Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
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 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 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 at 21:37
    
I think the translation makes no sense because the mathematical statement itself makes no sense. –  Newb Jan 31 at 21:39
    
@beginnerprogrammer regardless, I think your interpretation/translation came out priceless! :) –  Dror Jan 31 at 21:43

3 Answers 3

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 $k$.

share|improve this answer

@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"

share|improve this answer
    
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 at 22:44
    
@MPW your translation here is a serious candidate for my about me. –  Dror Jan 31 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 $k$."

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.