Prove that the dyadic rationals $D$ are dense in $mathbb{Q}$












1












$begingroup$



Prove that the dyadic rationals $D$ are dense in $mathbb{R}$.




I saw some proofs using showing that $D$ is dense on $[0,1]$.



I proved that if $A$ is dense in $B$ and $B$ is dense in $C$, then $A$ is dense in $C$. So, I want use it. My idea is show that $D$ is dense in $mathbb{Q}$. Can someone help me? Just a hint, not a solution.




Attempt. Let $D = left{frac{m}{2^{n}}mid n in mathbb{N}text{ and }m in mathbb{Z}right}$. Take $I = left(k-frac{epsilon}{2},k+frac{epsilon}{2}right)$ an open interval in $mathbb{Q}$. We can find a $n in mathbb{N}$ such that $epsilon > frac{1}{2^{n}}$. If there is no $frac{m}{2^{n}} in I$, there is $frac{m}{2^n}$ and $frac{m+1}{2^{n}}$ such that $frac{m}{2^{n}} < k - frac{epsilon}{2}$ and $k + frac{epsilon}{2} < frac{m+1}{2^{n}}$. Thus
$$frac{1}{2^{n}} =frac{m+1}{2^{n}} - frac{m}{2^n} > epsilon,$$
a contradiction.






real-analysis metric-spaces







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    Prove that the dyadic rationals $D$ are dense in $mathbb{R}$.




    I saw some proofs using showing that $D$ is dense on $[0,1]$.



    I proved that if $A$ is dense in $B$ and $B$ is dense in $C$, then $A$ is dense in $C$. So, I want use it. My idea is show that $D$ is dense in $mathbb{Q}$. Can someone help me? Just a hint, not a solution.




    Attempt. Let $D = left{frac{m}{2^{n}}mid n in mathbb{N}text{ and }m in mathbb{Z}right}$. Take $I = left(k-frac{epsilon}{2},k+frac{epsilon}{2}right)$ an open interval in $mathbb{Q}$. We can find a $n in mathbb{N}$ such that $epsilon > frac{1}{2^{n}}$. If there is no $frac{m}{2^{n}} in I$, there is $frac{m}{2^n}$ and $frac{m+1}{2^{n}}$ such that $frac{m}{2^{n}} < k - frac{epsilon}{2}$ and $k + frac{epsilon}{2} < frac{m+1}{2^{n}}$. Thus
    $$frac{1}{2^{n}} =frac{m+1}{2^{n}} - frac{m}{2^n} > epsilon,$$
    a contradiction.






    real-analysis metric-spaces







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Prove that the dyadic rationals $D$ are dense in $mathbb{R}$.




      I saw some proofs using showing that $D$ is dense on $[0,1]$.



      I proved that if $A$ is dense in $B$ and $B$ is dense in $C$, then $A$ is dense in $C$. So, I want use it. My idea is show that $D$ is dense in $mathbb{Q}$. Can someone help me? Just a hint, not a solution.




      Attempt. Let $D = left{frac{m}{2^{n}}mid n in mathbb{N}text{ and }m in mathbb{Z}right}$. Take $I = left(k-frac{epsilon}{2},k+frac{epsilon}{2}right)$ an open interval in $mathbb{Q}$. We can find a $n in mathbb{N}$ such that $epsilon > frac{1}{2^{n}}$. If there is no $frac{m}{2^{n}} in I$, there is $frac{m}{2^n}$ and $frac{m+1}{2^{n}}$ such that $frac{m}{2^{n}} < k - frac{epsilon}{2}$ and $k + frac{epsilon}{2} < frac{m+1}{2^{n}}$. Thus
      $$frac{1}{2^{n}} =frac{m+1}{2^{n}} - frac{m}{2^n} > epsilon,$$
      a contradiction.






      real-analysis metric-spaces







      share|cite|improve this question











      $endgroup$





      Prove that the dyadic rationals $D$ are dense in $mathbb{R}$.




      I saw some proofs using showing that $D$ is dense on $[0,1]$.



      I proved that if $A$ is dense in $B$ and $B$ is dense in $C$, then $A$ is dense in $C$. So, I want use it. My idea is show that $D$ is dense in $mathbb{Q}$. Can someone help me? Just a hint, not a solution.




