Prove $x geq 2$ implies $x^{n} geq 2^{n}$












0












$begingroup$


Prove $x geq 2$ implies $x^{n} geq 2^{n}$.



By induction. Clearly it holds for $n = 1$ by the assumption. Now assume $x geq 2$ and $x^{k} geq 2^{k}$ for some $k in mathbb{N}$.



Then combine the assumption inequality and $x geq 2$ to get $x cdot x^{k} geq 2 cdot 2^{k}$, which gives our result.



Is it correct?






induction







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$endgroup$








  • 1




    $begingroup$
    It is correct. If you wanted to be a little more thorough you could say $xcdot x^kgeq 2cdot x^kgeq 2cdot 2^k,$ but I personally don't think that step is necessary.
    $endgroup$
    – Melody
    Dec 3 '18 at 8:36












  • $begingroup$
    If you are a beginner at the university, you should :)
    $endgroup$
    – Stockfish
    Dec 3 '18 at 8:55
















0












$begingroup$


Prove $x geq 2$ implies $x^{n} geq 2^{n}$.



By induction. Clearly it holds for $n = 1$ by the assumption. Now assume $x geq 2$ and $x^{k} geq 2^{k}$ for some $k in mathbb{N}$.



Then combine the assumption inequality and $x geq 2$ to get $x cdot x^{k} geq 2 cdot 2^{k}$, which gives our result.



Is it correct?






induction







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It is correct. If you wanted to be a little more thorough you could say $xcdot x^kgeq 2cdot x^kgeq 2cdot 2^k,$ but I personally don't think that step is necessary.
    $endgroup$
    – Melody
    Dec 3 '18 at 8:36












  • $begingroup$
    If you are a beginner at the university, you should :)
    $endgroup$
    – Stockfish
    Dec 3 '18 at 8:55














0












0








0





$begingroup$


Prove $x geq 2$ implies $x^{n} geq 2^{n}$.



By induction. Clearly it holds for $n = 1$ by the assumption. Now assume $x geq 2$ and $x^{k} geq 2^{k}$ for some $k in mathbb{N}$.



Then combine the assumption inequality and $x geq 2$ to get $x cdot x^{k} geq 2 cdot 2^{k}$, which gives our result.



Is it correct?






induction







share|cite|improve this question









$endgroup$




Prove $x geq 2$ implies $x^{n} geq 2^{n}$.



By induction. Clearly it holds for $n = 1$ by the assumption. Now assume $x geq 2$ and $x^{k} geq 2^{k}$ for some $k in mathbb{N}$.



Then combine the assumption inequality and $x geq 2$ to get $x cdot x^{k} geq 2 cdot 2^{k}$, which gives our result.



Is it correct?






induction




induction






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 8:34









josephjoseph

500111




500111








  • 1




    $begingroup$
    It is correct. If you wanted to be a little more thorough you could say $xcdot x^kgeq 2cdot x^kgeq 2cdot 2^k,$ but I personally don't think that step is necessary.
    $endgroup$
    – Melody
    Dec 3 '18 at 8:36












  • $begingroup$
    If you are a beginner at the university, you should :)
    $endgroup$
    – Stockfish
    Dec 3 '18 at 8:55














  • 1




    $begingroup$
    It is correct. If you wanted to be a little more thorough you could say $xcdot x^kgeq 2cdot x^kgeq 2cdot 2^k,$ but I personally don't think that step is necessary.
    $endgroup$
    – Melody
    Dec 3 '18 at 8:36












  • $begingroup$
    If you are a beginner at the university, you should :)
    $endgroup$
    – Stockfish
    Dec 3 '18 at 8:55








1




1




$begingroup$
It is correct. If you wanted to be a little more thorough you could say $xcdot x^kgeq 2cdot x^kgeq 2cdot 2^k,$ but I personally don't think that step is necessary.
$endgroup$
– Melody
Dec 3 '18 at 8:36






$begingroup$
It is correct. If you wanted to be a little more thorough you could say $xcdot x^kgeq 2cdot x^kgeq 2cdot 2^k,$ but I personally don't think that step is necessary.
$endgroup$
– Melody
Dec 3 '18 at 8:36














$begingroup$
If you are a beginner at the university, you should :)
$endgroup$
– Stockfish
Dec 3 '18 at 8:55




$begingroup$
If you are a beginner at the university, you should :)
$endgroup$
– Stockfish
Dec 3 '18 at 8:55










2 Answers
2






active

oldest

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1












$begingroup$

Functions $F(x) = A^x$ are always increasing as long as $A > 1$. Thus, with $A=x$ and $x=n$, we have



$$ x geq 2 implies x^n geq 2^n $$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    begin{align}
    x^{k+1} &=x(x^k) \
    &ge x(2^k) text{, since } x>0.\
    &ge 2(2^k)text{, since } xge2 text{ and } 2^k >0\
    &ge 2^{k+1}
    end{align}






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
      2






      active

      oldest

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      active

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      active

      oldest

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      1












      $begingroup$

      Functions $F(x) = A^x$ are always increasing as long as $A > 1$. Thus, with $A=x$ and $x=n$, we have



      $$ x geq 2 implies x^n geq 2^n $$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Functions $F(x) = A^x$ are always increasing as long as $A > 1$. Thus, with $A=x$ and $x=n$, we have



        $$ x geq 2 implies x^n geq 2^n $$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Functions $F(x) = A^x$ are always increasing as long as $A > 1$. Thus, with $A=x$ and $x=n$, we have



          $$ x geq 2 implies x^n geq 2^n $$






          share|cite|improve this answer









          $endgroup$



          Functions $F(x) = A^x$ are always increasing as long as $A > 1$. Thus, with $A=x$ and $x=n$, we have



          $$ x geq 2 implies x^n geq 2^n $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 8:37









          Jimmy SabaterJimmy Sabater

          2,662321




          2,662321























              1












              $begingroup$

              begin{align}
              x^{k+1} &=x(x^k) \
              &ge x(2^k) text{, since } x>0.\
              &ge 2(2^k)text{, since } xge2 text{ and } 2^k >0\
              &ge 2^{k+1}
              end{align}






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                begin{align}
                x^{k+1} &=x(x^k) \
                &ge x(2^k) text{, since } x>0.\
                &ge 2(2^k)text{, since } xge2 text{ and } 2^k >0\
                &ge 2^{k+1}
                end{align}






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  begin{align}
                  x^{k+1} &=x(x^k) \
                  &ge x(2^k) text{, since } x>0.\
                  &ge 2(2^k)text{, since } xge2 text{ and } 2^k >0\
                  &ge 2^{k+1}
                  end{align}






                  share|cite|improve this answer









                  $endgroup$



                  begin{align}
                  x^{k+1} &=x(x^k) \
                  &ge x(2^k) text{, since } x>0.\
                  &ge 2(2^k)text{, since } xge2 text{ and } 2^k >0\
                  &ge 2^{k+1}
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 3 '18 at 8:39









                  Siong Thye GohSiong Thye Goh

                  100k1466117




                  100k1466117






























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