Proof of $2^n > n$ by Induction












3












$begingroup$


I'm new to induction and trying to prove $2^n > n$ for all natural numbers.



I've seen a couple of examples but am confused about the the case going from $k = 1$ to $k =2$.



So I show $2^1 > 1$ as the base case.



Then I assume $2^k > k$



Meaning that



$2.2^k > 2k$



i.e.



$2^{k+1} > 2k$



Or



$2^{k+1} > k + k > k + 1$



So it is considered proven.



But when $k = 1$, $k + k not> k + 1$



What am I missing please?



Do I need a special case for going from $k=1$ to $k=2$?






inequality induction







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+kgeq k+1$ for $kgeq1$ is probably nicer though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 9:45


















3












$begingroup$


I'm new to induction and trying to prove $2^n > n$ for all natural numbers.



I've seen a couple of examples but am confused about the the case going from $k = 1$ to $k =2$.



So I show $2^1 > 1$ as the base case.



Then I assume $2^k > k$



Meaning that



$2.2^k > 2k$



i.e.



$2^{k+1} > 2k$



Or



$2^{k+1} > k + k > k + 1$



So it is considered proven.



But when $k = 1$, $k + k not> k + 1$



What am I missing please?



Do I need a special case for going from $k=1$ to $k=2$?






inequality induction







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+kgeq k+1$ for $kgeq1$ is probably nicer though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 9:45
















3












3








3


0



$begingroup$


I'm new to induction and trying to prove $2^n > n$ for all natural numbers.



I've seen a couple of examples but am confused about the the case going from $k = 1$ to $k =2$.



So I show $2^1 > 1$ as the base case.



Then I assume $2^k > k$



Meaning that



$2.2^k > 2k$



i.e.



$2^{k+1} > 2k$



Or



$2^{k+1} > k + k > k + 1$



So it is considered proven.



But when $k = 1$, $k + k not> k + 1$



What am I missing please?



Do I need a special case for going from $k=1$ to $k=2$?






inequality induction







share|cite|improve this question









$endgroup$




I'm new to induction and trying to prove $2^n > n$ for all natural numbers.



I've seen a couple of examples but am confused about the the case going from $k = 1$ to $k =2$.



So I show $2^1 > 1$ as the base case.



Then I assume $2^k > k$



Meaning that



$2.2^k > 2k$



i.e.



$2^{k+1} > 2k$



Or



$2^{k+1} > k + k > k + 1$



So it is considered proven.



But when $k = 1$, $k + k not> k + 1$



What am I missing please?



Do I need a special case for going from $k=1$ to $k=2$?






inequality induction




inequality induction






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 9:42









RobinRobin

1577




1577








  • 1




    $begingroup$
    Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+kgeq k+1$ for $kgeq1$ is probably nicer though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 9:45
















  • 1




    $begingroup$
    Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+kgeq k+1$ for $kgeq1$ is probably nicer though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 9:45










1




1




$begingroup$
Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+kgeq k+1$ for $kgeq1$ is probably nicer though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:45






$begingroup$
Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+kgeq k+1$ for $kgeq1$ is probably nicer though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:45












2 Answers
2






active

oldest

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4












$begingroup$

Technically, the statement $$2^{k+1} > k+k>k+1$$ is wrong, precisely because $k$ could be $1$. You can, however, write
$$2^{k+1} > k+k geq k+1$$



and from that, you can still conclude that $2^{k+1}>k+1$. No special case needed, since $a>b$ and $bgeq c$ always implies $a>c$.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    No, you don't need a special case. $2^{k+1} >2k $ and $2k geq k+1$ together impliy $2^{k+1} >k+1 $






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Technically, the statement $$2^{k+1} > k+k>k+1$$ is wrong, precisely because $k$ could be $1$. You can, however, write
      $$2^{k+1} > k+k geq k+1$$



      and from that, you can still conclude that $2^{k+1}>k+1$. No special case needed, since $a>b$ and $bgeq c$ always implies $a>c$.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Technically, the statement $$2^{k+1} > k+k>k+1$$ is wrong, precisely because $k$ could be $1$. You can, however, write
        $$2^{k+1} > k+k geq k+1$$



        and from that, you can still conclude that $2^{k+1}>k+1$. No special case needed, since $a>b$ and $bgeq c$ always implies $a>c$.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Technically, the statement $$2^{k+1} > k+k>k+1$$ is wrong, precisely because $k$ could be $1$. You can, however, write
          $$2^{k+1} > k+k geq k+1$$



          and from that, you can still conclude that $2^{k+1}>k+1$. No special case needed, since $a>b$ and $bgeq c$ always implies $a>c$.






          share|cite|improve this answer









          $endgroup$



          Technically, the statement $$2^{k+1} > k+k>k+1$$ is wrong, precisely because $k$ could be $1$. You can, however, write
          $$2^{k+1} > k+k geq k+1$$



          and from that, you can still conclude that $2^{k+1}>k+1$. No special case needed, since $a>b$ and $bgeq c$ always implies $a>c$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 9:45









          5xum5xum

          91.8k394161




          91.8k394161























              4












              $begingroup$

              No, you don't need a special case. $2^{k+1} >2k $ and $2k geq k+1$ together impliy $2^{k+1} >k+1 $






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                No, you don't need a special case. $2^{k+1} >2k $ and $2k geq k+1$ together impliy $2^{k+1} >k+1 $






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  No, you don't need a special case. $2^{k+1} >2k $ and $2k geq k+1$ together impliy $2^{k+1} >k+1 $






                  share|cite|improve this answer









                  $endgroup$



                  No, you don't need a special case. $2^{k+1} >2k $ and $2k geq k+1$ together impliy $2^{k+1} >k+1 $







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 9:45









                  Kavi Rama MurthyKavi Rama Murthy

                  71.7k53170




                  71.7k53170






























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