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I have the following question:

Let $A=\{x:\exists m,n \in\mathbb{Z} \text{ such that } x=m+n\sqrt{p}\}$, where $p\in\mathbb{Z}$ is a fixed prime. Show that $A$ is denumerable.

I hope someone can solve this question.


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what does $p$ got to do with it? – Mud Apr 25 '13 at 23:57
@Mud: I suspect the OP means $m+n\sqrt{p}$. Anyway, what have you tried? – Clive Newstead Apr 25 '13 at 23:57
Check what you wrote: where does the prime $\,p\,$ appear in the definition of $\,A\,$ ...? – DonAntonio Apr 25 '13 at 23:58
tags are confusing me... is this a real analysis question? – user67133 Apr 26 '13 at 0:05
yes it is from real analysis about countable set – leena adam Apr 26 '13 at 0:07

2 Answers 2

up vote 2 down vote accepted

We define the map: $$f\colon \mathbb{Z}\times\mathbb{Z}\to A,\qquad (m,n)\to m+n\sqrt{p}$$ then $f$ is clearly surjective and since $\mathbb{Z}\times\mathbb{Z}$ is denumerable then $A$ is also denumerable.

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please explain more – leena adam Apr 26 '13 at 0:11
It's a theorem. If $X$ is denumerable and exists $f:X\rightarrow Y$ surjective, then $Y$ is denumerable. – Integral Apr 26 '13 at 0:12
Is "denumerable" the right word? I always thought it was "enumerable". – Integral Apr 26 '13 at 0:14
For all $x\in A$ there's by definition of the set $A$ a pair $(m,n)\in\mathbb{Z}^2$ s.t $f(m,n)=x$ and then the cardinality of $A$ is ''less than'' the cardinality of $\mathbb{Z}^2$ which is denumerable. – user63181 Apr 26 '13 at 0:15
@Integral I don't know if I can say this in english: $A$ is at most denumerable but this definition exists in the french jargon ''au plus dénombrable'' and it means $A$ is finite or denumerable – user63181 Apr 26 '13 at 0:20

Notice that $A= \bigcup\limits_{m \in \mathbb{Z}} \bigcup\limits_{n \in \mathbb{Z}} \{m+n\sqrt{p} \}$ is a countable union of singletons.

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thank you very much – leena adam Apr 28 '13 at 21:38

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