trying to figure out how to make sense of this. I was trying to follow an example that is somewhat similar in my book, but that is not working out correctly – beginnerprogrammer Jan 31 '14 at 21:32
Perhaps if you gave more context around this mysterious string, it would be helpful :). Also, are you sure you copied it correctly? – Dror Jan 31 '14 at 21:37
I think the translation makes no sense because the mathematical statement itself makes no sense. – Newb Jan 31 '14 at 21:39
@beginnerprogrammer regardless, I think your interpretation/translation came out priceless! :) – Dror Jan 31 '14 at 21:43

Let's take each part individually:

  • $\forall k \in 3 \mathbb{Z}$: By definition, $3\mathbb{Z} = \{3n : n \in \mathbb{Z}\} = \{..., -6, -3, 0, 3, 6, ...\}$ is the set of all multiples of $3$.

  • $\exists S \subseteq \mathbb{N}$: There exists a subset $S$ of $\mathbb{N}$, with properties to be specified in

  • $|S| = k$: The cardinality (number of elements of $S$) is equal to $k$.

Putting it all together, the translation is:

For every integer $k$ which is a multiple of $3$, there is a subset of $\mathbb{N}$ whose cardinality is equal to $k$.

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@T.Bongers already translated the profoundly boring statement, so I'm not going to.
Regarding the negation:
You can see the statement is false (Can the number of elements in a set be a negative number, like $-3$ for instance?), so It's negation being false would lead to a contradiction, so It's true and it reads:

"There exists an integer multiple of 3 ($k$), for which all subsets of $\mathbb{N}$ have cardinality different from k"

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In the spirit of your answer, the negation of my example would be "$\exists\text{(foo)},\forall\text{(bar)}, \text{(not blah)}$", which reads as "There is a foo such that not blah for every bar." – MPW Jan 31 '14 at 22:44
@MPW your translation here is a serious candidate for my about me. – Dror Jan 31 '14 at 22:57

For statements of the form "$\forall \text{(foo)},\exists \text{(bar)}, \text{(blah)}$" you can say "For every foo there is a bar such that blah". So in your case, maybe something like

"For every integral multiple $k$ of $3$ there is a set of natural numbers whose size is $k$."

You have misinterpreted the pieces of the statement, I believe. "$k\in 3\mathbb{Z}$" means $k$ is in the set $3\mathbb{Z}=\{\ldots -12, -9, -6, -3, 0, 3, 6, 9, 12, \ldots \}$. So the first part of the statement is "For every $k$ in $3\mathbb{Z}$", which can be said more plainly as "For every integral multiple of $3$".

The second part is "There is a subset $S$ of $\mathbb{N}$ such that $|S|$ is $k$." Said more plainly, this is "There is a set of natural numbers of size $k$."

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