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So I saw that $ \mathbb{Q}\sim\mathbb{N} $ and I tried to prove it.

The official proof was a function using prime factors.

I'm learning and in doing so I try to prove each theorem and corollary or example before I read the one in the script or book. But mine was a little funky as I tried long and hard to think of something, but unfortunately couldn't come up with anything better.

Would this work:

Every number in $ \mathbb{Q}$ can be interpreted as two integers $a$ and $b$. Then we can devise a function $f: \mathbb{Q} \rightarrow \mathbb{N} $ where $a$ is the first digit and $b$ is appended.

For example $f(5/4) = 54$ or $f(1/1) = 11$.

Is such a function even allowed? Would it be injective and bijective?

Also, if this is an allowed function. Would $0/4$ and $0/1$ count as two different rational numbers or the same? ( I was struggling finding a definition for the zero numbers if they count as different).

Sorry if this is complete nonsense!

Shaun
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oliver
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    $\frac{11}1$, $\frac{1}{11}$. – WhatsUp Dec 27 '19 at 16:18
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    $1/1=2/2$ but $f(1/1)=11$ and $f(2/2)=22$ so this isn't even a function, however you could define it that $a, b$ have to be coprime that would solve the issue and say that $f(0)=0$ and ignore the case $0/b$ – kingW3 Dec 27 '19 at 16:22
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    This is neither injective not surjective, assuming you fix it as kingW3 suggests. Then 22 is not in the image, and f(54/11)=f(541/1)=5411$. – jgon Dec 27 '19 at 16:26
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    Finding a bijection is indeed hard as none of a+b, ab, a/b, a^n, are bijective, I think you can't make a bijective function with elementary operations but thinking instead of numbering rational numbers might be easier to think about. – kingW3 Dec 27 '19 at 16:45
  • thank you all, you quickly came up with so many examples, I feel silly that I tried many myself and couldn't find similar ones! thanks again! – oliver Dec 27 '19 at 16:46
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    Use $\sim$ for $\sim$. – Shaun Dec 27 '19 at 18:16

3 Answers3

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As you probably know by now, your $f$ is not injective (if it were a function), but most importantly, it is actually not a function as it is not well defined.

That is, as you pointed out, $0=\frac{0}{1}=\frac{0}{4}=\frac{0}{n}$. Thus, $f(0)$ could actually be any integer, so $f$ is not even a function.

If you want a explicit bijection you can have a look here. It shows a bijection from rationals to naturals, but it is easy to modify it to get integers of course.

J. W. Tanner
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TheDoyduro
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  • Thank you for taking the time and the link, I'll check it out, it's a different proof from the one in my book, but it uses one I've used before (Z to N). That's super cool! – oliver Dec 27 '19 at 16:47
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I like Cantor's pairing function. You can look it up and get a formula. But it is utterly trivial from a geometric point of view. Roughly speaking, arrange the rationals in an array, and then just zig-zag your way from the upper left hand corner to put them in a list.

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Here is another route. It is quite common when trying to prove that two sets $A$ and $B$ have the same cardinality that you can fairly easily find a function $f:A \rightarrow B$ which is an injection but not a surjection and another $g:B \rightarrow A$ which is also an injection but not a surjection.

Intuitively, this says that $|A| \le |B|$ and $|B| \le |A|$ so $|A| = |B|$.

The Schröder–Bernstein theorem deals with this case: you can deduce that the sets have the same cardinality. It is worth studying that and then you can stop once you find such a pair of injections.

Even in a fairly simple case such as yours, an explicit bijection can be rather messy.

badjohn
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    Thank you, in my script this was mentioned but reduced to a simple line and not further commented. I seen it more detailed in How to prove it's new edition. I will check it out. Most proofs I've seen only show an injection and not a surjection. – oliver Dec 28 '19 at 06:59