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What's wrong with this bogus proof?


What's wrong with this random variable proof?What's wrong with this proof that all UFDs are Bezout?What's wrong with this proof by contradiction?A bogus proof of countable power setWhat's wrong with this proof $1=i^2=-1$What is wrong with this proofWhat's wrong with this proof of symmetry of equality?What's wrong with this 1 = -1 proof?What's wrong with this proof? (Regular languages)What's wrong in this proof of 1=2?













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











    $endgroup$













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      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 2 hours ago









                  J. W. Tanner

                  3,0981320




                  3,0981320










                  answered 3 hours ago









                  Martin ArgeramiMartin Argerami

                  128k1184184




                  128k1184184






























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