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I am not a student of core mathematics and hence as a result, all of my educational (engg.physics) background is based on the notion of applied mathematics rather than core mathematics, but I am attending a lecture in probability theory where the professor proved that the set of rational numbers is bijective to the set of natural numbers and hence the set of rational numbers $\mathbb{Q}$ is countable.

I am trying to prove it in a different way and would like the "core" mathematicians to check if I am making any logical fallacy in my arguments. Here it goes

Let me define $\mathbb{N}^2$ as $$\mathbb{N}^2=\left\{(x,y)|x,y\in\mathbb{N}\right\}$$ Hence, I claim that $\mathbb{N}^2$ is countable, since $f:\mathbb{N}\to\mathbb{N}^2$ is clearly bijective. Now, I define $$\mathbb{Q}=\left\{\frac{p}{q}|(p,q)\in\mathbb{N}^2\right\}$$ Since, $f:\mathbb{N}^2 \to \mathbb{Q}$ is clearly surjective and $\mathbb{N}^2$ is countable, that implies that $\mathbb{Q}$ is countable

Is this proof of mine logically correct? If not, please let me know if I have assumed something somewhere where I should not have assumed stuff. Your help is really appreciated.

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    $\begingroup$ You haven't defined the $f$'s? $\endgroup$
    – user1001001
    Commented Jul 25, 2020 at 20:01
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    $\begingroup$ As a general piece of advice, any time you set out to deepen your understanding of a mathematical theorem by giving your own rigorous proof, it's a good idea to avoid using the word "clearly." Stop instead and ask yourself just exactly what makes the thing you're saying so clear. In many cases (and this is one such case), the point that seems "clear" is the actual crux of the proof -- and sometimes isn't so clear after all! $\endgroup$
    – Barry Cipra
    Commented Jul 25, 2020 at 20:13
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    $\begingroup$ Let me add to my previous comment that your instinct to deepen your understanding by giving your own alternative proof is admirable and to be encouraged. Keep it up! $\endgroup$
    – Barry Cipra
    Commented Jul 25, 2020 at 20:20
  • $\begingroup$ Your $f: \mathbb{N}^2 \to \mathbb{Q}$ is not 'clearly surjective'. Rationals include negatives! $\endgroup$
    – layabout
    Commented Jul 25, 2020 at 20:47

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No, it is not correct, because:

  • you did not define any of the $f$'s;
  • the set $\left\{\frac pq\,\middle|\,(p,q)\in\Bbb N^2\right\}$ is not equal to $\Bbb Q$; it is equal to $\{q\in\Bbb Q\mid q>0\}$.
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  • $\begingroup$ ahaa....ok, I understand the second comment, since I don't take into account the negative rationals if $(p,q)\in\mathbb{N}^2$, but for the first comment, could you please elaborate how the definition of $f$'s could be done in either case in keeping with "core mathematical notations"?? $\endgroup$
    – Ghosal_C
    Commented Jul 25, 2020 at 20:08
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    $\begingroup$ If you want us to check the proof, then it's up to you to define the functions. $\endgroup$
    – José Carlos Santos
    Commented Jul 25, 2020 at 20:10
  • $\begingroup$ @ubuntu_noob Also, it's a bad idea to overload variable names: use $f$ for one function and, say, $g$ for the other. $\endgroup$
    – Noah Schweber
    Commented Jul 25, 2020 at 20:14
  • $\begingroup$ @NoahSchweber Could you help me by showing how the variable definition/mapping can be shown. I actually meant to index the values I got out of $\mathbb{Q}$ using $\mathbb{N}^2$ $\endgroup$
    – Ghosal_C
    Commented Jul 25, 2020 at 23:10
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    $\begingroup$ I don't even know what the question “Doesn't the mapping, $f\colon\Bbb N\longrightarrow\Bbb N^2$ define the function $f$?” means. If you want us to understand your work concerning a function $f$, then tell us how that function is defined. $\endgroup$
    – José Carlos Santos
    Commented Jul 26, 2020 at 6:59

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