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Are the two algebraic structures $(\mathbb{N},\leq,+)$ and $(\mathbb{Q}_{\geq 0},\leq,+)$ isomorphic?

I read that they are, but to prove that, I need to find an isomorphism between $\mathbb{N}$ and $\mathbb{Q}$.

If it were between $\mathbb{N}$ and $\mathbb{Z}$ it would be easy. The main issue I have with it is, that the rational numbers don't have a "next" rational number like the natural numbers have and it seems very counterintuitive to me that these two sets have the same cardinality.

How should I start with this proof? And how do I find an explicit bijection between the natural numbers and the rational numbers?

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    The algebraic structures with ordering and addition are certainly not isomorphic (one is well-ordered, and generated by a single element, for instance). But $\Bbb N$ and $\Bbb Q$ do have the same number of elements. Which one of those properties are you actually asking about? – 2017-01-18
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    How do you get around the fact that $\mathbb Q_{\ge0}$ is dense - given $x,y\in\mathbb Q_{\ge0}$, there is an element in between them - i.e. a $z\in\mathbb Q_{\ge0}$ with $x – 2017-01-18
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    @Arthur: I am talking about the isomorphism – 2017-01-18
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    It's simple enough to construct a bijection between the two. But it won't play nice, neither with $+$ nor $\leq$, and this not be an isomorphism. – 2017-01-18
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    In your first sentence you are asking if the structures $(\mathbb N,\leq,+)$ and $({\mathbb Q}_{\geq 0},\leq,+)$ are isomorphic, but in the third paragraph you talk about the cardinality of the two sets $\mathbb N$ and $\mathbb Q_{\geq 0}$. Those are different questions: the structures are not isomorphic, but the sets have the same cardinality. – 2017-01-18
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    @Magdiragdag: Yes, I asked it because I falsely believed they were isomorphic – 2017-01-18

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The main issue I have with it is, that the rational numbers don't have a "next" rational number like the natural numbers

And consequently they are not isomorphic.

The property of having a "next" element can be defined in terms of $\leq$; you don't even need the addition operation:

$y$ is the successor of $x$ if and only if $x < y$ and $\forall z: \neg(x < z < y)$

If you're not familiar with how this sort of thing works, here's a sample theorem: let $\varphi$ be an isomorphism between ordered sets $(X, \leq)$ and $(Y, \preceq)$. Then:

$y$ is the successor of $x$ in $X$ if and only if $\varphi(y)$ is the successor of $\varphi(x)$ in $Y$

To prove the forward direction, suppose $y$ is the successor of $x$. Because $\varphi$ (and $\varphi^{-1}$) are order preserving, for all $z \in Y$, we have $\varphi(x) \prec z \prec \varphi(y)$ if and only if $x < \varphi^{-1}(z) < y$. Since the latter is never true, we conclude the former is never true.

The other direction is similar. (and also follows by symmetry, since $\varphi^{-1}$ is an isomorphism too)

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    So basically not having a "next" element breaks the isomorphism? Is that really all? – 2017-01-18
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    @de_dust For algebraic structures, isomorphic implies "Entirely indistinguishable as far as any algebraic property is concerned" – 2017-01-18
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No. They are not isomorphic as posets, since $(\mathbb Q_{\ge0}.\le)$ has the property that given $x,y\in\mathbb Q_{\ge0}$, there is an element in between them - i.e. a $z\in\mathbb Q_{\ge0}$ with $$x

However, $(\mathbb N,\le)$ does not have this property.

And they are not isomorphic as monoids either, since $(\mathbb N$,+) is cyclic - it is generated by $1$, whereas $(\mathbb Q_{\ge0},+)$ is not.

On the other hand $\mathbb Q_{\ge0}$ and $\mathbb N$ are isomorphic as sets, since they have the same cardinality.