Why $frac {a}{b} = frac {c}{d} implies frac {a+c}{b+d} = frac {a}{b} = frac {c}{d}$?












4












$begingroup$


In geometry class, this property was quickly introduced:




$$dfrac {a}{b} = dfrac {c}{d} implies dfrac {a+c}{b+d} = dfrac {a}{b} = dfrac {c}{d} $$




It usually avoids some boring quadratic equations, so it's useful but I don't really get it. Seems trivial but I can't prove.



How to demonstrate it?






algebra-precalculus fractions







share|cite|improve this question











$endgroup$








  • 8




    $begingroup$
    That is generally untrue. For counterexample, consider $frac{1}{2}+frac{1}{3}=frac{5}{6}neq frac{2}{5}=frac{1+1}{2+3}$. Instead what is actually true is $frac{a}{b}+frac{c}{d}=frac{ad+cb}{bd}$
    $endgroup$
    – JMoravitz
    Apr 18 '17 at 2:07








  • 1




    $begingroup$
    This is not true @rodorgas. Take the L.C.M. and solve is simply you will got $ad^2+cb^2=0$ which is not given or true.
    $endgroup$
    – Harsh Kumar
    Apr 18 '17 at 2:12












  • $begingroup$
    @JMoravitz I'm sorry I typed it wrong. Please see my updated and corrected question.
    $endgroup$
    – rodorgas
    Apr 18 '17 at 2:14
















4












$begingroup$


In geometry class, this property was quickly introduced:




$$dfrac {a}{b} = dfrac {c}{d} implies dfrac {a+c}{b+d} = dfrac {a}{b} = dfrac {c}{d} $$




It usually avoids some boring quadratic equations, so it's useful but I don't really get it. Seems trivial but I can't prove.



How to demonstrate it?






algebra-precalculus fractions







share|cite|improve this question











$endgroup$








  • 8




    $begingroup$
    That is generally untrue. For counterexample, consider $frac{1}{2}+frac{1}{3}=frac{5}{6}neq frac{2}{5}=frac{1+1}{2+3}$. Instead what is actually true is $frac{a}{b}+frac{c}{d}=frac{ad+cb}{bd}$
    $endgroup$
    – JMoravitz
    Apr 18 '17 at 2:07








  • 1




    $begingroup$
    This is not true @rodorgas. Take the L.C.M. and solve is simply you will got $ad^2+cb^2=0$ which is not given or true.
    $endgroup$
    – Harsh Kumar
    Apr 18 '17 at 2:12












  • $begingroup$
    @JMoravitz I'm sorry I typed it wrong. Please see my updated and corrected question.
    $endgroup$
    – rodorgas
    Apr 18 '17 at 2:14














4












4








4





$begingroup$


In geometry class, this property was quickly introduced:




$$dfrac {a}{b} = dfrac {c}{d} implies dfrac {a+c}{b+d} = dfrac {a}{b} = dfrac {c}{d} $$




It usually avoids some boring quadratic equations, so it's useful but I don't really get it. Seems trivial but I can't prove.



How to demonstrate it?






algebra-precalculus fractions







share|cite|improve this question











$endgroup$




In geometry class, this property was quickly introduced:




$$dfrac {a}{b} = dfrac {c}{d} implies dfrac {a+c}{b+d} = dfrac {a}{b} = dfrac {c}{d} $$




It usually avoids some boring quadratic equations, so it's useful but I don't really get it. Seems trivial but I can't prove.



How to demonstrate it?






algebra-precalculus fractions




algebra-precalculus fractions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 16:46









Martin Sleziak

45k10123277




45k10123277










asked Apr 18 '17 at 2:04









rodorgasrodorgas

20116




20116








  • 8




    $begingroup$
    That is generally untrue. For counterexample, consider $frac{1}{2}+frac{1}{3}=frac{5}{6}neq frac{2}{5}=frac{1+1}{2+3}$. Instead what is actually true is $frac{a}{b}+frac{c}{d}=frac{ad+cb}{bd}$
    $endgroup$
    – JMoravitz
    Apr 18 '17 at 2:07