      Attempt. Let $D = left{frac{m}{2^{n}}mid n in mathbb{N}text{ and }m in mathbb{Z}right}$. Take $I = left(k-frac{epsilon}{2},k+frac{epsilon}{2}right)$ an open interval in $mathbb{Q}$. We can find a $n in mathbb{N}$ such that $epsilon > frac{1}{2^{n}}$. If there is no $frac{m}{2^{n}} in I$, there is $frac{m}{2^n}$ and $frac{m+1}{2^{n}}$ such that $frac{m}{2^{n}} < k - frac{epsilon}{2}$ and $k + frac{epsilon}{2} < frac{m+1}{2^{n}}$. Thus
      $$frac{1}{2^{n}} =frac{m+1}{2^{n}} - frac{m}{2^n} > epsilon,$$
      a contradiction.






      real-analysis metric-spaces




      real-analysis metric-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 29 '18 at 19:56







      Lucas Corrêa

















      asked Dec 29 '18 at 19:24









      Lucas CorrêaLucas Corrêa

      1,5601321




      1,5601321






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          Instead of reducing the statement to the density of $mathbb{Q}$, just use the same proof idea.
          Remainder: $mathbb{Q}$ is dense, because given an interval of length $varepsilon> 0$, it is possible to pick an integer $nin mathbb{N}$ such that $n> frac{1}{varepsilon}$. Equivalently, $varepsilon> frac{1}{n}$. So given the gap of length $frac{1}{n}$ (the interval), you cannot step over it if the length of your step is $frac{1}{n}$. Starting from zero, and walking towards the interval, you will step into it at a point. If this requires $k$ steps, then $frac{k}{n}$ is in the interval.



          Can you modify this proof to make it work for dyadic rationals?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I posted my attempt with this idea.
            $endgroup$
            – Lucas Corrêa
            Dec 29 '18 at 19:57






          • 2




            $begingroup$
            It is a correct proof.
            $endgroup$
            – A. Pongrácz
            Dec 29 '18 at 20:23










          • $begingroup$
            Thank you! ${}$
            $endgroup$
            – Lucas Corrêa
            Dec 29 '18 at 20:28



















          2












          $begingroup$

          Hint: There is no need to go through $mathbb Q$. Just prove that if $y>x$ there are integers $n,m$ such that $y-x>m2^{-n}$.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            The real number $mathbb{R}$ and hence all subsets of the real numbers have come with a notion of distance $left(a, bright) to left|a - bright|$ .



            In particular the rationals themselves form a metric space, see ProofWiki .



            This means we can use the definition of dense set from Wikipedia that requires the existence of a metric.



            Here's the definition of the closure of a set, slightly rewritten



            $$ bar{A} stackrel{def}{=} A cup left{lim_{n to infty} a_n text{ where } a in text{Seq}left[Aright] right} tag{1} $$



            Note that $bar{A}$ is constrained to be a subset of $mathbb{Q}$ because $mathbb{Q}$ is where we're taking limits.



            $$ A text{ is dense in } X stackrel{def}{iff} bar{A} = X tag{2} $$



            One way to show it directly is to come up with an explicit sequence of dyadic rationals that converges to a given rational number $frac{a}{b}$ for every such $frac{a}{b} in mathbb{Q}$ .






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That is my first idea, but I cannot find such sequence.
              $endgroup$
              – Lucas Corrêa
              Dec 29 '18 at 19:57










            • $begingroup$
              @LucasCorrêa ... what sort of hints are you okay with?
              $endgroup$
              – Gregory Nisbet
              Dec 29 '18 at 19:58










            • $begingroup$
              My problem was in the construction of the sequence. My idea was take $d_{n} in D$ such thath $|d_{n} - a| < 1/n$ for some $a in mathbb{Q}$. But $mathbb{Q}$ is not complete, how can I ensure that $(d_{n})$ converges?
              $endgroup$
              – Lucas Corrêa
              Dec 29 '18 at 20:06






