Proof that $e$ is irrational. (Proof verification)












2












$begingroup$


Here, $e = sum_{k=0}^infty 1/k!$. Define $S_n$ as:



begin{align}
S_n = n!e - n!sum_{k=0}^n frac{1}{k!}
end{align}
where $n!sum_{k=0}^n frac{1}{k!}$ is an integer. Write $e = 1/0!+1/1!+...+1/n!+1/(n+1)!+...$
begin{align}
S_n &= n!left(e - sum_{k=0}^n frac{1}{k!}right) = n!left( frac{1}{0!}+
frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - sum_{k=0}^n frac{1}{k!}right)\
&=n!left( frac{1}{0!}+
frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - frac{1}{0!} - frac{1}{1!} - ... - frac{1}{n!}
right)\
&=n!left( frac{1}{(n+1)!}+frac{1}{(n+2)!}+... right) \
&= frac{1}{(n+1)}+frac{1}{(n+2)(n+1)}+... < frac{1}{(n+1)}+frac{1}{(n+1)^2}+... = frac{1}{n}
end{align}



So, $0<S_n<1/n$. Now, if $e=p/q$ then we would have an integer between $0$ and $1$ for a large $n$.



$$0 < n!p - qn!sum_{k=0}^n frac{1}{k!}<frac{q}{n}$$



Is this correct?






proof-verification irrational-numbers







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Here, $e = sum_{k=0}^infty 1/k!$. Define $S_n$ as:



    begin{align}
    S_n = n!e - n!sum_{k=0}^n frac{1}{k!}
    end{align}
    where $n!sum_{k=0}^n frac{1}{k!}$ is an integer. Write $e = 1/0!+1/1!+...+1/n!+1/(n+1)!+...$
    begin{align}
    S_n &= n!left(e - sum_{k=0}^n frac{1}{k!}right) = n!left( frac{1}{0!}+
    frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - sum_{k=0}^n frac{1}{k!}right)\
    &=n!left( frac{1}{0!}+
    frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - frac{1}{0!} - frac{1}{1!} - ... - frac{1}{n!}
    right)\
    &=n!left( frac{1}{(n+1)!}+frac{1}{(n+2)!}+... right) \
    &= frac{1}{(n+1)}+frac{1}{(n+2)(n+1)}+... < frac{1}{(n+1)}+frac{1}{(n+1)^2}+... = frac{1}{n}
    end{align}



    So, $0<S_n<1/n$. Now, if $e=p/q$ then we would have an integer between $0$ and $1$ for a large $n$.



    $$0 < n!p - qn!sum_{k=0}^n frac{1}{k!}<frac{q}{n}$$



    Is this correct?






    proof-verification irrational-numbers







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      2



      $begingroup$


      Here, $e = sum_{k=0}^infty 1/k!$. Define $S_n$ as:



      begin{align}
      S_n = n!e - n!sum_{k=0}^n frac{1}{k!}
      end{align}
      where $n!sum_{k=0}^n frac{1}{k!}$ is an integer. Write $e = 1/0!+1/1!+...+1/n!+1/(n+1)!+...$
      begin{align}
      S_n &= n!left(e - sum_{k=0}^n frac{1}{k!}right) = n!left( frac{1}{0!}+
      frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - sum_{k=0}^n frac{1}{k!}right)\
      &=n!left( frac{1}{0!}+
      frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - frac{1}{0!} - frac{1}{1!} - ... - frac{1}{n!}
      right)\
      &=n!left( frac{1}{(n+1)!}+frac{1}{(n+2)!}+... right) \
      &= frac{1}{(n+1)}+frac{1}{(n+2)(n+1)}+... < frac{1}{(n+1)}+frac{1}{(n+1)^2}+... = frac{1}{n}
      end{align}



      So, $0<S_n<1/n$. Now, if $e=p/q$ then we would have an integer between $0$ and $1$ for a large $n$.



      $$0 < n!p - qn!sum_{k=0}^n frac{1}{k!}<frac{q}{n}$$



      Is this correct?






