Characterization of non-unique decimal expansions












12












$begingroup$


In proofs exhibiting bijections of $mathbb{R}$ for the sake of proving arguments about cardinality, care is generally taken to avoid the $.999!ldots=1.00!ldots$ problem. However, it is generally accepted without comment that this is the only way to make distinct decimal expansions that represent the same real number. Although it seems obvious, how can we be sure there are there no other corner cases to be concerned with?



Formally, if $x_d$ and $y_d$ are distinct sequences of integers between $0$ and $9$ representing decimal expansions of numbers $x$ and $y$ in $[0,1)$, is it true that $x=y$ iff there is an $Ninmathbb{N}$ such that




  • if $n<N$ then $x_n=y_n$

  • $x_N=y_N-1$, and

  • if $n>N$ then $x_n=9$ and $y_n=0$,


and if so, how does one prove this?






decimal-expansion







share|cite|improve this question









$endgroup$

















    12












    $begingroup$


    In proofs exhibiting bijections of $mathbb{R}$ for the sake of proving arguments about cardinality, care is generally taken to avoid the $.999!ldots=1.00!ldots$ problem. However, it is generally accepted without comment that this is the only way to make distinct decimal expansions that represent the same real number. Although it seems obvious, how can we be sure there are there no other corner cases to be concerned with?



    Formally, if $x_d$ and $y_d$ are distinct sequences of integers between $0$ and $9$ representing decimal expansions of numbers $x$ and $y$ in $[0,1)$, is it true that $x=y$ iff there is an $Ninmathbb{N}$ such that




    • if $n<N$ then $x_n=y_n$

    • $x_N=y_N-1$, and

    • if $n>N$ then $x_n=9$ and $y_n=0$,


    and if so, how does one prove this?






    decimal-expansion







    share|cite|improve this question









    $endgroup$















      12












      12








      12


      4



      $begingroup$


      In proofs exhibiting bijections of $mathbb{R}$ for the sake of proving arguments about cardinality, care is generally taken to avoid the $.999!ldots=1.00!ldots$ problem. However, it is generally accepted without comment that this is the only way to make distinct decimal expansions that represent the same real number. Although it seems obvious, how can we be sure there are there no other corner cases to be concerned with?



      Formally, if $x_d$ and $y_d$ are distinct sequences of integers between $0$ and $9$ representing decimal expansions of numbers $x$ and $y$ in $[0,1)$, is it true that $x=y$ iff there is an $Ninmathbb{N}$ such that




      • if $n<N$ then $x_n=y_n$

      • $x_N=y_N-1$, and

      • if $n>N$ then $x_n=9$ and $y_n=0$,


      and if so, how does one prove this?






      decimal-expansion







      share|cite|improve this question









      $endgroup$




      In proofs exhibiting bijections of $mathbb{R}$ for the sake of proving arguments about cardinality, care is generally taken to avoid the $.999!ldots=1.00!ldots$ problem. However, it is generally accepted without comment that this is the only way to make distinct decimal expansions that represent the same real number. Although it seems obvious, how can we be sure there are there no other corner cases to be concerned with?



      Formally, if $x_d$ and $y_d$ are distinct sequences of integers between $0$ and $9$ representing decimal expansions of numbers $x$ and $y$ in $[0,1)$, is it true that $x=y$ iff there is an $Ninmathbb{N}$ such that




      • if $n<N$ then $x_n=y_n$

      • $x_N=y_N-1$, and

      • if $n>N$ then $x_n=9$ and $y_n=0$,


      and if so, how does one prove this?






      decimal-expansion




      decimal-expansion






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 5 '13 at 19:54









      Eric StuckyEric Stucky

      10.3k32461




      10.3k32461






















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          Yes (up to switching $x$ and $y$). Let $N$ be the least integer such that $x_Nne y_N$. Assume WLOG that $y_N>$. If $y_N>x_N+1$ then since the remaining digits together contribute at most $10^{-N}$ we have $y-xge 2cdot 10^{-N}-10^{-N}>0$, so $xne y$. If $x_{N+1}ne 9$ or $y_{N+1}ne 0$, then $x_{N+1}-y_{N+1}le 8$ and since the remaining digits contribute at most $10^{-N-1}$ we have
          $$y-xge 10^{-N}-8cdot 10^{-N-1}-10^{-N-1}>0$$
          so $xne y$. Thus we must have $x_{N+1}=9$ and $y_{N+1}=0$. Continuing in this manner we see that it is true for all remaining digits.






