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There is a justification in my book that says :

Since $\mathbb Q$ contains $\mathbb N$, $\mathbb Q$ must be a denumerable set.

Why is this true?

Lucas Alanis
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    I don't know how true that statement is.. $\Bbb R $ contains $\Bbb N$ doesn't mean $\Bbb R $ is denumerable.. But sure $\Bbb Q$ is denumerable.. Try http://math.stackexchange.com/questions/12167/the-set-of-rationals-has-the-same-cardinality-as-the-set-of-integers and http://www.people.vcu.edu/~rhammack/BookOfProof/Cardinality.pdf – Ishfaaq Feb 05 '14 at 03:15
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    Which book is this mate? Maybe a title and a page number could get you a better answer. I suspect this was taken out of context. – Ishfaaq Feb 05 '14 at 03:27
  • My wild guess, without the book in front of me, is that the author has defined "denumerable" to mean "having the same cardinality as ${\mathbb N}$'', that he has already proved that the cardinality of ${\mathbb Q}$ is at most that of ${\mathbb N}$ and is therefore reduced to proving that ${\mathbb Q}$ has cardinality at least that of ${\mathbb N}$, for which this inclusion suffices. – WillO Feb 05 '14 at 03:35

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The quote above is incorrect. $\mathbb{Q}$ is denumerable, but not because it contains $\mathbb{N}$. Being denumerable means that a set is not too large; how can we get that information by knowing the size of a subset? As Ishfaaq pointed out in the comments, $\mathbb{R}$ is not denumerable but also contains $\mathbb{N}$.

A possible correction: $\mathbb{Q}$ can be thought of as a subset of $\mathbb{N} \times \mathbb{N}$, and since $\mathbb{N} \times \mathbb{N}$ is denumerable, so is $\mathbb{Q}$.

Elchanan Solomon
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    Or, if "it" refers to $\mathbb{N}$, and if $\mathbb{Q}$ has been shown to be denumerable, then the statement is true. I'm really reaching here... – MPW Feb 05 '14 at 03:25