  • 1




    $begingroup$
    This is not true @rodorgas. Take the L.C.M. and solve is simply you will got $ad^2+cb^2=0$ which is not given or true.
    $endgroup$
    – Harsh Kumar
    Apr 18 '17 at 2:12












  • $begingroup$
    @JMoravitz I'm sorry I typed it wrong. Please see my updated and corrected question.
    $endgroup$
    – rodorgas
    Apr 18 '17 at 2:14














  • 8




    $begingroup$
    That is generally untrue. For counterexample, consider $frac{1}{2}+frac{1}{3}=frac{5}{6}neq frac{2}{5}=frac{1+1}{2+3}$. Instead what is actually true is $frac{a}{b}+frac{c}{d}=frac{ad+cb}{bd}$
    $endgroup$
    – JMoravitz
    Apr 18 '17 at 2:07








  • 1




    $begingroup$
    This is not true @rodorgas. Take the L.C.M. and solve is simply you will got $ad^2+cb^2=0$ which is not given or true.
    $endgroup$
    – Harsh Kumar
    Apr 18 '17 at 2:12












  • $begingroup$
    @JMoravitz I'm sorry I typed it wrong. Please see my updated and corrected question.
    $endgroup$
    – rodorgas
    Apr 18 '17 at 2:14








8




8




$begingroup$
That is generally untrue. For counterexample, consider $frac{1}{2}+frac{1}{3}=frac{5}{6}neq frac{2}{5}=frac{1+1}{2+3}$. Instead what is actually true is $frac{a}{b}+frac{c}{d}=frac{ad+cb}{bd}$
$endgroup$
– JMoravitz
Apr 18 '17 at 2:07






$begingroup$
That is generally untrue. For counterexample, consider $frac{1}{2}+frac{1}{3}=frac{5}{6}neq frac{2}{5}=frac{1+1}{2+3}$. Instead what is actually true is $frac{a}{b}+frac{c}{d}=frac{ad+cb}{bd}$
$endgroup$
– JMoravitz
Apr 18 '17 at 2:07






1




1




$begingroup$
This is not true @rodorgas. Take the L.C.M. and solve is simply you will got $ad^2+cb^2=0$ which is not given or true.
$endgroup$
– Harsh Kumar
Apr 18 '17 at 2:12






$begingroup$
This is not true @rodorgas. Take the L.C.M. and solve is simply you will got $ad^2+cb^2=0$ which is not given or true.
$endgroup$
– Harsh Kumar
Apr 18 '17 at 2:12














$begingroup$
@JMoravitz I'm sorry I typed it wrong. Please see my updated and corrected question.
$endgroup$
– rodorgas
Apr 18 '17 at 2:14




$begingroup$
@JMoravitz I'm sorry I typed it wrong. Please see my updated and corrected question.
$endgroup$
– rodorgas
Apr 18 '17 at 2:14










3 Answers
3






active

oldest

votes


















5












$begingroup$

$$frac{a}{b}=frac{c}{d}Rightarrowfrac{a+c}{b+d}=frac{a+frac{ad}{b}}{b+d}=frac{ab+ad}{b(b+d)}=frac{a}{b}=frac{c}{d}$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Assume that none of the unknowns is zero.
    Then, let
    $$dfrac{a}{b}=dfrac{c}{d}=k$$
    This implies,
    $$ a=kb,qquad c=kd$$



    $$therefore{a+c}=k(b+d)$$
    $$implies dfrac{a+c}{b+d}=k$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I believe you mean (a+c)/(b+d) in the last expression. Nice method, though! Thanks
      $endgroup$
      – rodorgas
      Aug 24 '17 at 13:35



















    0












    $begingroup$

    Hint:
    $$
    frac{p}{q} = frac{r}{s}
    $$
    if and only if $ps = rq$.






    share|cite|improve this answer









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      $$frac{a}{b}=frac{c}{d}Rightarrowfrac{a+c}{b+d}=frac{a+frac{ad}{b}}{b+d}=frac{ab+ad}{b(b+d)}=frac{a}{b}=frac{c}{d}$$






