$ bigcup_{n=1}^{infty}{zinBbb C:z^n=1}={zinBbb C:|z|=1}$?












6












$begingroup$


Is $$ bigcup_{n=1}^{infty}{zinBbb C:z^n=1}={zinBbb C:|z|=1}$$my argument is that the argument of the elements of the first set are rational multiples of $pi$ whereas the second set also consists of elements with irrational multiples of $pi$.
But I am not so sure, there is still some doubt.










share|cite|improve this question











$endgroup$

















    6












    $begingroup$


    Is $$ bigcup_{n=1}^{infty}{zinBbb C:z^n=1}={zinBbb C:|z|=1}$$my argument is that the argument of the elements of the first set are rational multiples of $pi$ whereas the second set also consists of elements with irrational multiples of $pi$.
    But I am not so sure, there is still some doubt.










    share|cite|improve this question











    $endgroup$















      6












      6








      6





      $begingroup$


      Is $$ bigcup_{n=1}^{infty}{zinBbb C:z^n=1}={zinBbb C:|z|=1}$$my argument is that the argument of the elements of the first set are rational multiples of $pi$ whereas the second set also consists of elements with irrational multiples of $pi$.
      But I am not so sure, there is still some doubt.










      share|cite|improve this question











      $endgroup$




      Is $$ bigcup_{n=1}^{infty}{zinBbb C:z^n=1}={zinBbb C:|z|=1}$$my argument is that the argument of the elements of the first set are rational multiples of $pi$ whereas the second set also consists of elements with irrational multiples of $pi$.
      But I am not so sure, there is still some doubt.







      complex-numbers roots-of-unity






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 2 at 2:52









      user21820

      40.1k544162




      40.1k544162










      asked Feb 1 at 15:14









      KarthikKarthik

      13212




      13212






















          2 Answers
          2






          active

          oldest

          votes


















          17












          $begingroup$

          Those two sets are certainly not equal.



          $displaystylebigcup_{n=1}^{infty}{z,|,z^n=1,nin mathbb N}$ is countable as it is a countable union of finite sets, while ${z/|z|=1}$ has the cardinality of the continuum.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            What does "power of a continuum" mean?
            $endgroup$
            – Karthik
            Feb 1 at 15:21






          • 4




            $begingroup$
            I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
            $endgroup$
            – mathcounterexamples.net
            Feb 1 at 15:23






          • 1




            $begingroup$
            "power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
            $endgroup$
            – Paul Sinclair
            Feb 1 at 17:22






          • 1




            $begingroup$
            @PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
            $endgroup$
            – mathcounterexamples.net
            Feb 1 at 17:35










          • $begingroup$
            This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
            $endgroup$
            – drjpizzle
            Feb 1 at 18:22



















          11












          $begingroup$

          Your argument is correct.



          For example, $e^{sqrt{2} pi i}$ is not in the first set.






          share|cite|improve this answer









          $endgroup$














            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            17












            $begingroup$

            Those two sets are certainly not equal.



            $displaystylebigcup_{n=1}^{infty}{z,|,z^n=1,nin mathbb N}$ is countable as it is a countable union of finite sets, while ${z/|z|=1}$ has the cardinality of the continuum.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              What does "power of a continuum" mean?
              $endgroup$
              – Karthik
              Feb 1 at 15:21






            • 4




              $begingroup$
              I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
              $endgroup$
              – mathcounterexamples.net
              Feb 1 at 15:23






            • 1




              $begingroup$
              "power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
              $endgroup$
              – Paul Sinclair
              Feb 1 at 17:22






            • 1




              $begingroup$
              @PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
              $endgroup$
              – mathcounterexamples.net
              Feb 1 at 17:35










            • $begingroup$
              This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
              $endgroup$
              – drjpizzle
              Feb 1 at 18:22
















            17












            $begingroup$

            Those two sets are certainly not equal.



            $displaystylebigcup_{n=1}^{infty}{z,|,z^n=1,nin mathbb N}$ is countable as it is a countable union of finite sets, while ${z/|z|=1}$ has the cardinality of the continuum.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              What does "power of a continuum" mean?
              $endgroup$
              – Karthik
              Feb 1 at 15:21






            • 4




              $begingroup$
              I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
              $endgroup$
              – mathcounterexamples.net
              Feb 1 at 15:23






            • 1




              $begingroup$
              "power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
              $endgroup$
              – Paul Sinclair
              Feb 1 at 17:22






            • 1




              $begingroup$
              @PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
              $endgroup$
              – mathcounterexamples.net
              Feb 1 at 17:35










            • $begingroup$
              This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
              $endgroup$
              – drjpizzle
              Feb 1 at 18:22














            17












            17








            17





            $begingroup$

            Those two sets are certainly not equal.



