bijection between $mathbb{Q}$ and $mathbb{N}$ that preserve the order.












2












$begingroup$


I know there is a bijection between $mathbb{Q}$ and $mathbb{N}$. But is there a bijection $mathbb{Q}xrightarrow{f}mathbb{N}$ that preserves the order? Intuitively I think this is not possible. What I would think of is enumerating the rationals in a form ${q_n}_{n in mathbb{N}}$ where $q_n < q_m$ if $n<m$ (I think this is not possible but why?) and then map to elements of $mathbb{N}$.






rational-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 24 at 16:36








  • 1




    $begingroup$
    Your question is equivalent to this one, since defining an order-preserving bijection $mathbb{Q} to mathbb{N}$ is equivalent to listing the rational numbers in increasing order.
    $endgroup$
    – Clive Newstead
    Jan 24 at 16:37










  • $begingroup$
    Compare also with this question here.
    $endgroup$
    – Dietrich Burde
    Jan 24 at 16:38










  • $begingroup$
    @CliveNewstead Thanks, that was actually easyer than I thought.
    $endgroup$
    – roi_saumon
    Jan 24 at 16:46
















2












$begingroup$


I know there is a bijection between $mathbb{Q}$ and $mathbb{N}$. But is there a bijection $mathbb{Q}xrightarrow{f}mathbb{N}$ that preserves the order? Intuitively I think this is not possible. What I would think of is enumerating the rationals in a form ${q_n}_{n in mathbb{N}}$ where $q_n < q_m$ if $n<m$ (I think this is not possible but why?) and then map to elements of $mathbb{N}$.






rational-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 24 at 16:36








  • 1




    $begingroup$
    Your question is equivalent to this one, since defining an order-preserving bijection $mathbb{Q} to mathbb{N}$ is equivalent to listing the rational numbers in increasing order.
    $endgroup$
    – Clive Newstead
    Jan 24 at 16:37










  • $begingroup$
    Compare also with this question here.
    $endgroup$
    – Dietrich Burde
    Jan 24 at 16:38










  • $begingroup$
    @CliveNewstead Thanks, that was actually easyer than I thought.
    $endgroup$
    – roi_saumon
    Jan 24 at 16:46














2












2








2





$begingroup$


I know there is a bijection between $mathbb{Q}$ and $mathbb{N}$. But is there a bijection $mathbb{Q}xrightarrow{f}mathbb{N}$ that preserves the order? Intuitively I think this is not possible. What I would think of is enumerating the rationals in a form ${q_n}_{n in mathbb{N}}$ where $q_n < q_m$ if $n<m$ (I think this is not possible but why?) and then map to elements of $mathbb{N}$.






rational-numbers







share|cite|improve this question









$endgroup$




I know there is a bijection between $mathbb{Q}$ and $mathbb{N}$. But is there a bijection $mathbb{Q}xrightarrow{f}mathbb{N}$ that preserves the order? Intuitively I think this is not possible. What I would think of is enumerating the rationals in a form ${q_n}_{n in mathbb{N}}$ where $q_n < q_m$ if $n<m$ (I think this is not possible but why?) and then map to elements of $mathbb{N}$.






rational-numbers




rational-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 24 at 16:32









roi_saumonroi_saumon

59438




59438












  • $begingroup$
    And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 24 at 16:36








  • 1




    $begingroup$
    Your question is equivalent to this one, since defining an order-preserving bijection $mathbb{Q} to mathbb{N}$ is equivalent to listing the rational numbers in increasing order.
    $endgroup$
    – Clive Newstead
    Jan 24 at 16:37










  • $begingroup$
    Compare also with this question here.
    $endgroup$
    – Dietrich Burde
    Jan 24 at 16:38










  • $begingroup$
    @CliveNewstead Thanks, that was actually easyer than I thought.
    $endgroup$
    – roi_saumon
    Jan 24 at 16:46


















  • $begingroup$
    And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 24 at 16:36








  • 1




    $begingroup$
    Your question is equivalent to this one, since defining an order-preserving bijection $mathbb{Q} to mathbb{N}$ is equivalent to listing the rational numbers in increasing order.
    $endgroup$
    – Clive Newstead
    Jan 24 at 16:37










  • $begingroup$
    Compare also with this question here.
    $endgroup$
    – Dietrich Burde
    Jan 24 at 16:38










  • $begingroup$
    @CliveNewstead Thanks, that was actually easyer than I thought.
    $endgroup$
    – roi_saumon
    Jan 24 at 16:46
















$begingroup$
And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:36






$begingroup$
And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:36






1




1




$begingroup$
Your question is equivalent to this one, since defining an order-preserving bijection $mathbb{Q} to mathbb{N}$ is equivalent to listing the rational numbers in increasing order.
$endgroup$
– Clive Newstead
Jan 24 at 16:37




$begingroup$
Your question is equivalent to this one, since defining an order-preserving bijection $mathbb{Q} to mathbb{N}$ is equivalent to listing the rational numbers in increasing order.
$endgroup$
– Clive Newstead
Jan 24 at 16:37












$begingroup$
Compare also with this question here.
$endgroup$
– Dietrich Burde
Jan 24 at 16:38




$begingroup$
Compare also with this question here.
$endgroup$
– Dietrich Burde
Jan 24 at 16:38












$begingroup$
@CliveNewstead Thanks, that was actually easyer than I thought.
$endgroup$
– roi_saumon
Jan 24 at 16:46




$begingroup$
@CliveNewstead Thanks, that was actually easyer than I thought.
$endgroup$
– roi_saumon
Jan 24 at 16:46










2 Answers
2






active

oldest

votes


















1












$begingroup$

Let $n in mathbb{N}$ and $q_n,q_{n+1} in mathbb{Q}$ where $q_n < q_{n+1}$.