      proof-verification irrational-numbers







      share|cite|improve this question











      $endgroup$




      Here, $e = sum_{k=0}^infty 1/k!$. Define $S_n$ as:



      begin{align}
      S_n = n!e - n!sum_{k=0}^n frac{1}{k!}
      end{align}
      where $n!sum_{k=0}^n frac{1}{k!}$ is an integer. Write $e = 1/0!+1/1!+...+1/n!+1/(n+1)!+...$
      begin{align}
      S_n &= n!left(e - sum_{k=0}^n frac{1}{k!}right) = n!left( frac{1}{0!}+
      frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - sum_{k=0}^n frac{1}{k!}right)\
      &=n!left( frac{1}{0!}+
      frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - frac{1}{0!} - frac{1}{1!} - ... - frac{1}{n!}
      right)\
      &=n!left( frac{1}{(n+1)!}+frac{1}{(n+2)!}+... right) \
      &= frac{1}{(n+1)}+frac{1}{(n+2)(n+1)}+... < frac{1}{(n+1)}+frac{1}{(n+1)^2}+... = frac{1}{n}
      end{align}



      So, $0<S_n<1/n$. Now, if $e=p/q$ then we would have an integer between $0$ and $1$ for a large $n$.



      $$0 < n!p - qn!sum_{k=0}^n frac{1}{k!}<frac{q}{n}$$



      Is this correct?






      proof-verification irrational-numbers




      proof-verification irrational-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 7:59









      Eevee Trainer

      7,59721338




      7,59721338










      asked Jan 6 at 7:43









      PintecoPinteco

      731313




      731313






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          It's basically fine. To tie up the details, you should specify exactly how large $n$ needs to be.



          Also, writing "$0<S_n =$ (something that isn't equal to $S_n$) is bad notation. I'm not even sure exactly what you were trying to say there. If you're not sure how to write what you mean in symbols, use some words to clear things up.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
            $endgroup$
            – Eevee Trainer
            Jan 6 at 7:55










          • $begingroup$
            I've fixed that part, thanks.
            $endgroup$
            – Pinteco
            Jan 6 at 7:56



















          1












          $begingroup$

          Your proof seems sufficient enough, as far as I can tell. You did make a slight typo in the final inequality (multiplying through by $q$ would mean you have $qS_n$ there); this was touched on in another answer and you already edited your answer to rectify this.



          If I had to nitpick anything of significance...



          Personally it could use a lot of elaboration to explain various manipulations and results - where they come from and so on. I found myself having to fill in the holes myself; it can make it a bit difficult for someone to see where you're going.



          But this may just be a personal issue. If nothing else, at least everything seems to logically follow. But taking some time to explain everything for the reader will help. (In particular if this is a proof for a class or something: no sense in getting points taken off because you thought it was obvious what was going on.)






          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









            2












            $begingroup$

            It's basically fine. To tie up the details, you should specify exactly how large $n$ needs to be.



            Also, writing "$0<S_n =$ (something that isn't equal to $S_n$) is bad notation. I'm not even sure exactly what you were trying to say there. If you're not sure how to write what you mean in symbols, use some words to clear things up.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
              $endgroup$
              – Eevee Trainer
              Jan 6 at 7:55










            • $begingroup$
              I've fixed that part, thanks.
              $endgroup$
              – Pinteco
              Jan 6 at 7:56
















            2












            $begingroup$

            It's basically fine. To tie up the details, you should specify exactly how large $n$ needs to be.



            Also, writing "$0<S_n =$ (something that isn't equal to $S_n$) is bad notation. I'm not even sure exactly what you were trying to say there. If you're not sure how to write what you mean in symbols, use some words to clear things up.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
              $endgroup$
              – Eevee Trainer
              Jan 6 at 7:55










            • $begingroup$
              I've fixed that part, thanks.
              $endgroup$
              – Pinteco
              Jan 6 at 7:56














            2












            2








            2





            $begingroup$

            It's basically fine. To tie up the details, you should specify exactly how large $n$ needs to be.



            Also, writing "$0<S_n =$ (something that isn't equal to $S_n$) is bad notation. I'm not even sure exactly what you were trying to say there. If you're not sure how to write what you mean in symbols, use some words to clear things up.






            share|cite|improve this answer









            $endgroup$



            It's basically fine. To tie up the details, you should specify exactly how large $n$ needs to be.



