Another proof that dividing by 0 does not exist — is it right? The Next CEO of Stack OverflowHow do you explain to a 5th grader why division by zero is meaningless?Proof that $Bbb Z$ has no other subring than itselfProof that odd perfect numbers cannot consist of single unique factors?showing that no repunit is a square - proof verificationUsing induction, prove that $(3^2^n -1)$ is divisible by $2^n+2$ but not by $2^n+3$.Proof that $sqrt2$ is irrationalProve that between two unequal rational numbers there is another rationalMultiplicative inverse questionsIs my proof of $sqrt2 + sqrt3 + sqrt5$ is an irrational number valid?Proof of even numbersLeft and Right inverses - Proof by contradiction

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Another proof that dividing by 0 does not exist — is it right?



The Next CEO of Stack OverflowHow do you explain to a 5th grader why division by zero is meaningless?Proof that $Bbb Z$ has no other subring than itselfProof that odd perfect numbers cannot consist of single unique factors?showing that no repunit is a square - proof verificationUsing induction, prove that $(3^2^n -1)$ is divisible by $2^n+2$ but not by $2^n+3$.Proof that $sqrt2$ is irrationalProve that between two unequal rational numbers there is another rationalMultiplicative inverse questionsIs my proof of $sqrt2 + sqrt3 + sqrt5$ is an irrational number valid?Proof of even numbersLeft and Right inverses - Proof by contradiction










17












$begingroup$


Ok I am in grade 9 and I am maybe too young for this.



But I thought about this, why dividing by 0 is impossible.



Dividing by 0 is possible would mean 1/0 is possible, which would mean 0 has a multiplicative inverse.



So if we multiply a number by 0 then by 1/0 we get the same number.



But thats impossible because all numbers multiplied by 0 gives 0 therefore we can’t have an inverse for 0, as that gives us the initial number and thus division by 0 is impossible



Is this right?










share|cite|improve this question









New contributor




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







$endgroup$











  • $begingroup$
    Sometimes division by zero is defined, such as in the extended complex plane.
    $endgroup$
    – Shaun
    8 hours ago






  • 11




    $begingroup$
    Your answer is 100% correct and you should probably become a mathematician. These kinds of answers (mathematicians also call them proofs) are what mathematicians do all day long.
    $endgroup$
    – ErotemeObelus
    4 hours ago











  • $begingroup$
    math.stackexchange.com/questions/2883450/…
    $endgroup$
    – Maria Mazur
    2 hours ago















17












$begingroup$


Ok I am in grade 9 and I am maybe too young for this.



But I thought about this, why dividing by 0 is impossible.



Dividing by 0 is possible would mean 1/0 is possible, which would mean 0 has a multiplicative inverse.



So if we multiply a number by 0 then by 1/0 we get the same number.



But thats impossible because all numbers multiplied by 0 gives 0 therefore we can’t have an inverse for 0, as that gives us the initial number and thus division by 0 is impossible



Is this right?










share|cite|improve this question









New contributor




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







$endgroup$











  • $begingroup$
    Sometimes division by zero is defined, such as in the extended complex plane.
    $endgroup$
    – Shaun
    8 hours ago






  • 11




    $begingroup$
    Your answer is 100% correct and you should probably become a mathematician. These kinds of answers (mathematicians also call them proofs) are what mathematicians do all day long.
    $endgroup$
    – ErotemeObelus
    4 hours ago











  • $begingroup$
    math.stackexchange.com/questions/2883450/…
    $endgroup$
    – Maria Mazur
    2 hours ago













17












17








17


2



$begingroup$


Ok I am in grade 9 and I am maybe too young for this.



But I thought about this, why dividing by 0 is impossible.



Dividing by 0 is possible would mean 1/0 is possible, which would mean 0 has a multiplicative inverse.



So if we multiply a number by 0 then by 1/0 we get the same number.



But thats impossible because all numbers multiplied by 0 gives 0 therefore we can’t have an inverse for 0, as that gives us the initial number and thus division by 0 is impossible



Is this right?










share|cite|improve this question









New contributor




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







$endgroup$




Ok I am in grade 9 and I am maybe too young for this.



But I thought about this, why dividing by 0 is impossible.



Dividing by 0 is possible would mean 1/0 is possible, which would mean 0 has a multiplicative inverse.



So if we multiply a number by 0 then by 1/0 we get the same number.