            • 1




              $begingroup$
              I think a sequence of dyadic rationals that converges to an irrational number in $mathbb{R}$ simply does not converge to anything in $mathbb{Q}$ .
              $endgroup$
              – Gregory Nisbet
              Dec 29 '18 at 20:16











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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Instead of reducing the statement to the density of $mathbb{Q}$, just use the same proof idea.
            Remainder: $mathbb{Q}$ is dense, because given an interval of length $varepsilon> 0$, it is possible to pick an integer $nin mathbb{N}$ such that $n> frac{1}{varepsilon}$. Equivalently, $varepsilon> frac{1}{n}$. So given the gap of length $frac{1}{n}$ (the interval), you cannot step over it if the length of your step is $frac{1}{n}$. Starting from zero, and walking towards the interval, you will step into it at a point. If this requires $k$ steps, then $frac{k}{n}$ is in the interval.



            Can you modify this proof to make it work for dyadic rationals?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I posted my attempt with this idea.
              $endgroup$
              – Lucas Corrêa
              Dec 29 '18 at 19:57






            • 2




              $begingroup$
              It is a correct proof.
              $endgroup$
              – A. Pongrácz
              Dec 29 '18 at 20:23










            • $begingroup$
              Thank you! ${}$
              $endgroup$
              – Lucas Corrêa
              Dec 29 '18 at 20:28
















            3












            $begingroup$

            Instead of reducing the statement to the density of $mathbb{Q}$, just use the same proof idea.
            Remainder: $mathbb{Q}$ is dense, because given an interval of length $varepsilon> 0$, it is possible to pick an integer $nin mathbb{N}$ such that $n> frac{1}{varepsilon}$. Equivalently, $varepsilon> frac{1}{n}$. So given the gap of length $frac{1}{n}$ (the interval), you cannot step over it if the length of your step is $frac{1}{n}$. Starting from zero, and walking towards the interval, you will step into it at a point. If this requires $k$ steps, then $frac{k}{n}$ is in the interval.



            Can you modify this proof to make it work for dyadic rationals?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I posted my attempt with this idea.
              $endgroup$
              – Lucas Corrêa
              Dec 29 '18 at 19:57






            • 2




              $begingroup$
              It is a correct proof.
              $endgroup$
              – A. Pongrácz
              Dec 29 '18 at 20:23










            • $begingroup$
              Thank you! ${}$
              $endgroup$
              – Lucas Corrêa
              Dec 29 '18 at 20:28














            3












            3








            3





            $begingroup$

            Instead of reducing the statement to the density of $mathbb{Q}$, just use the same proof idea.
            Remainder: $mathbb{Q}$ is dense, because given an interval of length $varepsilon> 0$, it is possible to pick an integer $nin mathbb{N}$ such that $n> frac{1}{varepsilon}$. Equivalently, $varepsilon> frac{1}{n}$. So given the gap of length $frac{1}{n}$ (the interval), you cannot step over it if the length of your step is $frac{1}{n}$. Starting from zero, and walking towards the interval, you will step into it at a point. If this requires $k$ steps, then $frac{k}{n}$ is in the interval.



            Can you modify this proof to make it work for dyadic rationals?






            share|cite|improve this answer









            $endgroup$



            Instead of reducing the statement to the density of $mathbb{Q}$, just use the same proof idea.
            Remainder: $mathbb{Q}$ is dense, because given an interval of length $varepsilon> 0$, it is possible to pick an integer $nin mathbb{N}$ such that $n> frac{1}{varepsilon}$. Equivalently, $varepsilon> frac{1}{n}$. So given the gap of length $frac{1}{n}$ (the interval), you cannot step over it if the length of your step is $frac{1}{n}$. Starting from zero, and walking towards the interval, you will step into it at a point. If this requires $k$ steps, then $frac{k}{n}$ is in the interval.



