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












0














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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  • 1




    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.
    – Melody
    Dec 3 '18 at 8:36












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
















0














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


















  • 1




    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.
    – Melody
    Dec 3 '18 at 8:36












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














0












0








0







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













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






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share|cite|improve this question











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share|cite|improve this question










asked Dec 3 '18 at 8:34









joseph

4329




4329








  • 1




    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.
    – Melody
    Dec 3 '18 at 8:36












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














  • 1




    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.
    – Melody
    Dec 3 '18 at 8:36












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








1




1




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.
– Melody
Dec 3 '18 at 8:36






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.
– Melody
Dec 3 '18 at 8:36














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




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










2 Answers
2






active

oldest

votes


















1














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





























    1














    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





















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      2 Answers
      2






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      2 Answers
      2






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      active

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      active

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      1














      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


























        1














        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
























          1












          1








          1






          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












          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 Sabater

          1,951219




          1,951219























              1














              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


























                1














                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
























                  1












                  1








                  1






                  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












                  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 Goh

                  99.5k1464117




                  99.5k1464117






























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