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Why is the reciprocal used in fraction division?


How to make sense of fractions?How do I rewrite -100+1/2 as the mixed number -99 1/2?Fraction exponents in divisionfraction division understandingFraction and Decimal: Reciprocal of x's non-integerWhy was I taught to convert “improper fractions” into mixed numbers?The division of a fraction - Whole or Part?division by fraction proofWhen dividing by a fraction, why can you not take the reciprocal of term involving addition/subtraction?How to tell when a fraction does not end?Basic division problem: dividing a fraction by a fraction













2












$begingroup$


I don't know if this is a basic question or whatever, but I can't seem to find an answer.



As far as I understand the reciprocal of a number the inverse of that number, that still doesn't clarify why it is needed.



For many years I've only ever done math like if I were a robot. I just did it and never understood what I was doing. So when I went and divided fractions I just used the reciprocal, because "that was the way to do it". I want to understand math at a deeper level, especially subjects like probability, statistics, calculus, and linear algebra. To do that I have to understand the fundamentals however.



Any response is appreciated.










share|cite|improve this question









$endgroup$











  • $begingroup$
    This may be helpful. math.stackexchange.com/questions/1127483/…
    $endgroup$
    – Ethan Bolker
    2 hours ago










  • $begingroup$
    If you are asking this question, it probably means that you do not have enough experience with algebra
    $endgroup$
    – rash
    2 hours ago










  • $begingroup$
    Also see How to explain the flipping of division by a fraction? on Mathematics Educators, showcasing many attempts at an intuitive and elementary explanation.
    $endgroup$
    – Henning Makholm
    2 hours ago
















2












$begingroup$


I don't know if this is a basic question or whatever, but I can't seem to find an answer.



As far as I understand the reciprocal of a number the inverse of that number, that still doesn't clarify why it is needed.



For many years I've only ever done math like if I were a robot. I just did it and never understood what I was doing. So when I went and divided fractions I just used the reciprocal, because "that was the way to do it". I want to understand math at a deeper level, especially subjects like probability, statistics, calculus, and linear algebra. To do that I have to understand the fundamentals however.



Any response is appreciated.










share|cite|improve this question









$endgroup$











  • $begingroup$
    This may be helpful. math.stackexchange.com/questions/1127483/…
    $endgroup$
    – Ethan Bolker
    2 hours ago










  • $begingroup$
    If you are asking this question, it probably means that you do not have enough experience with algebra
    $endgroup$
    – rash
    2 hours ago










  • $begingroup$
    Also see How to explain the flipping of division by a fraction? on Mathematics Educators, showcasing many attempts at an intuitive and elementary explanation.
    $endgroup$
    – Henning Makholm
    2 hours ago














2












2








2


1



$begingroup$


I don't know if this is a basic question or whatever, but I can't seem to find an answer.



As far as I understand the reciprocal of a number the inverse of that number, that still doesn't clarify why it is needed.



For many years I've only ever done math like if I were a robot. I just did it and never understood what I was doing. So when I went and divided fractions I just used the reciprocal, because "that was the way to do it". I want to understand math at a deeper level, especially subjects like probability, statistics, calculus, and linear algebra. To do that I have to understand the fundamentals however.



Any response is appreciated.










share|cite|improve this question









$endgroup$




I don't know if this is a basic question or whatever, but I can't seem to find an answer.



As far as I understand the reciprocal of a number the inverse of that number, that still doesn't clarify why it is needed.



For many years I've only ever done math like if I were a robot. I just did it and never understood what I was doing. So when I went and divided fractions I just used the reciprocal, because "that was the way to do it". I want to understand math at a deeper level, especially subjects like probability, statistics, calculus, and linear algebra. To do that I have to understand the fundamentals however.



Any response is appreciated.







fractions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours ago









ArgusArgus

20418




20418











  • $begingroup$
    This may be helpful. math.stackexchange.com/questions/1127483/…
    $endgroup$
    – Ethan Bolker
    2 hours ago










  • $begingroup$
    If you are asking this question, it probably means that you do not have enough experience with algebra
    $endgroup$
    – rash
    2 hours ago










  • $begingroup$
    Also see How to explain the flipping of division by a fraction? on Mathematics Educators, showcasing many attempts at an intuitive and elementary explanation.
    $endgroup$
    – Henning Makholm
    2 hours ago

















  • $begingroup$
    This may be helpful. math.stackexchange.com/questions/1127483/…
    $endgroup$
    – Ethan Bolker
    2 hours ago










  • $begingroup$
    If you are asking this question, it probably means that you do not have enough experience with algebra
    $endgroup$
    – rash
    2 hours ago










