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












37












$begingroup$


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 that's impossible because all numbers multiplied by $0$ give $0$ 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 soft-question







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  • $begingroup$
    Sometimes division by zero is defined, such as in the extended complex plane.
    $endgroup$
    – Shaun
    Apr 1 at 19:57






  • 19




    $begingroup$
    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
    Apr 1 at 23:44








  • 1




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
















37












$begingroup$


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 that's impossible because all numbers multiplied by $0$ give $0$ 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 soft-question







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
    Apr 1 at 19:57






  • 19




    $begingroup$
    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
    Apr 1 at 23:44








  • 1




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














37












37








37


3



$begingroup$


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 that's impossible because all numbers multiplied by $0$ give $0$ 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 soft-question







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 that's impossible because all numbers multiplied by $0$ give $0$ 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 soft-question




proof-verification soft-question






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 yesterday









Jack

27.7k1782204




27.7k1782204






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asked Apr 1 at 19:54









Selim Jean ElliehSelim Jean Ellieh

19618




19618




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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
    Apr 1 at 19:57






  • 19




    $begingroup$
    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
    Apr 1 at 23:44








  • 1




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


















  • $begingroup$
    Sometimes division by zero is defined, such as in the extended complex plane.
    $endgroup$
    – Shaun
    Apr 1 at 19:57






  • 19




    $begingroup$
    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
    Apr 1 at 23:44








  • 1




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
















$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
Apr 1 at 19:57




$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
Apr 1 at 19:57




19




19




$begingroup$
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
Apr 1 at 23:44






$begingroup$
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
Apr 1 at 23:44






1




1




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




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










4 Answers
4






active

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45












$begingroup$

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









$endgroup$









  • 7




    $begingroup$
    +1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
    $endgroup$
    – Martin Argerami
    2 days ago



















10












$begingroup$

Yes . . . and no.



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



See Lemma 2 of the 1997 article "Wheels," by A. Setzer for tables describing addition, multiplication, and their inverses in what is called $R_bot^infty$, the wheel given by adjoining special symbols and rules to an arbitrary integral domain $R$ in order to allow division by zero, even $frac{0}{0}=:bot$.






share|cite|improve this answer











$endgroup$









  • 22




    $begingroup$
    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
    Apr 1 at 20:02






  • 3




    $begingroup$
    That's a fair comment, @Arthur. Thank you for the feedback.
    $endgroup$
    – Shaun
    Apr 1 at 20:03






  • 2




    $begingroup$
    What d'you think, @SelimJeanEllieh?
    $endgroup$
    – Shaun
    Apr 1 at 20:04






  • 2




    $begingroup$
    Oh: The OP has insufficient rep to comment. Nevermind.
    $endgroup$
    – Shaun
    Apr 1 at 20:06






  • 10




    $begingroup$
    @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
    Apr 1 at 22:34



















2












$begingroup$

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









$endgroup$









  • 7




    $begingroup$
    While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
    $endgroup$
    – Chef Cyanide
    2 days ago






  • 4




    $begingroup$
    In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
    $endgroup$
    – Vaelus
    2 days ago








  • 2




    $begingroup$
    @Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
    $endgroup$
    – Henning Makholm
    2 days ago






  • 1




    $begingroup$
    @HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
    $endgroup$
    – Sir Jective
    2 days ago






  • 1




    $begingroup$
    @HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
    $endgroup$
    – Sir Jective
    2 days ago



















0












$begingroup$

You are quite right.



There is a simpler way, though (which spares the concept of multiplicative inverse):



By definition, $q$ is the quotient of the division of $d$ by $0$ if the following equation is satisfied:



$$0cdot q=d.$$



But we know that $0cdot q=0$, so the equation has no solution (unless $d=0$).






share|cite|improve this answer









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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    45












    $begingroup$

    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









    $endgroup$









    • 7




      $begingroup$
      +1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
      $endgroup$
      – Martin Argerami
      2 days ago
















    45












    $begingroup$

    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









    $endgroup$









    • 7




      $begingroup$
      +1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
      $endgroup$
      – Martin Argerami
      2 days ago














    45












    45








    45





    $begingroup$

    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 Apr 1 at 20:00









    ArthurArthur

    122k7122210




    122k7122210








    • 7




      $begingroup$
      +1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
      $endgroup$
      – Martin Argerami
      2 days ago














    • 7




      $begingroup$
      +1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
      $endgroup$
      – Martin Argerami
      2 days ago








    7




    7




    $begingroup$
    +1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
    $endgroup$
    – Martin Argerami
    2 days ago




    $begingroup$
    +1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
    $endgroup$
    – Martin Argerami
    2 days ago











    10












    $begingroup$

    Yes . . . and no.