          share|cite|improve this answer









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            $begingroup$

            Yes (up to switching $x$ and $y$). Let $N$ be the least integer such that $x_Nne y_N$. Assume WLOG that $y_N>$. If $y_N>x_N+1$ then since the remaining digits together contribute at most $10^{-N}$ we have $y-xge 2cdot 10^{-N}-10^{-N}>0$, so $xne y$. If $x_{N+1}ne 9$ or $y_{N+1}ne 0$, then $x_{N+1}-y_{N+1}le 8$ and since the remaining digits contribute at most $10^{-N-1}$ we have
            $$y-xge 10^{-N}-8cdot 10^{-N-1}-10^{-N-1}>0$$
            so $xne y$. Thus we must have $x_{N+1}=9$ and $y_{N+1}=0$. Continuing in this manner we see that it is true for all remaining digits.






            share|cite|improve this answer









            $endgroup$


















              7












              $begingroup$

              Yes (up to switching $x$ and $y$). Let $N$ be the least integer such that $x_Nne y_N$. Assume WLOG that $y_N>$. If $y_N>x_N+1$ then since the remaining digits together contribute at most $10^{-N}$ we have $y-xge 2cdot 10^{-N}-10^{-N}>0$, so $xne y$. If $x_{N+1}ne 9$ or $y_{N+1}ne 0$, then $x_{N+1}-y_{N+1}le 8$ and since the remaining digits contribute at most $10^{-N-1}$ we have
              $$y-xge 10^{-N}-8cdot 10^{-N-1}-10^{-N-1}>0$$
              so $xne y$. Thus we must have $x_{N+1}=9$ and $y_{N+1}=0$. Continuing in this manner we see that it is true for all remaining digits.






              share|cite|improve this answer









              $endgroup$
















                7












                7








                7





                $begingroup$

                Yes (up to switching $x$ and $y$). Let $N$ be the least integer such that $x_Nne y_N$. Assume WLOG that $y_N>$. If $y_N>x_N+1$ then since the remaining digits together contribute at most $10^{-N}$ we have $y-xge 2cdot 10^{-N}-10^{-N}>0$, so $xne y$. If $x_{N+1}ne 9$ or $y_{N+1}ne 0$, then $x_{N+1}-y_{N+1}le 8$ and since the remaining digits contribute at most $10^{-N-1}$ we have
                $$y-xge 10^{-N}-8cdot 10^{-N-1}-10^{-N-1}>0$$
                so $xne y$. Thus we must have $x_{N+1}=9$ and $y_{N+1}=0$. Continuing in this manner we see that it is true for all remaining digits.






                share|cite|improve this answer









                $endgroup$



                Yes (up to switching $x$ and $y$). Let $N$ be the least integer such that $x_Nne y_N$. Assume WLOG that $y_N>$. If $y_N>x_N+1$ then since the remaining digits together contribute at most $10^{-N}$ we have $y-xge 2cdot 10^{-N}-10^{-N}>0$, so $xne y$. If $x_{N+1}ne 9$ or $y_{N+1}ne 0$, then $x_{N+1}-y_{N+1}le 8$ and since the remaining digits contribute at most $10^{-N-1}$ we have
                $$y-xge 10^{-N}-8cdot 10^{-N-1}-10^{-N-1}>0$$
                so $xne y$. Thus we must have $x_{N+1}=9$ and $y_{N+1}=0$. Continuing in this manner we see that it is true for all remaining digits.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 5 '13 at 20:10









                Alex BeckerAlex Becker

                49.1k698161




                49.1k698161






























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