  • $begingroup$
    Also see How to explain the flipping of division by a fraction? on Mathematics Educators, showcasing many attempts at an intuitive and elementary explanation.
    $endgroup$
    – Henning Makholm
    2 hours ago
















$begingroup$
This may be helpful. math.stackexchange.com/questions/1127483/…
$endgroup$
– Ethan Bolker
2 hours ago




$begingroup$
This may be helpful. math.stackexchange.com/questions/1127483/…
$endgroup$
– Ethan Bolker
2 hours ago












$begingroup$
If you are asking this question, it probably means that you do not have enough experience with algebra
$endgroup$
– rash
2 hours ago




$begingroup$
If you are asking this question, it probably means that you do not have enough experience with algebra
$endgroup$
– rash
2 hours ago












$begingroup$
Also see How to explain the flipping of division by a fraction? on Mathematics Educators, showcasing many attempts at an intuitive and elementary explanation.
$endgroup$
– Henning Makholm
2 hours ago





$begingroup$
Also see How to explain the flipping of division by a fraction? on Mathematics Educators, showcasing many attempts at an intuitive and elementary explanation.
$endgroup$
– Henning Makholm
2 hours ago











2 Answers
2






active

oldest

votes


















3












$begingroup$

I think you're asking why the rule for division of fractions,
$$fracpq div fracrs = fracpq cdot fracsr,$$
works.
And I'm assuming that you're already comfortable with how to multiply fractions.



We need to go back to what division is supposed to achieve in the first place. When we look into that, the answer is that $Adiv B$ means something that gives $A$ when we multiply it by $B$ -- or, written in symbols, $Adiv B$ means the $X$ that solves the equation $$ Xcdot B = A $$



When our $A$ and $B$ are fraction, the "reciprocal" division rule can be regarded as a trick that happens to produce an $X$ that works. It's easy enough to see that it does work: If we're dividing $frac pq div frac rs$ we need to solve the equation
$$ X cdot frac rs = frac pq $$
And indeed setting $X=frac pqcdot frac sr = fracpsqr$ does this:
$$ fracpsqrcdotfrac rs = fracpscdot rqrcdot s = fracpcdot srqcdot sr = frac pq$$
like we want. (I'm also assuming that you're comfortable with cancelling the common factor $sr$ in the middle fraction).



This computation hopefully also gives some ides why it works, at least part way. In $fracpsqr$ the $p$ and $q$ are what we want to end up with, and the $s$ and $r$ are there to "neutralize" the $r$ and $s$ we have but want to discard. By making sure that the product has exactly one $r$ and one $s$ on each side of the fraction bar they make sure we can cancel them away.



Writing the solution $fracpsqr$ as $frac pqcdot frac sr$ might be best understood as just an easy way to remember what goes where. But this memory trick itself then also serves as motivation for considering the reciprocal to be an interesting operation in its own right in higher algebra.






share|cite|improve this answer











$endgroup$




















    0












    $begingroup$

    Your question isn't completely clear but what I understood is that you don't get why $$fracfracabfraccd= fracab*fracdc$$ the answer it's located in the axioms of the real numbers, a number $b$ it's the reciprocal of a number $d$ if $$ d*b=1$$ now, let's see the definition of fraction $$e/f=e*f^-1$$ with $f^-1$ the reciprocal of $f$, therefore $$fracfracabfraccd=fracab(fraccd)^-1$$ and since $$fraccd*fracdc=1$$ we have $$fracfracabfraccd= fracab*fracdc$$ our result






    share|cite|improve this answer








    New contributor



    Mario Aldean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      I think you're asking why the rule for division of fractions,
      $$fracpq div fracrs = fracpq cdot fracsr,$$
      works.
      And I'm assuming that you're already comfortable with how to multiply fractions.



      We need to go back to what division is supposed to achieve in the first place. When we look into that, the answer is that $Adiv B$ means something that gives $A$ when we multiply it by $B$ -- or, written in symbols, $Adiv B$ means the $X$ that solves the equation $$ Xcdot B = A $$



      When our $A$ and $B$ are fraction, the "reciprocal" division rule can be regarded as a trick that happens to produce an $X$ that works. It's easy enough to see that it does work: If we're dividing $frac pq div frac rs$ we need to solve the equation
      $$ X cdot frac rs = frac pq $$
      And indeed setting $X=frac pqcdot frac sr = fracpsqr$ does this:
      $$ fracpsqrcdotfrac rs = fracpscdot rqrcdot s = fracpcdot srqcdot sr = frac pq$$
      like we want. (I'm also assuming that you're comfortable with cancelling the common factor $sr$ in the middle fraction).



