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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?

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  • $\begingroup$ Sometimes division by zero is defined, such as in the extended complex plane. $\endgroup$
    – Shaun
    Apr 1, 2019 at 19:57
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    $\begingroup$ 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. $\endgroup$
    – HolyKnowing
    Apr 1, 2019 at 23:44
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    $\begingroup$ math.stackexchange.com/questions/2883450/… $\endgroup$
    – nonuser
    Apr 2, 2019 at 1:28
  • $\begingroup$ @ErotemeObelus I think giving a career advice, such as "you should study this and this" based upon skill is extremely dangerous; I cook very successful bread of various kinds, should I be baker even though I cannot image a life without mathematics and physics ? $\endgroup$
    – Our
    Apr 20, 2019 at 4:59

4 Answers 4

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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 ($1\cdot a = a$ for any $a$)
  • What $0$ means ($0 \cdot a = 0$ for any $a$) (actually a consequence of $0+a=a$ and $(a+b)\cdot c=a\cdot c+a\cdot b$, two other Nice Things)
  • What division means ($\frac ab = c$ means $a = c\cdot b$)
  • $0\neq1$
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    $\begingroup$ +1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable). $\endgroup$
    – Martin Argerami
    Apr 2, 2019 at 19:02
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    $\begingroup$ @MartinArgerami Thank you. I finally decided to add that to my answer. $\endgroup$
    – Arthur
    Apr 7, 2019 at 10:57
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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 $\frac{0}{0}=:\bot$.

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    $\begingroup$ 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, 2019 at 20:02
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    $\begingroup$ That's a fair comment, @Arthur. Thank you for the feedback. $\endgroup$
    – Shaun
    Apr 1, 2019 at 20:03
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    $\begingroup$ Oh: The OP has insufficient rep to comment. Nevermind. $\endgroup$
    – Shaun
    Apr 1, 2019 at 20:06
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    $\begingroup$ @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, 2019 at 22:34
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    $\begingroup$ Something that might improve this answer is if you could provide some high points of Wheel Theory at an accessible level (I have a post-graduate level education in engineering, and I find the Wikipedia article intimidating). $\endgroup$
    – bob
    Apr 2, 2019 at 2:53
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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 $a,b,c\in F$,
- $a+b=b+a$,
- $(a+b)+c=a+(b+c)$,
- $e_++a=a$,
- there exists $a'$ such that $a'+a=e_+$,
- $(a\times b)\times c=a\times (b\times c)$,
- $e_\times\times a=a$,
- there exists $a''$ such that $a''\times a=e_\times$ if $a\ne e_+$.

Now verify that the rationals and the reals are fields.

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

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    $\begingroup$ 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). $\endgroup$
    – user303830
    Apr 2, 2019 at 4:49
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    $\begingroup$ 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
    Apr 2, 2019 at 5:40
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    $\begingroup$ @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. $\endgroup$
    – hmakholm left over Monica
    Apr 2, 2019 at 9:37
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    $\begingroup$ @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. $\endgroup$
    – silvascientist
    Apr 2, 2019 at 18:48
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    $\begingroup$ @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. $\endgroup$
    – silvascientist
    Apr 2, 2019 at 18:50
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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:

$$0\cdot q=d.$$

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

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  • $\begingroup$ but this whould lead to another question, 0/0 is indefinite and could be any real number. $\endgroup$
    – Larry Eppes
    Apr 7, 2019 at 7:28
  • $\begingroup$ @LarryEppes: or not be a real number. $\endgroup$
    – user65203
    Apr 7, 2019 at 8:08

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