If I want to say that n ∈ all of the known number sets do I have to write n ∈ $\mathbb{N} , \mathbb{Z} , \mathbb{Q} , \mathbb{R} , \mathbb{C}$ or should I just leave it blank?
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Someone please help :) – OpenHax May 31 '16 at 17:33
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Since quaternions exists, $\mathbb C$ is not enough. – Peter May 31 '16 at 17:33
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2Define "all the known number sets". – DylanSp May 31 '16 at 17:34
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@Peter Do not forget octonions and sedenions – Laurent Duval May 31 '16 at 17:41
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The expression "all of the known number sets" is unlucky because for every element $n$ , we can chose a number set NOT containing $n$. We could for example choose the negative numbers or the irrational numbers or ... – Peter May 31 '16 at 17:46
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I am working on a math theorem and I just want to say that they can choose any number n and it will still work in my formulas. – OpenHax May 31 '16 at 18:00
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1@OpenHax I sincerely doubt you can plug in any number. What about a sedenion? A 2-adic? A surreal? You should state the theorem you are trying to prove precisely, and not shoot for too great (and, almost certainly, false) generality. – Noah Schweber May 31 '16 at 18:05
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Well I already tested it with N,Z,Q,R,C . I formulated wrongly when I said any number. But still, should I write that n ∈ N,Z,Q,R,C or should I leave it blank ? – OpenHax May 31 '16 at 18:31
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@OpenHax Do you mean a set that big that it contains every number we know, including p-adics,quaternions and so on ? – Peter May 31 '16 at 19:38
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No. Sorry for the confusion. – OpenHax Jun 05 '16 at 08:55
4 Answers
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If $n\in \Bbb{N}$, it follows that $n$ belongs to all of the other sets you listed since $\Bbb{N}$ is a subset of all of them.
$$\Bbb{N}\subseteq\Bbb{Z}\subseteq\Bbb{Q}\subseteq\Bbb{R}\subseteq\Bbb{C}$$
MathematicianByMistake
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1) $n ∈ H$
2) $for all H ∈ \{\mathbb{N} , \mathbb{Z} , \mathbb{Q} , \mathbb{R} , \mathbb{C} \}$
observe that in step 2, I didn't use the inclusion, but H is indeed an element of this "weird" set of sets. example:
Using step 2, since H belongs to this set of sets, then it must be for example that $H=\mathbb{Q}$.
Then by step 1 $n ∈ \mathbb{Q}$
Lex
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I believe $\mathbb{Z}_2$ is a number set as well, and smaller than $\mathbb{N}$.
And the set of imaginary numbers $\mathbb{I}$ has nothing in common with $\mathbb{N}$.
mvw
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