Proving that $g(x)$ defined as $x^2$ on the rationals and $x^4$ on irrationals, is discontinuous at $2$












1












$begingroup$


Define $g: mathbb{R} to mathbb{R}$ by $$g(x) =begin{cases} x^2 & text{if } x text{ is rational}, \\ x^4 & text{if } x text{ is irrational}. end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.



Solution Attempt:



Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_{xto2}g(x)=4$;



so taking $epsilon=1$, there is a $delta>0$ such that if $0<|x-2|<delta$, then $big|g(x)-4big|<1$.



Therefore if $x$ is irrational and $2-delta<x<2$, then
$big|x^4-4big|<1$



$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac{1}{4} implies x < 2$ Contradiction






real-analysis







share|cite|improve this question









New contributor




SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
    $endgroup$
    – Julian Mejia
    2 days ago








  • 1




    $begingroup$
    $x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
    $endgroup$
    – Julian Mejia
    2 days ago
















1












$begingroup$


Define $g: mathbb{R} to mathbb{R}$ by $$g(x) =begin{cases} x^2 & text{if } x text{ is rational}, \\ x^4 & text{if } x text{ is irrational}. end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.



Solution Attempt:



Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_{xto2}g(x)=4$;



so taking $epsilon=1$, there is a $delta>0$ such that if $0<|x-2|<delta$, then $big|g(x)-4big|<1$.



Therefore if $x$ is irrational and $2-delta<x<2$, then
$big|x^4-4big|<1$



$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac{1}{4} implies x < 2$ Contradiction






real-analysis







share|cite|improve this question









New contributor




SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
    $endgroup$
    – Julian Mejia
    2 days ago








  • 1




    $begingroup$
    $x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
    $endgroup$
    – Julian Mejia
    2 days ago














1












1








1





$begingroup$


Define $g: mathbb{R} to mathbb{R}$ by $$g(x) =begin{cases} x^2 & text{if } x text{ is rational}, \\ x^4 & text{if } x text{ is irrational}. end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.



Solution Attempt:



Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_{xto2}g(x)=4$;



so taking $epsilon=1$, there is a $delta>0$ such that if $0<|x-2|<delta$, then $big|g(x)-4big|<1$.



Therefore if $x$ is irrational and $2-delta<x<2$, then
$big|x^4-4big|<1$



$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac{1}{4} implies x < 2$ Contradiction






real-analysis







share|cite|improve this question









New contributor




SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Define $g: mathbb{R} to mathbb{R}$ by $$g(x) =begin{cases} x^2 & text{if } x text{ is rational}, \\ x^4 & text{if } x text{ is irrational}. end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.



Solution Attempt:



Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_{xto2}g(x)=4$;



so taking $epsilon=1$, there is a $delta>0$ such that if $0<|x-2|<delta$, then $big|g(x)-4big|<1$.



Therefore if $x$ is irrational and $2-delta<x<2$, then
$big|x^4-4big|<1$



$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac{1}{4} implies x < 2$ Contradiction






real-analysis




real-analysis






share|cite|improve this question









New contributor




SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Asaf Karagila♦

308k33441774




308k33441774






New contributor




SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









SHajsSHajs

162




162




New contributor




SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
    $endgroup$
    – Julian Mejia
    2 days ago








  • 1




    $begingroup$
    $x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
    $endgroup$
    – Julian Mejia
    2 days ago














  • 1




    $begingroup$
    Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
    $endgroup$
    – Julian Mejia
    2 days ago








  • 1




    $begingroup$
    $x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
    $endgroup$
    – Julian Mejia
    2 days ago








1




1




$begingroup$
Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
$endgroup$
– Julian Mejia
2 days ago






$begingroup$
Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
$endgroup$
– Julian Mejia
2 days ago






1




1




$begingroup$
$x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
$endgroup$
– Julian Mejia
2 days ago




$begingroup$
$x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
$endgroup$
– Julian Mejia
2 days ago










4 Answers
4






active

oldest

votes


















1












$begingroup$

Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
$$-sqrt[4]{5} < x < sqrt[4]{5}.$$
If you had chosen instead an irrational $x$ such that
$$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



    Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




    Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.







    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      An option?



      Sequential definition of continuity .



      $x_n=2(4n)^{(1/(4n))}.$



      $lim_{n rightarrow infty}x_n=2.$



      $lim_{n rightarrow infty}g(x_n)=$



      $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



      $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



      $=2^4 not =g(2)$.






      share|cite|improve this answer











      $endgroup$





















        0












        $begingroup$

        You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



        $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



        Observe the above sequence is irrational:



        $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



        yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...