But thats impossible because all numbers multiplied by 0 gives 0 therefore we can’t have an inverse for 0, as that gives us the initial number and thus division by 0 is impossible



Is this right?







proof-verification






share|cite|improve this question









New contributor




Selim Jean Ellieh 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 question









New contributor




Selim Jean Ellieh 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 question




share|cite|improve this question








edited 2 hours ago









H Huang

401111




401111






New contributor




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









asked 8 hours ago









Selim Jean ElliehSelim Jean Ellieh

965




965




New contributor




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





New contributor





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






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











  • $begingroup$
    Sometimes division by zero is defined, such as in the extended complex plane.
    $endgroup$
    – Shaun
    8 hours ago






  • 11




    $begingroup$
    Your answer is 100% correct and you should probably become a mathematician. These kinds of answers (mathematicians also call them proofs) are what mathematicians do all day long.
    $endgroup$
    – ErotemeObelus
    4 hours ago











  • $begingroup$
    math.stackexchange.com/questions/2883450/…
    $endgroup$
    – Maria Mazur
    2 hours ago
















  • $begingroup$
    Sometimes division by zero is defined, such as in the extended complex plane.
    $endgroup$
    – Shaun
    8 hours ago






  • 11




    $begingroup$
    Your answer is 100% correct and you should probably become a mathematician. These kinds of answers (mathematicians also call them proofs) are what mathematicians do all day long.
    $endgroup$
    – ErotemeObelus
    4 hours ago











  • $begingroup$
    math.stackexchange.com/questions/2883450/…
    $endgroup$
    – Maria Mazur
    2 hours ago















$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
8 hours ago




$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
8 hours ago




11




11




$begingroup$
Your answer is 100% correct and you should probably become a mathematician. These kinds of answers (mathematicians also call them proofs) are what mathematicians do all day long.
$endgroup$
– ErotemeObelus
4 hours ago





$begingroup$
Your answer is 100% correct and you should probably become a mathematician. These kinds of answers (mathematicians also call them proofs) are what mathematicians do all day long.
$endgroup$
– ErotemeObelus
4 hours ago













$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 hours ago




$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 hours ago










3 Answers
3






active

oldest

votes


















22












$begingroup$

That's the most basic reason that division by $0$ is usually considered to be a Bad Thing, yes. Because if we did allow dividing by $0$, we would have to give up at least of one of the following things (these are usually considered Very Nice):



  • What $1$ means ($1cdot a = a$ for any $a$)

  • What $0$ means ($0 cdot a = 0$ for any $a$)

  • What division means ($frac ab = c$ means $a = ccdot b$)





share|cite|improve this answer









$endgroup$




















    6












    $begingroup$

    Yes . . . and no.



    You might be interested in, for example, Wheel Theory, where division by zero is defined.






    share|cite|improve this answer











    $endgroup$








    • 10




      $begingroup$
      You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
      $endgroup$
      – Arthur
      8 hours ago






    • 1




      $begingroup$
      That's a fair comment, @Arthur. Thank you for the feedback.
      $endgroup$
      – Shaun
      8 hours ago






    • 1




      $begingroup$
      What d'you think, @SelimJeanEllieh?
      $endgroup$
      – Shaun
      8 hours ago






    • 1




      $begingroup$
      Oh: The OP has insufficient rep to comment. Nevermind.
      $endgroup$
      – Shaun
      8 hours ago






    • 5




      $begingroup$
      @Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
      $endgroup$
      – YiFan
      5 hours ago


















    0












    $begingroup$

    That is quite right. However, I would like you to have a higher point of view.



    Mathematicians derive theorems from axioms and definitions. And here is the definition of a field.



    A field is a set $F$ equipped with two binary operations $+,times$, such that there exists $e_+, e_times$, such that for all $a,b,cin F$,

    - $a+b=b+a$,

    - $(a+b)+c=a+(b+c)$,

    - $e_++a=a$,

    - there exists $a'$ such that $a'+a=e_+$,

    - $(atimes b)times c=atimes (btimes c)$,

    - $e_timestimes a=a$,

    - there exists $a''$ such that $a''times a=e_times$ if $ane e_+$.



    Now verify that the rationals and the reals are fields.



    Try and prove that if there exists $x$ such that $xtimes e_+=e_times$, the set $F$ can only have one element.