      This computation hopefully also gives some ides why it works, at least part way. In $fracpsqr$ the $p$ and $q$ are what we want to end up with, and the $s$ and $r$ are there to "neutralize" the $r$ and $s$ we have but want to discard. By making sure that the product has exactly one $r$ and one $s$ on each side of the fraction bar they make sure we can cancel them away.



      Writing the solution $fracpsqr$ as $frac pqcdot frac sr$ might be best understood as just an easy way to remember what goes where. But this memory trick itself then also serves as motivation for considering the reciprocal to be an interesting operation in its own right in higher algebra.






      share|cite|improve this answer











      $endgroup$

















        3












        $begingroup$

        I think you're asking why the rule for division of fractions,
        $$fracpq div fracrs = fracpq cdot fracsr,$$
        works.
        And I'm assuming that you're already comfortable with how to multiply fractions.



        We need to go back to what division is supposed to achieve in the first place. When we look into that, the answer is that $Adiv B$ means something that gives $A$ when we multiply it by $B$ -- or, written in symbols, $Adiv B$ means the $X$ that solves the equation $$ Xcdot B = A $$



        When our $A$ and $B$ are fraction, the "reciprocal" division rule can be regarded as a trick that happens to produce an $X$ that works. It's easy enough to see that it does work: If we're dividing $frac pq div frac rs$ we need to solve the equation
        $$ X cdot frac rs = frac pq $$
        And indeed setting $X=frac pqcdot frac sr = fracpsqr$ does this:
        $$ fracpsqrcdotfrac rs = fracpscdot rqrcdot s = fracpcdot srqcdot sr = frac pq$$
        like we want. (I'm also assuming that you're comfortable with cancelling the common factor $sr$ in the middle fraction).



        This computation hopefully also gives some ides why it works, at least part way. In $fracpsqr$ the $p$ and $q$ are what we want to end up with, and the $s$ and $r$ are there to "neutralize" the $r$ and $s$ we have but want to discard. By making sure that the product has exactly one $r$ and one $s$ on each side of the fraction bar they make sure we can cancel them away.



        Writing the solution $fracpsqr$ as $frac pqcdot frac sr$ might be best understood as just an easy way to remember what goes where. But this memory trick itself then also serves as motivation for considering the reciprocal to be an interesting operation in its own right in higher algebra.






        share|cite|improve this answer











        $endgroup$















          3












          3








          3





          $begingroup$

          I think you're asking why the rule for division of fractions,
          $$fracpq div fracrs = fracpq cdot fracsr,$$
          works.
          And I'm assuming that you're already comfortable with how to multiply fractions.



          We need to go back to what division is supposed to achieve in the first place. When we look into that, the answer is that $Adiv B$ means something that gives $A$ when we multiply it by $B$ -- or, written in symbols, $Adiv B$ means the $X$ that solves the equation $$ Xcdot B = A $$



          When our $A$ and $B$ are fraction, the "reciprocal" division rule can be regarded as a trick that happens to produce an $X$ that works. It's easy enough to see that it does work: If we're dividing $frac pq div frac rs$ we need to solve the equation
          $$ X cdot frac rs = frac pq $$
          And indeed setting $X=frac pqcdot frac sr = fracpsqr$ does this:
          $$ fracpsqrcdotfrac rs = fracpscdot rqrcdot s = fracpcdot srqcdot sr = frac pq$$
          like we want. (I'm also assuming that you're comfortable with cancelling the common factor $sr$ in the middle fraction).



          This computation hopefully also gives some ides why it works, at least part way. In $fracpsqr$ the $p$ and $q$ are what we want to end up with, and the $s$ and $r$ are there to "neutralize" the $r$ and $s$ we have but want to discard. By making sure that the product has exactly one $r$ and one $s$ on each side of the fraction bar they make sure we can cancel them away.



          Writing the solution $fracpsqr$ as $frac pqcdot frac sr$ might be best understood as just an easy way to remember what goes where. But this memory trick itself then also serves as motivation for considering the reciprocal to be an interesting operation in its own right in higher algebra.






          share|cite|improve this answer











          $endgroup$



          I think you're asking why the rule for division of fractions,
          $$fracpq div fracrs = fracpq cdot fracsr,$$
          works.
          And I'm assuming that you're already comfortable with how to multiply fractions.



