Left and right inverses of functions.












1












$begingroup$


Let $X$ be the set of real transcendental numbers. Define the realtion $sim$ on $X$ by $xsim y$ iff $x-y in mathbb{Q}$ is an equivalence relation.



Let $Y$ denote the set of equivalence classes generated by $sim$ defined above.



Define the function $P:X rightarrow Y$ by $f(x) = [x]$ where $[x]$ denotes the equivalence class of $x$. Does the function P have:




  • a left inverse?

  • a right inverse?


My work so far:



Since $P$ is surjective by definition of equivalence classes, $P$ has a right inverse.



Now if $P$ is injective then it will have a left inverse. I think it's not injective as two real transcendental numbers can end up in the same equivalence class, so $P$ doesn't have a left inverse, but I am not sure how to show this rigorously.






functions inverse equivalence-relations







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $X$ be the set of real transcendental numbers. Define the realtion $sim$ on $X$ by $xsim y$ iff $x-y in mathbb{Q}$ is an equivalence relation.



    Let $Y$ denote the set of equivalence classes generated by $sim$ defined above.



    Define the function $P:X rightarrow Y$ by $f(x) = [x]$ where $[x]$ denotes the equivalence class of $x$. Does the function P have:




    • a left inverse?

    • a right inverse?


    My work so far:



    Since $P$ is surjective by definition of equivalence classes, $P$ has a right inverse.



    Now if $P$ is injective then it will have a left inverse. I think it's not injective as two real transcendental numbers can end up in the same equivalence class, so $P$ doesn't have a left inverse, but I am not sure how to show this rigorously.






    functions inverse equivalence-relations







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $X$ be the set of real transcendental numbers. Define the realtion $sim$ on $X$ by $xsim y$ iff $x-y in mathbb{Q}$ is an equivalence relation.



      Let $Y$ denote the set of equivalence classes generated by $sim$ defined above.



      Define the function $P:X rightarrow Y$ by $f(x) = [x]$ where $[x]$ denotes the equivalence class of $x$. Does the function P have:




      • a left inverse?

      • a right inverse?


      My work so far:



      Since $P$ is surjective by definition of equivalence classes, $P$ has a right inverse.



      Now if $P$ is injective then it will have a left inverse. I think it's not injective as two real transcendental numbers can end up in the same equivalence class, so $P$ doesn't have a left inverse, but I am not sure how to show this rigorously.






      functions inverse equivalence-relations







      share|cite|improve this question









      $endgroup$




      Let $X$ be the set of real transcendental numbers. Define the realtion $sim$ on $X$ by $xsim y$ iff $x-y in mathbb{Q}$ is an equivalence relation.



      Let $Y$ denote the set of equivalence classes generated by $sim$ defined above.



      Define the function $P:X rightarrow Y$ by $f(x) = [x]$ where $[x]$ denotes the equivalence class of $x$. Does the function P have:




      • a left inverse?

      • a right inverse?


      My work so far:



      Since $P$ is surjective by definition of equivalence classes, $P$ has a right inverse.



      Now if $P$ is injective then it will have a left inverse. I think it's not injective as two real transcendental numbers can end up in the same equivalence class, so $P$ doesn't have a left inverse, but I am not sure how to show this rigorously.






      functions inverse equivalence-relations




      functions inverse equivalence-relations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 30 '18 at 13:02









      xAlyxAly

      404




      404






















          1 Answer
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          active

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          1












          $begingroup$

          Hint 1: prove that if $xin X$, then also $x+1in X$. Since $x+1sim x$, the map $P$ is not injective



          Hint 2: the sum of algebraic numbers is algebraic.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
            $endgroup$
            – xAly
            Dec 30 '18 at 19:22










          • $begingroup$
            @xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
            $endgroup$
            – egreg
            Dec 30 '18 at 20:10












          • $begingroup$
            But $X$ is the set of real transcendental numbers, not algebraic numbers?
            $endgroup$
            – xAly
            Dec 30 '18 at 21:16












          • $begingroup$
            @xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
            $endgroup$
            – egreg
            Dec 30 '18 at 21:52










          • $begingroup$
            Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
            $endgroup$
            – xAly
            Dec 30 '18 at 22:02











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Hint 1: prove that if $xin X$, then also $x+1in X$. Since $x+1sim x$, the map $P$ is not injective