    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









      22












      $begingroup$

      That's the most basic reason that division by $0$ is usually considered to be a Bad Thing, yes. Because if we did allow dividing by $0$, we would have to give up at least of one of the following things (these are usually considered Very Nice):



      • What $1$ means ($1cdot a = a$ for any $a$)

      • What $0$ means ($0 cdot a = 0$ for any $a$)

      • What division means ($frac ab = c$ means $a = ccdot b$)





      share|cite|improve this answer









      $endgroup$

















        22












        $begingroup$

        That's the most basic reason that division by $0$ is usually considered to be a Bad Thing, yes. Because if we did allow dividing by $0$, we would have to give up at least of one of the following things (these are usually considered Very Nice):



        • What $1$ means ($1cdot a = a$ for any $a$)

        • What $0$ means ($0 cdot a = 0$ for any $a$)

        • What division means ($frac ab = c$ means $a = ccdot b$)





        share|cite|improve this answer









        $endgroup$















          22












          22








          22





          $begingroup$

          That's the most basic reason that division by $0$ is usually considered to be a Bad Thing, yes. Because if we did allow dividing by $0$, we would have to give up at least of one of the following things (these are usually considered Very Nice):



          • What $1$ means ($1cdot a = a$ for any $a$)

          • What $0$ means ($0 cdot a = 0$ for any $a$)

          • What division means ($frac ab = c$ means $a = ccdot b$)





          share|cite|improve this answer









          $endgroup$



          That's the most basic reason that division by $0$ is usually considered to be a Bad Thing, yes. Because if we did allow dividing by $0$, we would have to give up at least of one of the following things (these are usually considered Very Nice):



          • What $1$ means ($1cdot a = a$ for any $a$)

          • What $0$ means ($0 cdot a = 0$ for any $a$)

          • What division means ($frac ab = c$ means $a = ccdot b$)






          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 8 hours ago









          ArthurArthur

          121k7121208




          121k7121208





















              6












              $begingroup$

              Yes . . . and no.



              You might be interested in, for example, Wheel Theory, where division by zero is defined.






              share|cite|improve this answer











              $endgroup$








              • 10




                $begingroup$
                You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
                $endgroup$
                – Arthur
                8 hours ago






              • 1




                $begingroup$
                That's a fair comment, @Arthur. Thank you for the feedback.
                $endgroup$
                – Shaun
                8 hours ago






              • 1




                $begingroup$
                What d'you think, @SelimJeanEllieh?
                $endgroup$
                – Shaun
                8 hours ago






              • 1




                $begingroup$
                Oh: The OP has insufficient rep to comment. Nevermind.
                $endgroup$
                – Shaun
                8 hours ago






              • 5




                $begingroup$
                @Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
                $endgroup$
                – YiFan
                5 hours ago















              6












              $begingroup$

              Yes . . . and no.



              You might be interested in, for example, Wheel Theory, where division by zero is defined.






              share|cite|improve this answer











              $endgroup$








              • 10




                $begingroup$
                You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
                $endgroup$
                – Arthur
                8 hours ago






              • 1




                $begingroup$
                That's a fair comment, @Arthur. Thank you for the feedback.
                $endgroup$
                – Shaun
                8 hours ago






              • 1




                $begingroup$
                What d'you think, @SelimJeanEllieh?
                $endgroup$
                – Shaun
                8 hours ago






              • 1




                $begingroup$
                Oh: The OP has insufficient rep to comment. Nevermind.
                $endgroup$
                – Shaun
                8 hours ago






              • 5




                $begingroup$
                @Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
                $endgroup$
                – YiFan
                5 hours ago













              6












              6








              6





              $begingroup$

              Yes . . . and no.



              You might be interested in, for example, Wheel Theory, where division by zero is defined.






              share|cite|improve this answer











              $endgroup$



              Yes . . . and no.



              You might be interested in, for example, Wheel Theory, where division by zero is defined.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 8 hours ago

























              answered 8 hours ago









              ShaunShaun

              9,933113684




              9,933113684







              • 10




                $begingroup$
                You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
                $endgroup$
                – Arthur
                8 hours ago






              • 1




                $begingroup$
                That's a fair comment, @Arthur. Thank you for the feedback.
                $endgroup$
                – Shaun
                8 hours ago






              • 1




                $begingroup$
                What d'you think, @SelimJeanEllieh?
                $endgroup$
                – Shaun
                8 hours ago






              • 1




                $begingroup$
                Oh: The OP has insufficient rep to comment. Nevermind.
                $endgroup$
                – Shaun
                8 hours ago






              • 5




                $begingroup$
                @Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
                $endgroup$
                – YiFan
                5 hours ago












              • 10




                $begingroup$
                You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
                $endgroup$
                – Arthur
                8 hours ago






              • 1




                $begingroup$
                That's a fair comment, @Arthur. Thank you for the feedback.
                $endgroup$
                – Shaun
                8 hours ago






              • 1




                $begingroup$
                What d'you think, @SelimJeanEllieh?
                $endgroup$
                – Shaun
                8 hours ago






              • 1




                $begingroup$
                Oh: The OP has insufficient rep to comment. Nevermind.
                $endgroup$
                – Shaun
                8 hours ago






              • 5




                $begingroup$
                @Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
                $endgroup$
                – YiFan
                5 hours ago







