Is it true for any real $a$ and $b$ that if $a*b > 0$ => $a/b>0$?












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Is it true for any real $a$ and $b$ that if $a*b > 0$ => $a/b>0$ ?






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  • 3




    $begingroup$
    Multiply by $1/b^2$ for $bneq 0$.
    $endgroup$
    – Dietrich Burde
    Nov 30 '18 at 19:15
















-2












$begingroup$


Is it true for any real $a$ and $b$ that if $a*b > 0$ => $a/b>0$ ?






inequality







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Multiply by $1/b^2$ for $bneq 0$.
    $endgroup$
    – Dietrich Burde
    Nov 30 '18 at 19:15














-2












-2








-2





$begingroup$


Is it true for any real $a$ and $b$ that if $a*b > 0$ => $a/b>0$ ?






inequality







share|cite|improve this question









$endgroup$




Is it true for any real $a$ and $b$ that if $a*b > 0$ => $a/b>0$ ?






inequality




inequality






share|cite|improve this question













share|cite|improve this question











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asked Nov 30 '18 at 19:12









caasdadscaasdads

1107




1107








  • 3




    $begingroup$
    Multiply by $1/b^2$ for $bneq 0$.
    $endgroup$
    – Dietrich Burde
    Nov 30 '18 at 19:15














  • 3




    $begingroup$
    Multiply by $1/b^2$ for $bneq 0$.
    $endgroup$
    – Dietrich Burde
    Nov 30 '18 at 19:15








3




3




$begingroup$
Multiply by $1/b^2$ for $bneq 0$.
$endgroup$
– Dietrich Burde
Nov 30 '18 at 19:15




$begingroup$
Multiply by $1/b^2$ for $bneq 0$.
$endgroup$
– Dietrich Burde
Nov 30 '18 at 19:15










2 Answers
2






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oldest

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0












$begingroup$

$bneq0$ because otherwise, from the given we obtain $0>0$, which is wrong.



Thus, $b^2>0$ and from the given again we obtain:
$$frac{a}{b}=frac{ab}{b^2}>0.$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $a,b not =0$, real.



    Assume $ab>0 Rightarrow a/b<0.$



    Then



    $(ab)(a/b)<0$



    (sign change since $(a/b) <0$).



    $a^2 <0.$



    Since $a$ real a contradiction.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






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      0












      $begingroup$

      $bneq0$ because otherwise, from the given we obtain $0>0$, which is wrong.



      Thus, $b^2>0$ and from the given again we obtain:
      $$frac{a}{b}=frac{ab}{b^2}>0.$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        $bneq0$ because otherwise, from the given we obtain $0>0$, which is wrong.



        Thus, $b^2>0$ and from the given again we obtain:
        $$frac{a}{b}=frac{ab}{b^2}>0.$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          $bneq0$ because otherwise, from the given we obtain $0>0$, which is wrong.



          Thus, $b^2>0$ and from the given again we obtain:
          $$frac{a}{b}=frac{ab}{b^2}>0.$$






          share|cite|improve this answer









          $endgroup$



          $bneq0$ because otherwise, from the given we obtain $0>0$, which is wrong.



          Thus, $b^2>0$ and from the given again we obtain:
          $$frac{a}{b}=frac{ab}{b^2}>0.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 19:31









          Michael RozenbergMichael Rozenberg

          97.8k1590188




          97.8k1590188























              0












              $begingroup$

              $a,b not =0$, real.



              Assume $ab>0 Rightarrow a/b<0.$



              Then



              $(ab)(a/b)<0$



              (sign change since $(a/b) <0$).



              $a^2 <0.$



              Since $a$ real a contradiction.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                $a,b not =0$, real.



                Assume $ab>0 Rightarrow a/b<0.$



                Then



                $(ab)(a/b)<0$



                (sign change since $(a/b) <0$).



                $a^2 <0.$



                Since $a$ real a contradiction.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $a,b not =0$, real.



                  Assume $ab>0 Rightarrow a/b<0.$



                  Then



                  $(ab)(a/b)<0$



                  (sign change since $(a/b) <0$).



                  $a^2 <0.$



                  Since $a$ real a contradiction.






                  share|cite|improve this answer











                  $endgroup$



                  $a,b not =0$, real.



                  Assume $ab>0 Rightarrow a/b<0.$



                  Then



                  $(ab)(a/b)<0$



                  (sign change since $(a/b) <0$).



                  $a^2 <0.$



                  Since $a$ real a contradiction.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 '18 at 20:20

























                  answered Nov 30 '18 at 19:42









                  Peter SzilasPeter Szilas

                  10.9k2720




                  10.9k2720






























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