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Is there an injective function from the set of natural numbers N to the set of rational numbers Q, and viceversa.

What are these two functions if they exist?

Please I would appreciate easy examples just using set theory without using cardinality or other complex notation.

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    Injection from one to the other is obvious. – 2017-02-06
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    @Wojowu If it was obvious to me I would have not posted the question. – 2017-02-06
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    @JJAb Remember that $\mathbb{N}\subset\mathbb{Q}$: every natural number is rational. If $A$ is a subset of $B$, do you see an injection from $A$ into $B$? (HINT: what's the easiest way to take in an element of $A$ and spit out an element of $B$, if every element of $A$ is an element of $B$?) – 2017-02-06
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    @JJAb I think Wojowu means the $\mathbb{N}\to\mathbb{Q}$ is obvious. – 2017-02-06
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    @NoahSchweber is it $f: n -> n $ ? – 2017-02-06
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    @JJAb Yup, exactly! The other one - finding an injection from $\mathbb{Q}$ to $\mathbb{N}$ - is going to be trickier, since *not* every rational is a natural number! So in this case one has to do some tricky thing, see e.g. Especially Lime's answer below. – 2017-02-06
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    @NoahSchweber thank you – 2017-02-06

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An injection from the naturals to the rationals is just the identity function (every natural is a rational).

For an injection from the rationals to the naturals, do the following. If $x\in\mathbb Q$ then $x=p/q$ for some $p,q$ with no common factor, $p\in\mathbb Z$ and $q\in\mathbb Z+$, and these values of $p$ and $q$ are uniquely determined. Write $f(x)=2^p\times 3^q$ if $p\geq 0$ and $f(x)=2^{-p}\times 3^q\times 5$ if $p<0$. This is an injection from $\mathbb Q$ to $\mathbb N$.

It is possible to define a bijection between the two, but it is more fiddly to do so.

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    I had a question, what do you mean when you say p and q are uniquely determined? – 2017-02-06
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    I meant that once you choose $x\in Q$ there's only one possible way of writing it as $p/q$ that satisfies those conditions (i.e. lowest terms with $q>0$). So what I've written does actually define a function (in other words if we both start from the same $x$ and apply this definition we'll both get the same value of $f(x)$). – 2017-02-06