              10




              10




              $begingroup$
              You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
              $endgroup$
              – Arthur
              8 hours ago




              $begingroup$
              You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
              $endgroup$
              – Arthur
              8 hours ago




              1




              1




              $begingroup$
              That's a fair comment, @Arthur. Thank you for the feedback.
              $endgroup$
              – Shaun
              8 hours ago




              $begingroup$
              That's a fair comment, @Arthur. Thank you for the feedback.
              $endgroup$
              – Shaun
              8 hours ago




              1




              1




              $begingroup$
              What d'you think, @SelimJeanEllieh?
              $endgroup$
              – Shaun
              8 hours ago




              $begingroup$
              What d'you think, @SelimJeanEllieh?
              $endgroup$
              – Shaun
              8 hours ago




              1




              1




              $begingroup$
              Oh: The OP has insufficient rep to comment. Nevermind.
              $endgroup$
              – Shaun
              8 hours ago




              $begingroup$
              Oh: The OP has insufficient rep to comment. Nevermind.
              $endgroup$
              – Shaun
              8 hours ago




              5




              5




              $begingroup$
              @Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
              $endgroup$
              – YiFan
              5 hours ago




              $begingroup$
              @Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
              $endgroup$
              – YiFan
              5 hours ago











              0












              $begingroup$

              That is quite right. However, I would like you to have a higher point of view.



              Mathematicians derive theorems from axioms and definitions. And here is the definition of a field.



              A field is a set $F$ equipped with two binary operations $+,times$, such that there exists $e_+, e_times$, such that for all $a,b,cin F$,

              - $a+b=b+a$,

              - $(a+b)+c=a+(b+c)$,

              - $e_++a=a$,

              - there exists $a'$ such that $a'+a=e_+$,

              - $(atimes b)times c=atimes (btimes c)$,

              - $e_timestimes a=a$,

              - there exists $a''$ such that $a''times a=e_times$ if $ane e_+$.



              Now verify that the rationals and the reals are fields.



              Try and prove that if there exists $x$ such that $xtimes e_+=e_times$, the set $F$ can only have one element.






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                That is quite right. However, I would like you to have a higher point of view.



                Mathematicians derive theorems from axioms and definitions. And here is the definition of a field.



                A field is a set $F$ equipped with two binary operations $+,times$, such that there exists $e_+, e_times$, such that for all $a,b,cin F$,

                - $a+b=b+a$,

                - $(a+b)+c=a+(b+c)$,

                - $e_++a=a$,

                - there exists $a'$ such that $a'+a=e_+$,

                - $(atimes b)times c=atimes (btimes c)$,

                - $e_timestimes a=a$,

                - there exists $a''$ such that $a''times a=e_times$ if $ane e_+$.



                Now verify that the rationals and the reals are fields.



                Try and prove that if there exists $x$ such that $xtimes e_+=e_times$, the set $F$ can only have one element.






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  That is quite right. However, I would like you to have a higher point of view.



                  Mathematicians derive theorems from axioms and definitions. And here is the definition of a field.



                  A field is a set $F$ equipped with two binary operations $+,times$, such that there exists $e_+, e_times$, such that for all $a,b,cin F$,

                  - $a+b=b+a$,

                  - $(a+b)+c=a+(b+c)$,

                  - $e_++a=a$,

                  - there exists $a'$ such that $a'+a=e_+$,

                  - $(atimes b)times c=atimes (btimes c)$,

                  - $e_timestimes a=a$,

                  - there exists $a''$ such that $a''times a=e_times$ if $ane e_+$.



                  Now verify that the rationals and the reals are fields.



                  Try and prove that if there exists $x$ such that $xtimes e_+=e_times$, the set $F$ can only have one element.






                  share|cite|improve this answer









                  $endgroup$



                  That is quite right. However, I would like you to have a higher point of view.



                  Mathematicians derive theorems from axioms and definitions. And here is the definition of a field.



                  A field is a set $F$ equipped with two binary operations $+,times$, such that there exists $e_+, e_times$, such that for all $a,b,cin F$,

                  - $a+b=b+a$,

                  - $(a+b)+c=a+(b+c)$,

                  - $e_++a=a$,

                  - there exists $a'$ such that $a'+a=e_+$,

                  - $(atimes b)times c=atimes (btimes c)$,

                  - $e_timestimes a=a$,

                  - there exists $a''$ such that $a''times a=e_times$ if $ane e_+$.



                  Now verify that the rationals and the reals are fields.



                  Try and prove that if there exists $x$ such that $xtimes e_+=e_times$, the set $F$ can only have one element.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 56 mins ago









                  TreborTrebor

                  95815




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