          Hint 2: the sum of algebraic numbers is algebraic.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
            $endgroup$
            – xAly
            Dec 30 '18 at 19:22










          • $begingroup$
            @xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
            $endgroup$
            – egreg
            Dec 30 '18 at 20:10












          • $begingroup$
            But $X$ is the set of real transcendental numbers, not algebraic numbers?
            $endgroup$
            – xAly
            Dec 30 '18 at 21:16












          • $begingroup$
            @xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
            $endgroup$
            – egreg
            Dec 30 '18 at 21:52










          • $begingroup$
            Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
            $endgroup$
            – xAly
            Dec 30 '18 at 22:02
















          1












          $begingroup$

          Hint 1: prove that if $xin X$, then also $x+1in X$. Since $x+1sim x$, the map $P$ is not injective



          Hint 2: the sum of algebraic numbers is algebraic.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
            $endgroup$
            – xAly
            Dec 30 '18 at 19:22










          • $begingroup$
            @xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
            $endgroup$
            – egreg
            Dec 30 '18 at 20:10












          • $begingroup$
            But $X$ is the set of real transcendental numbers, not algebraic numbers?
            $endgroup$
            – xAly
            Dec 30 '18 at 21:16












          • $begingroup$
            @xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
            $endgroup$
            – egreg
            Dec 30 '18 at 21:52










          • $begingroup$
            Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
            $endgroup$
            – xAly
            Dec 30 '18 at 22:02














          1












          1








          1





          $begingroup$

          Hint 1: prove that if $xin X$, then also $x+1in X$. Since $x+1sim x$, the map $P$ is not injective



          Hint 2: the sum of algebraic numbers is algebraic.






          share|cite|improve this answer









          $endgroup$



          Hint 1: prove that if $xin X$, then also $x+1in X$. Since $x+1sim x$, the map $P$ is not injective



          Hint 2: the sum of algebraic numbers is algebraic.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 13:49









          egregegreg

          185k1486206




          185k1486206












          • $begingroup$
            I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
            $endgroup$
            – xAly
            Dec 30 '18 at 19:22










          • $begingroup$
            @xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
            $endgroup$
            – egreg
            Dec 30 '18 at 20:10












          • $begingroup$
            But $X$ is the set of real transcendental numbers, not algebraic numbers?
            $endgroup$
            – xAly
            Dec 30 '18 at 21:16












          • $begingroup$
            @xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
            $endgroup$
            – egreg
            Dec 30 '18 at 21:52










          • $begingroup$
            Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
            $endgroup$
            – xAly
            Dec 30 '18 at 22:02


















          • $begingroup$
            I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
            $endgroup$
            – xAly
            Dec 30 '18 at 19:22










          • $begingroup$
            @xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
            $endgroup$
            – egreg
            Dec 30 '18 at 20:10












          • $begingroup$
            But $X$ is the set of real transcendental numbers, not algebraic numbers?
            $endgroup$
            – xAly
            Dec 30 '18 at 21:16












          • $begingroup$
            @xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
            $endgroup$
            – egreg
            Dec 30 '18 at 21:52










          • $begingroup$
            Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
            $endgroup$
            – xAly
            Dec 30 '18 at 22:02
















          $begingroup$
          I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
          $endgroup$
          – xAly
          Dec 30 '18 at 19:22




          $begingroup$
          I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
          $endgroup$
          – xAly
          Dec 30 '18 at 19:22












          $begingroup$
          @xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
          $endgroup$
          – egreg
          Dec 30 '18 at 20:10






          $begingroup$
          @xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
          $endgroup$
          – egreg
          Dec 30 '18 at 20:10














          $begingroup$
          But $X$ is the set of real transcendental numbers, not algebraic numbers?
          $endgroup$
          – xAly
          Dec 30 '18 at 21:16






          $begingroup$
          But $X$ is the set of real transcendental numbers, not algebraic numbers?
          $endgroup$
          – xAly
          Dec 30 '18 at 21:16














          $begingroup$
          @xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
          $endgroup$
          – egreg
          Dec 30 '18 at 21:52




          $begingroup$
          @xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
          $endgroup$
          – egreg
          Dec 30 '18 at 21:52












          $begingroup$
          Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
          $endgroup$
          – xAly
          Dec 30 '18 at 22:02




          $begingroup$
          Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
          $endgroup$
          – xAly
          Dec 30 '18 at 22:02


















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