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



    See Lemma 2 of the 1997 article "Wheels," by A. Setzer for tables describing addition, multiplication, and their inverses in what is called $R_bot^infty$, the wheel given by adjoining special symbols and rules to an arbitrary integral domain $R$ in order to allow division by zero, even $frac{0}{0}=:bot$.






    share|cite|improve this answer











    $endgroup$









    • 22




      $begingroup$
      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
      Apr 1 at 20:02






    • 3




      $begingroup$
      That's a fair comment, @Arthur. Thank you for the feedback.
      $endgroup$
      – Shaun
      Apr 1 at 20:03






    • 2




      $begingroup$
      What d'you think, @SelimJeanEllieh?
      $endgroup$
      – Shaun
      Apr 1 at 20:04






    • 2




      $begingroup$
      Oh: The OP has insufficient rep to comment. Nevermind.
      $endgroup$
      – Shaun
      Apr 1 at 20:06






    • 10




      $begingroup$
      @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
      Apr 1 at 22:34
















    10












    $begingroup$

    Yes . . . and no.



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



    See Lemma 2 of the 1997 article "Wheels," by A. Setzer for tables describing addition, multiplication, and their inverses in what is called $R_bot^infty$, the wheel given by adjoining special symbols and rules to an arbitrary integral domain $R$ in order to allow division by zero, even $frac{0}{0}=:bot$.






    share|cite|improve this answer











    $endgroup$









    • 22




      $begingroup$
      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
      Apr 1 at 20:02






    • 3




      $begingroup$
      That's a fair comment, @Arthur. Thank you for the feedback.
      $endgroup$
      – Shaun
      Apr 1 at 20:03






    • 2




      $begingroup$
      What d'you think, @SelimJeanEllieh?
      $endgroup$
      – Shaun
      Apr 1 at 20:04






    • 2




      $begingroup$
      Oh: The OP has insufficient rep to comment. Nevermind.
      $endgroup$
      – Shaun
      Apr 1 at 20:06






    • 10




      $begingroup$
      @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
      Apr 1 at 22:34














    10












    10








    10





    $begingroup$

    Yes . . . and no.



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



    See Lemma 2 of the 1997 article "Wheels," by A. Setzer for tables describing addition, multiplication, and their inverses in what is called $R_bot^infty$, the wheel given by adjoining special symbols and rules to an arbitrary integral domain $R$ in order to allow division by zero, even $frac{0}{0}=:bot$.






    share|cite|improve this answer











    $endgroup$



    Yes . . . and no.



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



    See Lemma 2 of the 1997 article "Wheels," by A. Setzer for tables describing addition, multiplication, and their inverses in what is called $R_bot^infty$, the wheel given by adjoining special symbols and rules to an arbitrary integral domain $R$ in order to allow division by zero, even $frac{0}{0}=:bot$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago

























    answered Apr 1 at 19:59









    ShaunShaun

    10.3k113686




    10.3k113686








    • 22




      $begingroup$
      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
      Apr 1 at 20:02






    • 3




      $begingroup$
      That's a fair comment, @Arthur. Thank you for the feedback.
      $endgroup$
      – Shaun
      Apr 1 at 20:03






    • 2




      $begingroup$
      What d'you think, @SelimJeanEllieh?
      $endgroup$
      – Shaun
      Apr 1 at 20:04






    • 2




      $begingroup$
      Oh: The OP has insufficient rep to comment. Nevermind.
      $endgroup$
      – Shaun
      Apr 1 at 20:06






    • 10




      $begingroup$
      @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
      Apr 1 at 22:34














    • 22




      $begingroup$
      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
      Apr 1 at 20:02






    • 3




      $begingroup$
      That's a fair comment, @Arthur. Thank you for the feedback.
      $endgroup$
      – Shaun
      Apr 1 at 20:03






    • 2




      $begingroup$
      What d'you think, @SelimJeanEllieh?
      $endgroup$
      – Shaun
      Apr 1 at 20:04






    • 2




      $begingroup$
      Oh: The OP has insufficient rep to comment. Nevermind.
      $endgroup$
      – Shaun
      Apr 1 at 20:06






    • 10




      $begingroup$
      @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
      Apr 1 at 22:34








    22




    22




    $begingroup$
    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
    Apr 1 at 20:02




    $begingroup$
    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
    Apr 1 at 20:02




    3




    3




    $begingroup$
    That's a fair comment, @Arthur. Thank you for the feedback.
    $endgroup$
    – Shaun
    Apr 1 at 20:03




    $begingroup$
    That's a fair comment, @Arthur. Thank you for the feedback.
    $endgroup$
    – Shaun
    Apr 1 at 20:03




    2




    2




    $begingroup$
    What d'you think, @SelimJeanEllieh?
    $endgroup$
    – Shaun
    Apr 1 at 20:04




    $begingroup$
    What d'you think, @SelimJeanEllieh?
    $endgroup$
    – Shaun
    Apr 1 at 20:04




    2




    2




    $begingroup$
    Oh: The OP has insufficient rep to comment. Nevermind.
    $endgroup$
    – Shaun
    Apr 1 at 20:06




    $begingroup$
    Oh: The OP has insufficient rep to comment. Nevermind.
    $endgroup$
    – Shaun
    Apr 1 at 20:06




    10




    10




    $begingroup$
    @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
    Apr 1 at 22:34




    $begingroup$
    @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
    Apr 1 at 22:34