          We need to go back to what division is supposed to achieve in the first place. When we look into that, the answer is that $Adiv B$ means something that gives $A$ when we multiply it by $B$ -- or, written in symbols, $Adiv B$ means the $X$ that solves the equation $$ Xcdot B = A $$



          When our $A$ and $B$ are fraction, the "reciprocal" division rule can be regarded as a trick that happens to produce an $X$ that works. It's easy enough to see that it does work: If we're dividing $frac pq div frac rs$ we need to solve the equation
          $$ X cdot frac rs = frac pq $$
          And indeed setting $X=frac pqcdot frac sr = fracpsqr$ does this:
          $$ fracpsqrcdotfrac rs = fracpscdot rqrcdot s = fracpcdot srqcdot sr = frac pq$$
          like we want. (I'm also assuming that you're comfortable with cancelling the common factor $sr$ in the middle fraction).



          This computation hopefully also gives some ides why it works, at least part way. In $fracpsqr$ the $p$ and $q$ are what we want to end up with, and the $s$ and $r$ are there to "neutralize" the $r$ and $s$ we have but want to discard. By making sure that the product has exactly one $r$ and one $s$ on each side of the fraction bar they make sure we can cancel them away.



          Writing the solution $fracpsqr$ as $frac pqcdot frac sr$ might be best understood as just an easy way to remember what goes where. But this memory trick itself then also serves as motivation for considering the reciprocal to be an interesting operation in its own right in higher algebra.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 hours ago

























          answered 2 hours ago









          Henning MakholmHenning Makholm

          246k17316561




          246k17316561





















              0












              $begingroup$

              Your question isn't completely clear but what I understood is that you don't get why $$fracfracabfraccd= fracab*fracdc$$ the answer it's located in the axioms of the real numbers, a number $b$ it's the reciprocal of a number $d$ if $$ d*b=1$$ now, let's see the definition of fraction $$e/f=e*f^-1$$ with $f^-1$ the reciprocal of $f$, therefore $$fracfracabfraccd=fracab(fraccd)^-1$$ and since $$fraccd*fracdc=1$$ we have $$fracfracabfraccd= fracab*fracdc$$ our result






              share|cite|improve this answer








              New contributor



              Mario Aldean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.





              $endgroup$

















                0












                $begingroup$

                Your question isn't completely clear but what I understood is that you don't get why $$fracfracabfraccd= fracab*fracdc$$ the answer it's located in the axioms of the real numbers, a number $b$ it's the reciprocal of a number $d$ if $$ d*b=1$$ now, let's see the definition of fraction $$e/f=e*f^-1$$ with $f^-1$ the reciprocal of $f$, therefore $$fracfracabfraccd=fracab(fraccd)^-1$$ and since $$fraccd*fracdc=1$$ we have $$fracfracabfraccd= fracab*fracdc$$ our result






                share|cite|improve this answer








                New contributor



                Mario Aldean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  Your question isn't completely clear but what I understood is that you don't get why $$fracfracabfraccd= fracab*fracdc$$ the answer it's located in the axioms of the real numbers, a number $b$ it's the reciprocal of a number $d$ if $$ d*b=1$$ now, let's see the definition of fraction $$e/f=e*f^-1$$ with $f^-1$ the reciprocal of $f$, therefore $$fracfracabfraccd=fracab(fraccd)^-1$$ and since $$fraccd*fracdc=1$$ we have $$fracfracabfraccd= fracab*fracdc$$ our result






                  share|cite|improve this answer








                  New contributor



                  Mario Aldean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





                  $endgroup$



                  Your question isn't completely clear but what I understood is that you don't get why $$fracfracabfraccd= fracab*fracdc$$ the answer it's located in the axioms of the real numbers, a number $b$ it's the reciprocal of a number $d$ if $$ d*b=1$$ now, let's see the definition of fraction $$e/f=e*f^-1$$ with $f^-1$ the reciprocal of $f$, therefore $$fracfracabfraccd=fracab(fraccd)^-1$$ and since $$fraccd*fracdc=1$$ we have $$fracfracabfraccd= fracab*fracdc$$ our result







                  share|cite|improve this answer








                  New contributor



                  Mario Aldean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.








                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor



                  Mario Aldean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.








                  answered 2 hours ago









                  Mario AldeanMario Aldean

                  135




                  135




                  New contributor



                  Mario Aldean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.




                  New contributor




                  Mario Aldean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





























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                      Черчино Становништво Референце Спољашње везе Мени за навигацију46°09′29″ СГШ; 9°30′29″ ИГД / 46.15809° СГШ; 9.50814° ИГД / 46.15809; 9.5081446°09′29″ СГШ; 9°30′29″ ИГД / 46.15809° СГШ; 9.50814° ИГД / 46.15809; 9.508143179111„The GeoNames geographical database”„Istituto Nazionale di Statistica”Званични веб-сајтпроширитиуу