What's wrong with this bogus proof?












4












$begingroup$


enter image description here



What is the mistake here? Is it matter of the unit?






discrete-mathematics proof-verification







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    Yes, the units don’t match across the 2nd equals sign
    $endgroup$
    – Alex
    3 hours ago






  • 2




    $begingroup$
    Yes, you have to square the unit. Conversion of squared units is different: if 100 cents is a dollar, then $100^2$ cents squared is a dollar squred.
    $endgroup$
    – Dean Young
    3 hours ago


















4












$begingroup$


enter image description here



What is the mistake here? Is it matter of the unit?






discrete-mathematics proof-verification







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    Yes, the units don’t match across the 2nd equals sign
    $endgroup$
    – Alex
    3 hours ago






  • 2




    $begingroup$
    Yes, you have to square the unit. Conversion of squared units is different: if 100 cents is a dollar, then $100^2$ cents squared is a dollar squred.
    $endgroup$
    – Dean Young
    3 hours ago
















4












4








4





$begingroup$


enter image description here



What is the mistake here? Is it matter of the unit?






discrete-mathematics proof-verification







share|cite|improve this question









$endgroup$




enter image description here



What is the mistake here? Is it matter of the unit?






discrete-mathematics proof-verification




discrete-mathematics proof-verification






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









ShinobuIsMyWifeShinobuIsMyWife

413




413








  • 5




    $begingroup$
    Yes, the units don’t match across the 2nd equals sign
    $endgroup$
    – Alex
    3 hours ago






  • 2




    $begingroup$
    Yes, you have to square the unit. Conversion of squared units is different: if 100 cents is a dollar, then $100^2$ cents squared is a dollar squred.
    $endgroup$
    – Dean Young
    3 hours ago
















  • 5




    $begingroup$
    Yes, the units don’t match across the 2nd equals sign
    $endgroup$
    – Alex
    3 hours ago






  • 2




    $begingroup$
    Yes, you have to square the unit. Conversion of squared units is different: if 100 cents is a dollar, then $100^2$ cents squared is a dollar squred.
    $endgroup$
    – Dean Young
    3 hours ago










5




5




$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
3 hours ago




$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
3 hours ago




2




2




$begingroup$
Yes, you have to square the unit. Conversion of squared units is different: if 100 cents is a dollar, then $100^2$ cents squared is a dollar squred.
$endgroup$
– Dean Young
3 hours ago






$begingroup$
Yes, you have to square the unit. Conversion of squared units is different: if 100 cents is a dollar, then $100^2$ cents squared is a dollar squred.
$endgroup$
– Dean Young
3 hours ago












2 Answers
2






active

oldest

votes


















5












$begingroup$

$$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    You can clearly see the fallacy if you keep track of the units:




    • In the second equality, $$0.01 = $0.1times $0.1$ is not true, if you are doing units.


    • Even if the second equality were true, the third one gives problems: since $c=$/100$, you have
      $$
      ($0.1)^2=left(frac c{100},0.1right)^2=frac{c^2}{100}timesfrac1{10}=frac{c^2}{1000}.
      $$
      This is not $(10c)^2=100c^2$.



    In conclusion, two equalities are bogus, and so is the argument.






    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%2f3144460%2fwhats-wrong-with-this-bogus-proof%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









      5












      $begingroup$

      $$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        $$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          $$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.






          share|cite|improve this answer









          $endgroup$



          $$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          ArthurArthur

          117k7116200




          117k7116200























              2












              $begingroup$

              You can clearly see the fallacy if you keep track of the units:




              • In the second equality, $$0.01 = $0.1times $0.1$ is not true, if you are doing units.


              • Even if the second equality were true, the third one gives problems: since $c=$/100$, you have
                $$
                ($0.1)^2=left(frac c{100},0.1right)^2=frac{c^2}{100}timesfrac1{10}=frac{c^2}{1000}.
                $$
                This is not $(10c)^2=100c^2$.



              In conclusion, two equalities are bogus, and so is the argument.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                You can clearly see the fallacy if you keep track of the units:




                • In the second equality, $$0.01 = $0.1times $0.1$ is not true, if you are doing units.