            Can you modify this proof to make it work for dyadic rationals?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 29 '18 at 19:32









            A. PongráczA. Pongrácz

            5,9381929




            5,9381929












            • $begingroup$
              I posted my attempt with this idea.
              $endgroup$
              – Lucas Corrêa
              Dec 29 '18 at 19:57






            • 2




              $begingroup$
              It is a correct proof.
              $endgroup$
              – A. Pongrácz
              Dec 29 '18 at 20:23










            • $begingroup$
              Thank you! ${}$
              $endgroup$
              – Lucas Corrêa
              Dec 29 '18 at 20:28


















            • $begingroup$
              I posted my attempt with this idea.
              $endgroup$
              – Lucas Corrêa
              Dec 29 '18 at 19:57






            • 2




              $begingroup$
              It is a correct proof.
              $endgroup$
              – A. Pongrácz
              Dec 29 '18 at 20:23










            • $begingroup$
              Thank you! ${}$
              $endgroup$
              – Lucas Corrêa
              Dec 29 '18 at 20:28
















            $begingroup$
            I posted my attempt with this idea.
            $endgroup$
            – Lucas Corrêa
            Dec 29 '18 at 19:57




            $begingroup$
            I posted my attempt with this idea.
            $endgroup$
            – Lucas Corrêa
            Dec 29 '18 at 19:57




            2




            2




            $begingroup$
            It is a correct proof.
            $endgroup$
            – A. Pongrácz
            Dec 29 '18 at 20:23




            $begingroup$
            It is a correct proof.
            $endgroup$
            – A. Pongrácz
            Dec 29 '18 at 20:23












            $begingroup$
            Thank you! ${}$
            $endgroup$
            – Lucas Corrêa
            Dec 29 '18 at 20:28




            $begingroup$
            Thank you! ${}$
            $endgroup$
            – Lucas Corrêa
            Dec 29 '18 at 20:28











            2












            $begingroup$

            Hint: There is no need to go through $mathbb Q$. Just prove that if $y>x$ there are integers $n,m$ such that $y-x>m2^{-n}$.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Hint: There is no need to go through $mathbb Q$. Just prove that if $y>x$ there are integers $n,m$ such that $y-x>m2^{-n}$.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Hint: There is no need to go through $mathbb Q$. Just prove that if $y>x$ there are integers $n,m$ such that $y-x>m2^{-n}$.






                share|cite|improve this answer











                $endgroup$



                Hint: There is no need to go through $mathbb Q$. Just prove that if $y>x$ there are integers $n,m$ such that $y-x>m2^{-n}$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 29 '18 at 19:34

























                answered Dec 29 '18 at 19:29









                MatematletaMatematleta

                10.2k2918




                10.2k2918























                    1












                    $begingroup$

                    The real number $mathbb{R}$ and hence all subsets of the real numbers have come with a notion of distance $left(a, bright) to left|a - bright|$ .



                    In particular the rationals themselves form a metric space, see ProofWiki .



                    This means we can use the definition of dense set from Wikipedia that requires the existence of a metric.



                    Here's the definition of the closure of a set, slightly rewritten



                    $$ bar{A} stackrel{def}{=} A cup left{lim_{n to infty} a_n text{ where } a in text{Seq}left[Aright] right} tag{1} $$



                    Note that $bar{A}$ is constrained to be a subset of $mathbb{Q}$ because $mathbb{Q}$ is where we're taking limits.



                    $$ A text{ is dense in } X stackrel{def}{iff} bar{A} = X tag{2} $$



                    One way to show it directly is to come up with an explicit sequence of dyadic rationals that converges to a given rational number $frac{a}{b}$ for every such $frac{a}{b} in mathbb{Q}$ .






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      That is my first idea, but I cannot find such sequence.
                      $endgroup$
                      – Lucas Corrêa
                      Dec 29 '18 at 19:57










                    • $begingroup$
                      @LucasCorrêa ... what sort of hints are you okay with?
                      $endgroup$
                      – Gregory Nisbet
                      Dec 29 '18 at 19:58










                    • $begingroup$
                      My problem was in the construction of the sequence. My idea was take $d_{n} in D$ such thath $|d_{n} - a| < 1/n$ for some $a in mathbb{Q}$. But $mathbb{Q}$ is not complete, how can I ensure that $(d_{n})$ converges?
                      $endgroup$
                      – Lucas Corrêa
                      Dec 29 '18 at 20:06






                    • 1




                      $begingroup$
                      I think a sequence of dyadic rationals that converges to an irrational number in $mathbb{R}$ simply does not converge to anything in $mathbb{Q}$ .
                      $endgroup$
                      – Gregory Nisbet
                      Dec 29 '18 at 20:16
















                    1












                    $begingroup$

                    The real number $mathbb{R}$ and hence all subsets of the real numbers have come with a notion of distance $left(a, bright) to left|a - bright|$ .



