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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?

Selim Jean Ellieh
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  • Sometimes division by zero is defined, such as in the extended complex plane. – Shaun Apr 01 '19 at 19:57
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    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. – Fomalhaut Apr 01 '19 at 23:44
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    https://math.stackexchange.com/questions/2883450/how-do-you-explain-to-a-5th-grader-why-division-by-zero-is-meaningless/2883454#2883454 – nonuser Apr 02 '19 at 01:28
  • @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 ? – Our Apr 20 '19 at 04:59

4 Answers4

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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 $a = c\cdot b$)
  • $0\neq1$
Arthur
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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 $\frac{0}{0}=:\bot$.

Shaun
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    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? – Arthur Apr 01 '19 at 20:02
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    That's a fair comment, @Arthur. Thank you for the feedback. – Shaun Apr 01 '19 at 20:03
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    What d'you think, @SelimJeanEllieh? – Shaun Apr 01 '19 at 20:04
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    Oh: The OP has insufficient rep to comment. Nevermind. – Shaun Apr 01 '19 at 20:06
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    Ye I'm understanding but i did not understand anything from thos wheel theory – Selim Jean Ellieh Apr 01 '19 at 20:23
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    @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. – YiFan Tey Apr 01 '19 at 22:34
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    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). – bob Apr 02 '19 at 02:53
  • The Wikipedia article links to a 1997 paper. There, Cayley tables are provided for $+, \times, \frac{1}{\cdot}, $ etc. I'll edit this answer later, @bob, to include them. – Shaun Apr 02 '19 at 03:28
  • Is that better, @bob? – Shaun Apr 02 '19 at 10:21
  • Yes--it gives some hints that there's some higher math where it is defined, while I think showing that for the math that OP is currently doing (and will be for at least the time being), OP is spot on. – bob Apr 02 '19 at 12:29
  • Sorry for the noisy comment, but this is SO cool! – Easymode44 Apr 02 '19 at 13:42
2

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.

Trebor
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    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). –  Apr 02 '19 at 04:49
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    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 $$. – Vaelus Apr 02 '19 at 05:40
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    @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. – hmakholm left over Monica Apr 02 '19 at 09:37
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    @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. – silvascientist Apr 02 '19 at 18:48
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    @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. – silvascientist Apr 02 '19 at 18:50
  • @SirJective: This is a 9th-grader who spontaneously starts sentences with "Consider". You will not be able to convince me that standard mathematical turns of phrase hold any terror for them. – hmakholm left over Monica Apr 02 '19 at 18:51
  • @HenningMakholm I would try to reason about why certain mathematical rules were true in high school too, but it was always informally, because I didn't yet have the background to do it formally. Introducing formality is good, but it should actually be introduced. The meaning of "consider" in math contexts is the same as in plain English. The meaning of "There exists an $x$ such that for all $y$, $x \sim y$" is much more specific, so it's not as easy to pick up without an explicit explanation, even for adults. – Vaelus Apr 03 '19 at 00:25
  • In China, we are supposed to understand most of the jargon you listed by the 9th Grade, and it rapidly gets more formal in the 10th Grade. – Trebor Apr 03 '19 at 05:27
1

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 $d=0$).