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?










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












$begingroup$


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?










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?










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






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

votes


















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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023784%2fprove-x-geq-2-implies-xn-geq-2n%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023784%2fprove-x-geq-2-implies-xn-geq-2n%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Plaza Victoria

                      Puebla de Zaragoza

                      Musa