Assume there does not exist another rational number $q_k$ where $q_n<q_k<q_{n+1}$ (i.e., we can order the rationals). Since $mathbb{Q}subset mathbb{R}$, this implies that the rationals are not dense in the reals, which is false. Therefore, our assumption is false.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The reson is that if you have natural number n then you can find the NEXT number (=n+1), but if you have Rational number q then you cannot find NEXT number - because between any two different rational numbers is infinite number of other rational numbers.



    If such bijection $f$ exists then when you find $q=f(n)$ then you shoud be able to find next rational number $p=f(n+1)$ but because $qneq p$ then between $q$ and $p$ exist infinite number of other rational numbers e.g. $r=(p+q)/2$






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Let $n in mathbb{N}$ and $q_n,q_{n+1} in mathbb{Q}$ where $q_n < q_{n+1}$.



      Assume there does not exist another rational number $q_k$ where $q_n<q_k<q_{n+1}$ (i.e., we can order the rationals). Since $mathbb{Q}subset mathbb{R}$, this implies that the rationals are not dense in the reals, which is false. Therefore, our assumption is false.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let $n in mathbb{N}$ and $q_n,q_{n+1} in mathbb{Q}$ where $q_n < q_{n+1}$.



        Assume there does not exist another rational number $q_k$ where $q_n<q_k<q_{n+1}$ (i.e., we can order the rationals). Since $mathbb{Q}subset mathbb{R}$, this implies that the rationals are not dense in the reals, which is false. Therefore, our assumption is false.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $n in mathbb{N}$ and $q_n,q_{n+1} in mathbb{Q}$ where $q_n < q_{n+1}$.



          Assume there does not exist another rational number $q_k$ where $q_n<q_k<q_{n+1}$ (i.e., we can order the rationals). Since $mathbb{Q}subset mathbb{R}$, this implies that the rationals are not dense in the reals, which is false. Therefore, our assumption is false.






          share|cite|improve this answer









          $endgroup$



          Let $n in mathbb{N}$ and $q_n,q_{n+1} in mathbb{Q}$ where $q_n < q_{n+1}$.



          Assume there does not exist another rational number $q_k$ where $q_n<q_k<q_{n+1}$ (i.e., we can order the rationals). Since $mathbb{Q}subset mathbb{R}$, this implies that the rationals are not dense in the reals, which is false. Therefore, our assumption is false.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 24 at 16:44









          BeyBey

          1464




          1464























              0












              $begingroup$

              The reson is that if you have natural number n then you can find the NEXT number (=n+1), but if you have Rational number q then you cannot find NEXT number - because between any two different rational numbers is infinite number of other rational numbers.



              If such bijection $f$ exists then when you find $q=f(n)$ then you shoud be able to find next rational number $p=f(n+1)$ but because $qneq p$ then between $q$ and $p$ exist infinite number of other rational numbers e.g. $r=(p+q)/2$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                The reson is that if you have natural number n then you can find the NEXT number (=n+1), but if you have Rational number q then you cannot find NEXT number - because between any two different rational numbers is infinite number of other rational numbers.



                If such bijection $f$ exists then when you find $q=f(n)$ then you shoud be able to find next rational number $p=f(n+1)$ but because $qneq p$ then between $q$ and $p$ exist infinite number of other rational numbers e.g. $r=(p+q)/2$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The reson is that if you have natural number n then you can find the NEXT number (=n+1), but if you have Rational number q then you cannot find NEXT number - because between any two different rational numbers is infinite number of other rational numbers.



                  If such bijection $f$ exists then when you find $q=f(n)$ then you shoud be able to find next rational number $p=f(n+1)$ but because $qneq p$ then between $q$ and $p$ exist infinite number of other rational numbers e.g. $r=(p+q)/2$






                  share|cite|improve this answer











                  $endgroup$



                  The reson is that if you have natural number n then you can find the NEXT number (=n+1), but if you have Rational number q then you cannot find NEXT number - because between any two different rational numbers is infinite number of other rational numbers.



                  If such bijection $f$ exists then when you find $q=f(n)$ then you shoud be able to find next rational number $p=f(n+1)$ but because $qneq p$ then between $q$ and $p$ exist infinite number of other rational numbers e.g. $r=(p+q)/2$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 24 at 16:49

























                  answered Jan 24 at 16:36









                  Kamil KiełczewskiKamil Kiełczewski

                  1267




                  1267






























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