            Also, writing "$0<S_n =$ (something that isn't equal to $S_n$) is bad notation. I'm not even sure exactly what you were trying to say there. If you're not sure how to write what you mean in symbols, use some words to clear things up.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 7:53









            jmerryjmerry

            12.8k1628




            12.8k1628












            • $begingroup$
              I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
              $endgroup$
              – Eevee Trainer
              Jan 6 at 7:55










            • $begingroup$
              I've fixed that part, thanks.
              $endgroup$
              – Pinteco
              Jan 6 at 7:56


















            • $begingroup$
              I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
              $endgroup$
              – Eevee Trainer
              Jan 6 at 7:55










            • $begingroup$
              I've fixed that part, thanks.
              $endgroup$
              – Pinteco
              Jan 6 at 7:56
















            $begingroup$
            I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
            $endgroup$
            – Eevee Trainer
            Jan 6 at 7:55




            $begingroup$
            I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
            $endgroup$
            – Eevee Trainer
            Jan 6 at 7:55












            $begingroup$
            I've fixed that part, thanks.
            $endgroup$
            – Pinteco
            Jan 6 at 7:56




            $begingroup$
            I've fixed that part, thanks.
            $endgroup$
            – Pinteco
            Jan 6 at 7:56











            1












            $begingroup$

            Your proof seems sufficient enough, as far as I can tell. You did make a slight typo in the final inequality (multiplying through by $q$ would mean you have $qS_n$ there); this was touched on in another answer and you already edited your answer to rectify this.



            If I had to nitpick anything of significance...



            Personally it could use a lot of elaboration to explain various manipulations and results - where they come from and so on. I found myself having to fill in the holes myself; it can make it a bit difficult for someone to see where you're going.



            But this may just be a personal issue. If nothing else, at least everything seems to logically follow. But taking some time to explain everything for the reader will help. (In particular if this is a proof for a class or something: no sense in getting points taken off because you thought it was obvious what was going on.)






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Your proof seems sufficient enough, as far as I can tell. You did make a slight typo in the final inequality (multiplying through by $q$ would mean you have $qS_n$ there); this was touched on in another answer and you already edited your answer to rectify this.



              If I had to nitpick anything of significance...



              Personally it could use a lot of elaboration to explain various manipulations and results - where they come from and so on. I found myself having to fill in the holes myself; it can make it a bit difficult for someone to see where you're going.



              But this may just be a personal issue. If nothing else, at least everything seems to logically follow. But taking some time to explain everything for the reader will help. (In particular if this is a proof for a class or something: no sense in getting points taken off because you thought it was obvious what was going on.)






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Your proof seems sufficient enough, as far as I can tell. You did make a slight typo in the final inequality (multiplying through by $q$ would mean you have $qS_n$ there); this was touched on in another answer and you already edited your answer to rectify this.



                If I had to nitpick anything of significance...



                Personally it could use a lot of elaboration to explain various manipulations and results - where they come from and so on. I found myself having to fill in the holes myself; it can make it a bit difficult for someone to see where you're going.



                But this may just be a personal issue. If nothing else, at least everything seems to logically follow. But taking some time to explain everything for the reader will help. (In particular if this is a proof for a class or something: no sense in getting points taken off because you thought it was obvious what was going on.)






                share|cite|improve this answer









                $endgroup$



                Your proof seems sufficient enough, as far as I can tell. You did make a slight typo in the final inequality (multiplying through by $q$ would mean you have $qS_n$ there); this was touched on in another answer and you already edited your answer to rectify this.



                If I had to nitpick anything of significance...



                Personally it could use a lot of elaboration to explain various manipulations and results - where they come from and so on. I found myself having to fill in the holes myself; it can make it a bit difficult for someone to see where you're going.



                But this may just be a personal issue. If nothing else, at least everything seems to logically follow. But taking some time to explain everything for the reader will help. (In particular if this is a proof for a class or something: no sense in getting points taken off because you thought it was obvious what was going on.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 7:57









                Eevee TrainerEevee Trainer

                7,59721338




                7,59721338






























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