Determining the cardinality of rational set












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I have two questions, but I don't even know where I can start to solve it, can you give me a hint?



The question is like (Forgive me if MathJax is going wrong):



Determine the cardinality of these sets



a) If X = ${x in mathbb{R} | 1 leqslant x leqslant 3}$



b) Be $mathbb{Q}$ like $mathbb{Q} = { p/q | p,q in mathbb{Z} q gt 0}$










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    $begingroup$
    What do you know about cardinalities?
    $endgroup$
    – ervx
    Dec 3 '18 at 13:57










  • $begingroup$
    that is a beautiful question, loved when I had those! They are really enlighting. I recommend you build yourself bijective maps into either $mathbb{N}$ or $mathbb{R}$, since you know the cardinality of those
    $endgroup$
    – Enkidu
    Dec 3 '18 at 14:11
















0












$begingroup$


I have two questions, but I don't even know where I can start to solve it, can you give me a hint?



The question is like (Forgive me if MathJax is going wrong):



Determine the cardinality of these sets



a) If X = ${x in mathbb{R} | 1 leqslant x leqslant 3}$



b) Be $mathbb{Q}$ like $mathbb{Q} = { p/q | p,q in mathbb{Z} q gt 0}$










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What do you know about cardinalities?
    $endgroup$
    – ervx
    Dec 3 '18 at 13:57










  • $begingroup$
    that is a beautiful question, loved when I had those! They are really enlighting. I recommend you build yourself bijective maps into either $mathbb{N}$ or $mathbb{R}$, since you know the cardinality of those
    $endgroup$
    – Enkidu
    Dec 3 '18 at 14:11














0












0








0





$begingroup$


I have two questions, but I don't even know where I can start to solve it, can you give me a hint?



The question is like (Forgive me if MathJax is going wrong):



Determine the cardinality of these sets



a) If X = ${x in mathbb{R} | 1 leqslant x leqslant 3}$



b) Be $mathbb{Q}$ like $mathbb{Q} = { p/q | p,q in mathbb{Z} q gt 0}$










share|cite|improve this question











$endgroup$




I have two questions, but I don't even know where I can start to solve it, can you give me a hint?



The question is like (Forgive me if MathJax is going wrong):



Determine the cardinality of these sets



a) If X = ${x in mathbb{R} | 1 leqslant x leqslant 3}$



b) Be $mathbb{Q}$ like $mathbb{Q} = { p/q | p,q in mathbb{Z} q gt 0}$







discrete-mathematics elementary-set-theory






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edited Dec 3 '18 at 13:57









ervx

10.3k31338




10.3k31338










asked Dec 3 '18 at 13:56









Will_UWill_U

11




11








  • 1




    $begingroup$
    What do you know about cardinalities?
    $endgroup$
    – ervx
    Dec 3 '18 at 13:57










  • $begingroup$
    that is a beautiful question, loved when I had those! They are really enlighting. I recommend you build yourself bijective maps into either $mathbb{N}$ or $mathbb{R}$, since you know the cardinality of those
    $endgroup$
    – Enkidu
    Dec 3 '18 at 14:11














  • 1




    $begingroup$
    What do you know about cardinalities?
    $endgroup$
    – ervx
    Dec 3 '18 at 13:57










  • $begingroup$
    that is a beautiful question, loved when I had those! They are really enlighting. I recommend you build yourself bijective maps into either $mathbb{N}$ or $mathbb{R}$, since you know the cardinality of those
    $endgroup$
    – Enkidu
    Dec 3 '18 at 14:11








1




1




$begingroup$
What do you know about cardinalities?
$endgroup$
– ervx
Dec 3 '18 at 13:57




$begingroup$
What do you know about cardinalities?
$endgroup$
– ervx
Dec 3 '18 at 13:57












$begingroup$
that is a beautiful question, loved when I had those! They are really enlighting. I recommend you build yourself bijective maps into either $mathbb{N}$ or $mathbb{R}$, since you know the cardinality of those
$endgroup$
– Enkidu
Dec 3 '18 at 14:11




$begingroup$
that is a beautiful question, loved when I had those! They are really enlighting. I recommend you build yourself bijective maps into either $mathbb{N}$ or $mathbb{R}$, since you know the cardinality of those
$endgroup$
– Enkidu
Dec 3 '18 at 14:11










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For one you'll need to make use of the fact that $omega$ (=$mathbb{N}$) is bijectively equivalent to $omegatimesomega$. A second useful fact is that every real number, except for countably many, has a unique binary expansion.



The last tool in you arsenal is the Cantor-Schröder-Bernstein theorem.






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    $begingroup$

    For one you'll need to make use of the fact that $omega$ (=$mathbb{N}$) is bijectively equivalent to $omegatimesomega$. A second useful fact is that every real number, except for countably many, has a unique binary expansion.



    The last tool in you arsenal is the Cantor-Schröder-Bernstein theorem.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      For one you'll need to make use of the fact that $omega$ (=$mathbb{N}$) is bijectively equivalent to $omegatimesomega$. A second useful fact is that every real number, except for countably many, has a unique binary expansion.



      The last tool in you arsenal is the Cantor-Schröder-Bernstein theorem.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        For one you'll need to make use of the fact that $omega$ (=$mathbb{N}$) is bijectively equivalent to $omegatimesomega$. A second useful fact is that every real number, except for countably many, has a unique binary expansion.



        The last tool in you arsenal is the Cantor-Schröder-Bernstein theorem.






        share|cite|improve this answer









        $endgroup$



        For one you'll need to make use of the fact that $omega$ (=$mathbb{N}$) is bijectively equivalent to $omegatimesomega$. A second useful fact is that every real number, except for countably many, has a unique binary expansion.



        The last tool in you arsenal is the Cantor-Schröder-Bernstein theorem.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 14:55









        Jean-Pierre de VilliersJean-Pierre de Villiers

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