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        $$frac{a}{b}=frac{c}{d}Rightarrowfrac{a+c}{b+d}=frac{a+frac{ad}{b}}{b+d}=frac{ab+ad}{b(b+d)}=frac{a}{b}=frac{c}{d}$$






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          $$frac{a}{b}=frac{c}{d}Rightarrowfrac{a+c}{b+d}=frac{a+frac{ad}{b}}{b+d}=frac{ab+ad}{b(b+d)}=frac{a}{b}=frac{c}{d}$$






          share|cite|improve this answer









          $endgroup$



          $$frac{a}{b}=frac{c}{d}Rightarrowfrac{a+c}{b+d}=frac{a+frac{ad}{b}}{b+d}=frac{ab+ad}{b(b+d)}=frac{a}{b}=frac{c}{d}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 18 '17 at 2:18









          John DoeJohn Doe

          12.1k11339




          12.1k11339























              2












              $begingroup$

              Assume that none of the unknowns is zero.
              Then, let
              $$dfrac{a}{b}=dfrac{c}{d}=k$$
              This implies,
              $$ a=kb,qquad c=kd$$



              $$therefore{a+c}=k(b+d)$$
              $$implies dfrac{a+c}{b+d}=k$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I believe you mean (a+c)/(b+d) in the last expression. Nice method, though! Thanks
                $endgroup$
                – rodorgas
                Aug 24 '17 at 13:35
















              2












              $begingroup$

              Assume that none of the unknowns is zero.
              Then, let
              $$dfrac{a}{b}=dfrac{c}{d}=k$$
              This implies,
              $$ a=kb,qquad c=kd$$



              $$therefore{a+c}=k(b+d)$$
              $$implies dfrac{a+c}{b+d}=k$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I believe you mean (a+c)/(b+d) in the last expression. Nice method, though! Thanks
                $endgroup$
                – rodorgas
                Aug 24 '17 at 13:35














              2












              2








              2





              $begingroup$

              Assume that none of the unknowns is zero.
              Then, let
              $$dfrac{a}{b}=dfrac{c}{d}=k$$
              This implies,
              $$ a=kb,qquad c=kd$$



              $$therefore{a+c}=k(b+d)$$
              $$implies dfrac{a+c}{b+d}=k$$






              share|cite|improve this answer











              $endgroup$



              Assume that none of the unknowns is zero.
              Then, let
              $$dfrac{a}{b}=dfrac{c}{d}=k$$
              This implies,
              $$ a=kb,qquad c=kd$$



              $$therefore{a+c}=k(b+d)$$
              $$implies dfrac{a+c}{b+d}=k$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Aug 25 '17 at 3:02

























              answered Aug 24 '17 at 5:29









              ArchimedesprincipleArchimedesprinciple

              34418




              34418












              • $begingroup$
                I believe you mean (a+c)/(b+d) in the last expression. Nice method, though! Thanks
                $endgroup$
                – rodorgas
                Aug 24 '17 at 13:35


















              • $begingroup$
                I believe you mean (a+c)/(b+d) in the last expression. Nice method, though! Thanks
                $endgroup$
                – rodorgas
                Aug 24 '17 at 13:35
















              $begingroup$
              I believe you mean (a+c)/(b+d) in the last expression. Nice method, though! Thanks
              $endgroup$
              – rodorgas
              Aug 24 '17 at 13:35




              $begingroup$
              I believe you mean (a+c)/(b+d) in the last expression. Nice method, though! Thanks
              $endgroup$
              – rodorgas
              Aug 24 '17 at 13:35











              0












              $begingroup$

              Hint:
              $$
              frac{p}{q} = frac{r}{s}
              $$
              if and only if $ps = rq$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint:
                $$
                frac{p}{q} = frac{r}{s}
                $$
                if and only if $ps = rq$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint:
                  $$
                  frac{p}{q} = frac{r}{s}
                  $$
                  if and only if $ps = rq$.






                  share|cite|improve this answer









                  $endgroup$



                  Hint:
                  $$
                  frac{p}{q} = frac{r}{s}
                  $$
                  if and only if $ps = rq$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 18 '17 at 2:18









                  Ethan BolkerEthan Bolker

                  46.2k554121




                  46.2k554121






























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