    2












    $begingroup$

    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









    $endgroup$









    • 7




      $begingroup$
      While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
      $endgroup$
      – Chef Cyanide
      2 days ago






    • 4




      $begingroup$
      In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
      $endgroup$
      – Vaelus
      2 days ago








    • 2




      $begingroup$
      @Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
      $endgroup$
      – Henning Makholm
      2 days ago






    • 1




      $begingroup$
      @HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
      $endgroup$
      – Sir Jective
      2 days ago






    • 1




      $begingroup$
      @HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
      $endgroup$
      – Sir Jective
      2 days ago
















    2












    $begingroup$

    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









    $endgroup$









    • 7




      $begingroup$
      While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
      $endgroup$
      – Chef Cyanide
      2 days ago






    • 4




      $begingroup$
      In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
      $endgroup$
      – Vaelus
      2 days ago








    • 2




      $begingroup$
      @Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
      $endgroup$
      – Henning Makholm
      2 days ago






    • 1




      $begingroup$
      @HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
      $endgroup$
      – Sir Jective
      2 days ago






    • 1




      $begingroup$
      @HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
      $endgroup$
      – Sir Jective
      2 days ago














    2












    2








    2





    $begingroup$

    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 2 days ago









    TreborTrebor

    99815




    99815








    • 7




      $begingroup$
      While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
      $endgroup$
      – Chef Cyanide
      2 days ago






    • 4




      $begingroup$
      In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
      $endgroup$
      – Vaelus
      2 days ago








    • 2




      $begingroup$
      @Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
      $endgroup$
      – Henning Makholm
      2 days ago






    • 1




      $begingroup$
      @HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
      $endgroup$
      – Sir Jective
      2 days ago






    • 1




      $begingroup$
      @HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
      $endgroup$
      – Sir Jective
      2 days ago














    • 7




      $begingroup$
      While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
      $endgroup$
      – Chef Cyanide
      2 days ago






    • 4




      $begingroup$
      In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
      $endgroup$
      – Vaelus
      2 days ago








    • 2




      $begingroup$
      @Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
      $endgroup$
      – Henning Makholm
      2 days ago






    • 1




      $begingroup$
      @HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
      $endgroup$
      – Sir Jective
      2 days ago






    • 1




      $begingroup$
      @HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
      $endgroup$
      – Sir Jective
      2 days ago








    7




    7




    $begingroup$
    While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
    $endgroup$
    – Chef Cyanide
    2 days ago




    $begingroup$
    While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
    $endgroup$
    – Chef Cyanide
    2 days ago




    4




    4




    $begingroup$
    In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
    $endgroup$
    – Vaelus
    2 days ago






    $begingroup$
    In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
    $endgroup$
    – Vaelus
    2 days ago






    2




    2




    $begingroup$
    @Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
    $endgroup$
    – Henning Makholm
    2 days ago




    $begingroup$
    @Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
    $endgroup$
    – Henning Makholm
    2 days ago




    1




    1




    $begingroup$
    @HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
    $endgroup$
    – Sir Jective
    2 days ago




    $begingroup$
    @HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
    $endgroup$
    – Sir Jective
    2 days ago




    1




    1




    $begingroup$
    @HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
    $endgroup$
    – Sir Jective
    2 days ago




    $begingroup$
    @HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
    $endgroup$
    – Sir Jective
    2 days ago











    0












    $begingroup$

    You are quite right.



    There is a simpler way, though (which spares the concept of multiplicative inverse):



    By definition, $q$ is the quotient of the division of $d$ by $0$ if the following equation is satisfied:



    $$0cdot q=d.$$



    But we know that $0cdot q=0$, so the equation has no solution (unless $d=0$).






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You are quite right.



      There is a simpler way, though (which spares the concept of multiplicative inverse):



      By definition, $q$ is the quotient of the division of $d$ by $0$ if the following equation is satisfied:



      $$0cdot q=d.$$



      But we know that $0cdot q=0$, so the equation has no solution (unless $d=0$).






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You are quite right.



        There is a simpler way, though (which spares the concept of multiplicative inverse):



        By definition, $q$ is the quotient of the division of $d$ by $0$ if the following equation is satisfied:



        $$0cdot q=d.$$



        But we know that $0cdot q=0$, so the equation has no solution (unless $d=0$).






        share|cite|improve this answer









        $endgroup$



        You are quite right.



        There is a simpler way, though (which spares the concept of multiplicative inverse):



        By definition, $q$ is the quotient of the division of $d$ by $0$ if the following equation is satisfied:



        $$0cdot q=d.$$



        But we know that $0cdot q=0$, so the equation has no solution (unless $d=0$).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        Yves DaoustYves Daoust

        132k676229




        132k676229






















            Selim Jean Ellieh is a new contributor. Be nice, and check out our Code of Conduct.










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            Selim Jean Ellieh is a new contributor. Be nice, and check out our Code of Conduct.













            Selim Jean Ellieh is a new contributor. Be nice, and check out our Code of Conduct.












            Selim Jean Ellieh is a new contributor. Be nice, and check out our Code of Conduct.
















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