        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
          });


          }
          });






          SHajs is a new contributor. Be nice, and check out our Code of Conduct.










          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3177808%2fproving-that-gx-defined-as-x2-on-the-rationals-and-x4-on-irrationals%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
          $$-sqrt[4]{5} < x < sqrt[4]{5}.$$
          If you had chosen instead an irrational $x$ such that
          $$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
          then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.






          share|cite|improve this answer









          $endgroup$


















            1












            $begingroup$

            Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
            $$-sqrt[4]{5} < x < sqrt[4]{5}.$$
            If you had chosen instead an irrational $x$ such that
            $$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
            then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.






            share|cite|improve this answer









            $endgroup$
















              1












              1








              1





              $begingroup$

              Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
              $$-sqrt[4]{5} < x < sqrt[4]{5}.$$
              If you had chosen instead an irrational $x$ such that
              $$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
              then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.






              share|cite|improve this answer









              $endgroup$



              Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
              $$-sqrt[4]{5} < x < sqrt[4]{5}.$$
              If you had chosen instead an irrational $x$ such that
              $$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
              then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 2 days ago









              Theo BenditTheo Bendit

              20.8k12355




              20.8k12355























                  1












                  $begingroup$

                  I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



                  Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




                  Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.







                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



                    Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




                    Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.







                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



                      Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




                      Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.







                      share|cite|improve this answer









                      $endgroup$



                      I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



                      Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




                      Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.








                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 days ago









                      zongxiang yizongxiang yi

                      352110




                      352110























                          1












                          $begingroup$

                          An option?



                          Sequential definition of continuity .



                          $x_n=2(4n)^{(1/(4n))}.$



                          $lim_{n rightarrow infty}x_n=2.$



                          $lim_{n rightarrow infty}g(x_n)=$



                          $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



                          $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



                          $=2^4 not =g(2)$.






                          share|cite|improve this answer











                          $endgroup$


















                            1












                            $begingroup$

                            An option?



                            Sequential definition of continuity .



                            $x_n=2(4n)^{(1/(4n))}.$



                            $lim_{n rightarrow infty}x_n=2.$



                            $lim_{n rightarrow infty}g(x_n)=$



                            $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



                            $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



                            $=2^4 not =g(2)$.






                            share|cite|improve this answer











                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              An option?



                              Sequential definition of continuity .



                              $x_n=2(4n)^{(1/(4n))}.$



                              $lim_{n rightarrow infty}x_n=2.$



                              $lim_{n rightarrow infty}g(x_n)=$



                              $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



                              $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



                              $=2^4 not =g(2)$.






                              share|cite|improve this answer











                              $endgroup$



                              An option?



                              Sequential definition of continuity .



                              $x_n=2(4n)^{(1/(4n))}.$



                              $lim_{n rightarrow infty}x_n=2.$



                              $lim_{n rightarrow infty}g(x_n)=$



                              $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



                              $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



                              $=2^4 not =g(2)$.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 2 days ago

























                              answered 2 days ago









                              Peter SzilasPeter Szilas

                              11.9k2822




                              11.9k2822























                                  0












                                  $begingroup$

                                  You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



                                  $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



                                  Observe the above sequence is irrational:



                                  $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



                                  yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



                                    $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



                                    Observe the above sequence is irrational:



                                    $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



                                    yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



                                      $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



                                      Observe the above sequence is irrational:



                                      $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



                                      yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...






                                      share|cite|improve this answer









                                      $endgroup$



                                      You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



                                      $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



                                      Observe the above sequence is irrational:



                                      $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



                                      yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 2 days ago









                                      DonAntonioDonAntonio

                                      180k1494233




                                      180k1494233






















                                          SHajs is a new contributor. Be nice, and check out our Code of Conduct.










                                          draft saved

                                          draft discarded


















                                          SHajs is a new contributor. Be nice, and check out our Code of Conduct.













                                          SHajs is a new contributor. Be nice, and check out our Code of Conduct.












                                          SHajs is a new contributor. Be nice, and check out our Code of Conduct.
















                                          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%2f3177808%2fproving-that-gx-defined-as-x2-on-the-rationals-and-x4-on-irrationals%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

                                          數位音樂下載

                                          Image
                                          數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
                                          Read more

                                          When can things happen in Etherscan, such as the picture below?

                                          Image
                                          2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
                                          Read more

                                          格利澤436b

                                          Image
                                          格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
                                          Read more