            $displaystylebigcup_{n=1}^{infty}{z,|,z^n=1,nin mathbb N}$ is countable as it is a countable union of finite sets, while ${z/|z|=1}$ has the cardinality of the continuum.






            share|cite|improve this answer











            $endgroup$



            Those two sets are certainly not equal.



            $displaystylebigcup_{n=1}^{infty}{z,|,z^n=1,nin mathbb N}$ is countable as it is a countable union of finite sets, while ${z/|z|=1}$ has the cardinality of the continuum.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Feb 1 at 15:56









            El borito

            664216




            664216










            answered Feb 1 at 15:19









            mathcounterexamples.netmathcounterexamples.net

            26.9k22158




            26.9k22158








            • 1




              $begingroup$
              What does "power of a continuum" mean?
              $endgroup$
              – Karthik
              Feb 1 at 15:21






            • 4




              $begingroup$
              I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
              $endgroup$
              – mathcounterexamples.net
              Feb 1 at 15:23






            • 1




              $begingroup$
              "power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
              $endgroup$
              – Paul Sinclair
              Feb 1 at 17:22






            • 1




              $begingroup$
              @PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
              $endgroup$
              – mathcounterexamples.net
              Feb 1 at 17:35










            • $begingroup$
              This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
              $endgroup$
              – drjpizzle
              Feb 1 at 18:22














            • 1




              $begingroup$
              What does "power of a continuum" mean?
              $endgroup$
              – Karthik
              Feb 1 at 15:21






            • 4




              $begingroup$
              I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
              $endgroup$
              – mathcounterexamples.net
              Feb 1 at 15:23






            • 1




              $begingroup$
              "power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
              $endgroup$
              – Paul Sinclair
              Feb 1 at 17:22






            • 1




              $begingroup$
              @PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
              $endgroup$
              – mathcounterexamples.net
              Feb 1 at 17:35










            • $begingroup$
              This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
              $endgroup$
              – drjpizzle
              Feb 1 at 18:22








            1




            1




            $begingroup$
            What does "power of a continuum" mean?
            $endgroup$
            – Karthik
            Feb 1 at 15:21




            $begingroup$
            What does "power of a continuum" mean?
            $endgroup$
            – Karthik
            Feb 1 at 15:21




            4




            4




            $begingroup$
            I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
            $endgroup$
            – mathcounterexamples.net
            Feb 1 at 15:23




            $begingroup$
            I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
            $endgroup$
            – mathcounterexamples.net
            Feb 1 at 15:23




            1




            1




            $begingroup$
            "power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
            $endgroup$
            – Paul Sinclair
            Feb 1 at 17:22




            $begingroup$
            "power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
            $endgroup$
            – Paul Sinclair
            Feb 1 at 17:22




            1




            1




            $begingroup$
            @PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
            $endgroup$
            – mathcounterexamples.net
            Feb 1 at 17:35




            $begingroup$
            @PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
            $endgroup$
            – mathcounterexamples.net
            Feb 1 at 17:35












            $begingroup$
            This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
            $endgroup$
            – drjpizzle
            Feb 1 at 18:22




            $begingroup$
            This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
            $endgroup$
            – drjpizzle
            Feb 1 at 18:22











            11












            $begingroup$

            Your argument is correct.



            For example, $e^{sqrt{2} pi i}$ is not in the first set.






            share|cite|improve this answer









            $endgroup$


















              11












              $begingroup$

              Your argument is correct.



              For example, $e^{sqrt{2} pi i}$ is not in the first set.






              share|cite|improve this answer









              $endgroup$
















                11












                11








                11





                $begingroup$

                Your argument is correct.



                For example, $e^{sqrt{2} pi i}$ is not in the first set.






                share|cite|improve this answer









                $endgroup$



                Your argument is correct.



                For example, $e^{sqrt{2} pi i}$ is not in the first set.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 1 at 15:22









                DubsDubs

                59426




                59426






























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