                • Even if the second equality were true, the third one gives problems: since $c=$/100$, you have
                  $$
                  ($0.1)^2=left(frac c{100},0.1right)^2=frac{c^2}{100}timesfrac1{10}=frac{c^2}{1000}.
                  $$
                  This is not $(10c)^2=100c^2$.



                In conclusion, two equalities are bogus, and so is the argument.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  You can clearly see the fallacy if you keep track of the units:




                  • In the second equality, $$0.01 = $0.1times $0.1$ is not true, if you are doing units.


                  • Even if the second equality were true, the third one gives problems: since $c=$/100$, you have
                    $$
                    ($0.1)^2=left(frac c{100},0.1right)^2=frac{c^2}{100}timesfrac1{10}=frac{c^2}{1000}.
                    $$
                    This is not $(10c)^2=100c^2$.



                  In conclusion, two equalities are bogus, and so is the argument.






                  share|cite|improve this answer











                  $endgroup$



                  You can clearly see the fallacy if you keep track of the units:




                  • In the second equality, $$0.01 = $0.1times $0.1$ is not true, if you are doing units.


                  • Even if the second equality were true, the third one gives problems: since $c=$/100$, you have
                    $$
                    ($0.1)^2=left(frac c{100},0.1right)^2=frac{c^2}{100}timesfrac1{10}=frac{c^2}{1000}.
                    $$
                    This is not $(10c)^2=100c^2$.



                  In conclusion, two equalities are bogus, and so is the argument.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 3 hours ago









                  J. W. Tanner

                  3,1081320




                  3,1081320










                  answered 3 hours ago









                  Martin ArgeramiMartin Argerami

                  128k1184184




                  128k1184184






























                      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%2f3144460%2fwhats-wrong-with-this-bogus-proof%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

                      Contact image not getting when fetch all contact list from iPhone by CNContact

                      Image
                      up vote 0 down vote favorite I know this question already asked but not getting solution. From this code I will get all the information from the contact but image not found when open vcf files on mac os, also not getting when share this file. I use this stackoverflow link here but It's not help full. var contacts = [CNContact]() let keys = [CNContactVCardSerialization.descriptorForRequiredKeys() ] as [Any] let request = CNContactFetchRequest(keysToFetch: keys as! [CNKeyDescriptor]) do { try self.contactStore.enumerateContacts(with: request) { (contact, stop) in // Array containing all unified contacts from everywhere contacts.append(contact) } } catch { print("unable to fetch contacts") } do { let data = try
                      Read more

                      count number of partitions of a set with n elements into k subsets

                      Image
                      up vote 6 down vote favorite 1 This program is for count number of partitions of a set with n elements into k subsets I am confusing here return k*countP(n-1, k) + countP(n-1, k-1); can some one explain what is happening here? why we are multiplying with k? NOTE->I know this is not the best way to calculate number of partitions that would be DP // A C++ program to count number of partitions // of a set with n elements into k subsets #include<iostream> using namespace std; // Returns count of different partitions of n // elements in k subsets int countP(int n, int k) { // Base cases if (n == 0 || k == 0 || k > n) return 0; if (k == 1 || k == n) return 1; // S(n+1, k) = k*S(n, k) + S(n, k-1) return k*countP(n-1, k) + countP(n-1, k-1); } // Driver program int ma
                      Read more

                      A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks

                      Image
                      0 I found a lot of questions abount appendices and ToC. Many users want appendices to be grouped in an Appendix part, however some problems arise with ToC, hyperref, PDF viewer bookmarks, and so on. There are different solutions which require extra packages, command patching and other extra code, however none of them satisfies me. I almost found an easy way to accomplish a good result, where appendices are added to bookmarks in the right way and hyperref links point to the right page. However, the number of the "Appendix" part page is wrong (it's the number of appendix A). Is there any EASY way to fix that? This is a MWE: documentclass{book} usepackage[nottoc,notlot,notlof]{tocbibind} usepackage{hyperref} begin{document} frontmatter tableofcontents mainmatter part{First} chapter{
                      Read more