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My question is $\frac{0}{0}*0=?$ I think it should be zero. Beause $\frac00$ can be any number (both real or imaginary). And I think any number multiplied by $0$ should be $0$. I know the proof like this: x*0=y*0 so (0/0)=(x/y) Thus (x/y) can be any number. If this is a common question, please do not downvote?

Asif Iqubal
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  • Why is $0/0$ any number? Why is it even a number? – Michael Burr Feb 05 '18 at 14:58
  • By 'any number' I think he means that form is indeterminate, or as some would say, a variable. – Allawonder Feb 05 '18 at 15:01
  • @Allawonder That presupposes that $0/0$ is a number, albeit unknown. – Michael Burr Feb 05 '18 at 15:05
  • The word "indeterminate" usually comes up in calculus, in the context of limits. Never have I seen "indeterminate" used to mean "both real or imaginary", whatever that means. – G Tony Jacobs Feb 05 '18 at 15:15
  • @MichaelBurr No. It doesn't presuppose it, for $0/0$ is indeed some arbitrary complex number (assuming we're talking about this particular field). I realise many don't know why $0/0$ is usually not allowed in basic arithmetic, but it's not because it's meaningless. The form $a/0$ is meaningless only when $a$ is nonzero (or in general, not the additive identity of a field). – Allawonder Feb 05 '18 at 15:38
  • @GTonyJacobs As I've said below, I should have used non-determinate instead since indeterminate may lead to associations about limits of indeterminate forms. – Allawonder Feb 05 '18 at 15:40
  • No, 0/0 does not represent an arbitrary element of a field. It is simply undefined; you can't divide by 0. – Théophile Feb 05 '18 at 17:11
  • Related, almost a duplicate : https://math.stackexchange.com/questions/26445/division-by-0 – Arnaud D. Feb 05 '18 at 17:17
  • @Théophile LOL. Merely asserting something, even with fervour akin to that of a religious zealot, doesn't make it true. I never said division by zero is defined, so calm down. What we're saying is that the symbols $0/0$ represent an arbitrary member of some field. That doesn't mean I've defined division by zero or something. Think. Understand. – Allawonder Feb 05 '18 at 17:29
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    @Allawonder, “Merely asserting something.... doesn’t make it true”. Physician, heal thyself – G Tony Jacobs Feb 05 '18 at 17:52
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    Maybe in your private mathematics, $0/0$ represents an arbitrary field element, but to the worldwide community of mathematicians, it does not. – G Tony Jacobs Feb 05 '18 at 17:53
  • @GTonyJacobs LOL. That's irrelevant to anything here. – Allawonder Feb 05 '18 at 17:53
  • @ArnaudD. Good dig. I'd recommend this for those who find it hard to comprehend that something can be meaningful or possible and yet be undefined. – Allawonder Feb 05 '18 at 17:56

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It is not $0$. And it is not different from $0$. It is just meaningless.

José Carlos Santos
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  • It is not meaningless. Any algebraic extension of arithmetic requires that an indeterminate multiplied by zero be zero. – Allawonder Feb 05 '18 at 14:59
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    @Allawonder But no algebraic extension of arithmetic can make sense of $0/0$. – Arnaud D. Feb 05 '18 at 15:55
  • @ArnaudD. That's not what I'm saying. If $0/0=x$ for some $x\in F$, where $F$ is some field, then $0\cdot x=0$. This holds for all $x\in F$. – Allawonder Feb 05 '18 at 16:01
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    I do agree that this implication is logically true, but only because the hypothesis is never satisfied. – Arnaud D. Feb 05 '18 at 16:07
  • @ArnaudD. Nice reply! – José Carlos Santos Feb 05 '18 at 16:08
  • @ArnaudD. LOL, actually it's stronger than an implication; it's an equivalence. But consider the inverse implication: If $0\cdot x=0$ for some $x$ in some field as before, then $x=0/0$. Is the hypothesis also not satisfied in this case? :) Indeed in the reverse case, whether the hypothesis is satisfied is irrelevant to whether the implication is true. This is basic logic. You can only prove an implication is false iff the consequent is false and the hypothesis true. So any implication 'If $A$ then $B$' is true so long as 'then $B$' isn't false whenever 'If $A$' is true. – Allawonder Feb 05 '18 at 16:45
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My question is (0/0)*0=?

It's not defined because $0/0$ is not defined

Beause (0/0) is any number

No it's not.

And any $\color{red}{\text{number}}$ multiplied by 0 should be 0

True, but as said before, $0/0$ is not a number.

5xum
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  • I think what he meant to say is that $0/0$ is an indeterminate. In that case he is indeed right. – Allawonder Feb 05 '18 at 15:00
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    @Allawonder, Why should an indeterminate times $0$ equal $0$? I can think of situations involving limits where that's not true. – G Tony Jacobs Feb 05 '18 at 15:07
  • @Allawonder Then they should replace "any number" with "indeterminate" in his question. "indeterminate" and "any number" is not the same thing, and I prefer to answer the questions that are asked, not those that maybe were and maybe weren't meant to be asked. – 5xum Feb 05 '18 at 15:08
  • Are we using the words "intermediate" and "indeterminate" interchangeably? – G Tony Jacobs Feb 05 '18 at 15:09
  • @GTonyJacobs It's a case of my fingers being quicker than my brain... – 5xum Feb 05 '18 at 15:10
  • @5xum, it happens, lol. I'm still confused about the definition of "indeterminate" that Allawonder is using. – G Tony Jacobs Feb 05 '18 at 15:11
  • @GTonyJacobs By indeterminate I simply mean a symbol which represents no particular number, matrix, or some other mathematical object, but is used to represent arbitrarily any such object. Most people call these variables or unknowns. – Allawonder Feb 05 '18 at 15:15
  • Thank you. When we work with a variable, we usually specify what type of object it represents. Thus "real variable", "complex variable", etc. If a variable is to represent an arbitrary matrix, we say: "Consider an arbitrary matrix $A$". In this case, what kind of variable can be used to replace the expression $0/0$, which is not any real number? – G Tony Jacobs Feb 05 '18 at 15:17
  • @GTonyJacobs As I said, calling it a variable is inappropriate outside the context of functions. Thus I prefer the word unknown or non-determinate. And no, it's unnecessary to specify what type of object you're working with as long as it belongs to some field. In this case, I think the OP has the complex field in mind, so that $0/0$ represents an arbitrary complex number. However, this applies to all fields. Thus the form $0/0$ is a somewhat cumbersome representation of an arbitrary member of some field. – Allawonder Feb 05 '18 at 15:45
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Common problem in mathematics to think about expressions as computational steps which you carry out in your head, so you can talk about steps being undefined. Mathematics is all about statements and the rules which connect these statements. When you write $\frac{0}{0}$ you say "the number which is 0 multiplied by the inverse of 0" at this instant you introduced an object the existence of which you have not proven (the inverse of zero) so your question can be rephrased as:

"Assuming the existence of the inverse of 0 and denoting it with $\frac{1}{0}$ what $\frac{0}{0}0$ is equal to?" Since your assumption leads to contradiction the answer is everything.

Adam
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