                    In particular the rationals themselves form a metric space, see ProofWiki .



                    This means we can use the definition of dense set from Wikipedia that requires the existence of a metric.



                    Here's the definition of the closure of a set, slightly rewritten



                    $$ bar{A} stackrel{def}{=} A cup left{lim_{n to infty} a_n text{ where } a in text{Seq}left[Aright] right} tag{1} $$



                    Note that $bar{A}$ is constrained to be a subset of $mathbb{Q}$ because $mathbb{Q}$ is where we're taking limits.



                    $$ A text{ is dense in } X stackrel{def}{iff} bar{A} = X tag{2} $$



                    One way to show it directly is to come up with an explicit sequence of dyadic rationals that converges to a given rational number $frac{a}{b}$ for every such $frac{a}{b} in mathbb{Q}$ .






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      That is my first idea, but I cannot find such sequence.
                      $endgroup$
                      – Lucas Corrêa
                      Dec 29 '18 at 19:57










                    • $begingroup$
                      @LucasCorrêa ... what sort of hints are you okay with?
                      $endgroup$
                      – Gregory Nisbet
                      Dec 29 '18 at 19:58










                    • $begingroup$
                      My problem was in the construction of the sequence. My idea was take $d_{n} in D$ such thath $|d_{n} - a| < 1/n$ for some $a in mathbb{Q}$. But $mathbb{Q}$ is not complete, how can I ensure that $(d_{n})$ converges?
                      $endgroup$
                      – Lucas Corrêa
                      Dec 29 '18 at 20:06






                    • 1




                      $begingroup$
                      I think a sequence of dyadic rationals that converges to an irrational number in $mathbb{R}$ simply does not converge to anything in $mathbb{Q}$ .
                      $endgroup$
                      – Gregory Nisbet
                      Dec 29 '18 at 20:16














                    1












                    1








                    1





                    $begingroup$

                    The real number $mathbb{R}$ and hence all subsets of the real numbers have come with a notion of distance $left(a, bright) to left|a - bright|$ .



                    In particular the rationals themselves form a metric space, see ProofWiki .



                    This means we can use the definition of dense set from Wikipedia that requires the existence of a metric.



                    Here's the definition of the closure of a set, slightly rewritten



                    $$ bar{A} stackrel{def}{=} A cup left{lim_{n to infty} a_n text{ where } a in text{Seq}left[Aright] right} tag{1} $$



                    Note that $bar{A}$ is constrained to be a subset of $mathbb{Q}$ because $mathbb{Q}$ is where we're taking limits.



                    $$ A text{ is dense in } X stackrel{def}{iff} bar{A} = X tag{2} $$



                    One way to show it directly is to come up with an explicit sequence of dyadic rationals that converges to a given rational number $frac{a}{b}$ for every such $frac{a}{b} in mathbb{Q}$ .






                    share|cite|improve this answer









                    $endgroup$



                    The real number $mathbb{R}$ and hence all subsets of the real numbers have come with a notion of distance $left(a, bright) to left|a - bright|$ .



                    In particular the rationals themselves form a metric space, see ProofWiki .



                    This means we can use the definition of dense set from Wikipedia that requires the existence of a metric.



                    Here's the definition of the closure of a set, slightly rewritten



                    $$ bar{A} stackrel{def}{=} A cup left{lim_{n to infty} a_n text{ where } a in text{Seq}left[Aright] right} tag{1} $$



                    Note that $bar{A}$ is constrained to be a subset of $mathbb{Q}$ because $mathbb{Q}$ is where we're taking limits.



                    $$ A text{ is dense in } X stackrel{def}{iff} bar{A} = X tag{2} $$



                    One way to show it directly is to come up with an explicit sequence of dyadic rationals that converges to a given rational number $frac{a}{b}$ for every such $frac{a}{b} in mathbb{Q}$ .







