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
    Mar 12 at 0:20






  • 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
    Mar 12 at 0:20


















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
    Mar 12 at 0:20






  • 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
    Mar 12 at 0:20
















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 Mar 12 at 0:18









王文军 or Wenjun Wang王文军 or Wenjun Wang

514




514








  • 5




    $begingroup$
    Yes, the units don’t match across the 2nd equals sign
    $endgroup$
    – Alex
    Mar 12 at 0:20






  • 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
    Mar 12 at 0:20
















  • 5




    $begingroup$
    Yes, the units don’t match across the 2nd equals sign
    $endgroup$
    – Alex
    Mar 12 at 0:20






  • 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
    Mar 12 at 0:20










5




5




$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
Mar 12 at 0:20




$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
Mar 12 at 0:20




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
Mar 12 at 0:20






$begingroup$
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
Mar 12 at 0:20












2 Answers
2






active

oldest

votes


















7












$begingroup$

$$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











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

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






      share|cite|improve this answer









      $endgroup$


















        7












        $begingroup$

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






        share|cite|improve this answer









        $endgroup$
















          7












          7








          7





          $begingroup$

          $$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 Mar 12 at 0:24









          ArthurArthur

          119k7118202




          119k7118202























              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 Mar 12 at 0:43









                  J. W. Tanner

                  3,4751320




                  3,4751320










                  answered Mar 12 at 0:25









                  Martin ArgeramiMartin Argerami

                  128k1184184




                  128k1184184






























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