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 29 '18 at 19:53









                    Gregory NisbetGregory Nisbet

                    561312




                    561312












                    • $begingroup$
                      That is my first idea, but I cannot find such sequence.
                      $endgroup$
                      – Lucas Corrêa
                      Dec 29 '18 at 19:57










                    • $begingroup$
                      @LucasCorrêa ... what sort of hints are you okay with?
                      $endgroup$
                      – Gregory Nisbet
                      Dec 29 '18 at 19:58










                    • $begingroup$
                      My problem was in the construction of the sequence. My idea was take $d_{n} in D$ such thath $|d_{n} - a| < 1/n$ for some $a in mathbb{Q}$. But $mathbb{Q}$ is not complete, how can I ensure that $(d_{n})$ converges?
                      $endgroup$
                      – Lucas Corrêa
                      Dec 29 '18 at 20:06






                    • 1




                      $begingroup$
                      I think a sequence of dyadic rationals that converges to an irrational number in $mathbb{R}$ simply does not converge to anything in $mathbb{Q}$ .
                      $endgroup$
                      – Gregory Nisbet
                      Dec 29 '18 at 20:16


















                    • $begingroup$
                      That is my first idea, but I cannot find such sequence.
                      $endgroup$
                      – Lucas Corrêa
                      Dec 29 '18 at 19:57










                    • $begingroup$
                      @LucasCorrêa ... what sort of hints are you okay with?
                      $endgroup$
                      – Gregory Nisbet
                      Dec 29 '18 at 19:58










                    • $begingroup$
                      My problem was in the construction of the sequence. My idea was take $d_{n} in D$ such thath $|d_{n} - a| < 1/n$ for some $a in mathbb{Q}$. But $mathbb{Q}$ is not complete, how can I ensure that $(d_{n})$ converges?
                      $endgroup$
                      – Lucas Corrêa
                      Dec 29 '18 at 20:06






                    • 1




                      $begingroup$
                      I think a sequence of dyadic rationals that converges to an irrational number in $mathbb{R}$ simply does not converge to anything in $mathbb{Q}$ .
                      $endgroup$
                      – Gregory Nisbet
                      Dec 29 '18 at 20:16
















                    $begingroup$
                    That is my first idea, but I cannot find such sequence.
                    $endgroup$
                    – Lucas Corrêa
                    Dec 29 '18 at 19:57




                    $begingroup$
                    That is my first idea, but I cannot find such sequence.
                    $endgroup$
                    – Lucas Corrêa
                    Dec 29 '18 at 19:57












                    $begingroup$
                    @LucasCorrêa ... what sort of hints are you okay with?
                    $endgroup$
                    – Gregory Nisbet
                    Dec 29 '18 at 19:58




                    $begingroup$
                    @LucasCorrêa ... what sort of hints are you okay with?
                    $endgroup$
                    – Gregory Nisbet
                    Dec 29 '18 at 19:58












                    $begingroup$
                    My problem was in the construction of the sequence. My idea was take $d_{n} in D$ such thath $|d_{n} - a| < 1/n$ for some $a in mathbb{Q}$. But $mathbb{Q}$ is not complete, how can I ensure that $(d_{n})$ converges?
                    $endgroup$
                    – Lucas Corrêa
                    Dec 29 '18 at 20:06




                    $begingroup$
                    My problem was in the construction of the sequence. My idea was take $d_{n} in D$ such thath $|d_{n} - a| < 1/n$ for some $a in mathbb{Q}$. But $mathbb{Q}$ is not complete, how can I ensure that $(d_{n})$ converges?
                    $endgroup$
                    – Lucas Corrêa
                    Dec 29 '18 at 20:06




                    1




                    1




                    $begingroup$
                    I think a sequence of dyadic rationals that converges to an irrational number in $mathbb{R}$ simply does not converge to anything in $mathbb{Q}$ .
                    $endgroup$
                    – Gregory Nisbet
                    Dec 29 '18 at 20:16




                    $begingroup$
                    I think a sequence of dyadic rationals that converges to an irrational number in $mathbb{R}$ simply does not converge to anything in $mathbb{Q}$ .
                    $endgroup$
                    – Gregory Nisbet
                    Dec 29 '18 at 20:16


















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