Talk:Division by zero/Archive 2

Directional Limits and 0+

 * When one is evaluating the limit lim x->0+ (1/x) = +infinity, although it's intuitively reasonable to think of the limit as lim x->0+ (1/x) = 1/0+ = +infinity, that is not technically correct, since 0+ is not a number, and so you can't divide by it. I ran across someone today who was misled by this, so I corrected the page to fix it. Twiffy (talk) 02:07, 10 June 2011 (UTC)

1/0+ is no more than a shorthand way of writing lim x->0+ (1/x). But you may be right that this notation can confuse some readers, and is best omitted. FilipeS (talk) 14:21, 6 February 2012 (UTC)

0/0 = 1, obviously...
The function x/x = 1 for every x on the complex plane, with a tiny, point sized hole of WTF? at zero. That says you can get arbitrarily close to zero, and the value of x/x is still 1. Madkaugh (talk) 21:32, 10 August 2009 (UTC)


 * The function (2x)/x = 2 for every x on the complex plane, with a tiny, point sized hole of WTF? at zero. That says you can get arbitrarily close to zero, and the value of (2x)/x is still 2. — JAO • T • C 05:05, 11 August 2009 (UTC)


 * Indeed it is. But you introduced an additional factor:


 * (2x)/x = 2(x/x) =* 2(1) = 2


 * (allowing that you can get arbitrarily close ...)


 * I said one is one, you said two times one is two. I say tomato, you say catsup is a vegetable ... Madkaugh (talk) 00:27, 12 August 2009 (UTC)


 * My point is that a fraction is not guaranteed to approach 1 as x approaches 0, just because the numerator and denominator both approach 0. For the fraction to approach 1, we need the numerator and denominator to approach 0 in the same way (like x/x), and there's no such information in the expression "0/0". Taking 0/0 to mean the limit of x/x, discarding any other limit of the form 0/0, is completely arbitrary. (We have an article about this at indeterminate form.) — JAO • T • C 05:14, 12 August 2009 (UTC)


 * Good point. Madkaugh (talk) 02:06, 13 August 2009 (UTC)

Also the function f(x=)sin(x)/tan(x) proves that 0/0=1. —Preceding unsigned comment added by 96.255.72.164 (talk) 18:39, 18 October 2009 (UTC)


 * Actually, the function f(x)=sinx/tanx can be rearranged into f(x)=(sinx)*(cosx/sinx), at which point is simplified to f(x)=cosx. The only thing you're proving is that cos(0)=1. — Preceding unsigned comment added by 68.62.43.18 (talk) 21:28, 16 December 2011 (UTC)


 * As mentioned in the article, see L'Hopital's Rule for how to, in general, handle two functions g(x)/h(x), when both g(x) and h(x) yield 0 at x=0. The solution has been around for over 300 years. Bill Wvbailey (talk) 23:06, 16 December 2011 (UTC)


 * hmm... lets assume z=0. if z/z=x, then in theory x=1. yet if z=0, then 0/z=x, and then x=0. so, x would equal 1 OR x would equal 0, in the same manner as y equals 2 OR -2 when y equals the square root of 4 Didlybob123 (talk) 03:43, 27 February 2013 (UTC)

You can divide by zero
.. when it becomes necessary to do so. The universe does it all the time. Here's an article: Dirac delta function - it crops up in discrete probabilities expressed as integrals and in signal processing. Madkaugh (talk) 21:32, 10 August 2009 (UTC)


 * Evidently Chuck Norris does it all the time too.
 * (I actually went to this article looking for that tidbit, but found it on Chuck Norris facts).


 * Actually, the Dirac delta function doesn't give you a way to divide by zero. That's simply a zero-width function having a finite integral. It's a construct. There are plenty of seeminly contradictory things in mathematics; for example if you rotate the curve y=1/x around the x axis from 1 to infinity, you get a trumpet-shaped surface that has a finite volume but infinite area; in other words you can fill it with paint but you can't paint it.


 * There are also examples like y=sin(x)/x that blow up mathematically when x=0, but actually have a value there. sin(x)/x=1 as x goes to 0. ~Amatulić (talk) 19:58, 1 December 2009 (UTC)


 * ...are you suggesting that we mention these things (black hole, Dirac delta function) in the article? If so, that's a good idea; feel free to go ahead and add them.


 * As for the Chuck Norris fact, I think it was considered not notable enough for this article. --Zarel (talk) 10:40, 2 December 2009 (UTC)

It is possible to divide by zero in real life situations, let us say that I have 5 sweets. When explaining division you use the sharing example, right? So, if I were to divide by one, a single person would gain the whole quantity of the object/s. When dividing by zero, you still have to give away all of the quantity/objects but there is not a person! It's almost as if you're putting it in the trash, but in maths' formulae you cannot get rid of the complete quantity without subtraction, so the calculator cannot process it. In other words dividing by zero is sharing with nobody but you're still willing to share so you keep none. Hope that clarified everything.80.194.229.243 (talk) 20:44, 26 July 2012 (UTC)

I'm not sure that's quite accurate. the truth is, we don't really know a real life aplication for divided by zero concepts, but that doesn't mean their isn't one. I believe that dividing by zero is possible in some cases, but the idea that dividing 5 cookies into 0 groups is the same as throwing away the cookies wouldn't be dividing by zero, but more along the lines of giving your cookies to the trash can. this situation would be 5/1, not 5/0. 12.207.89.80 (talk) 02:28, 28 February 2013 (UTC)

zero divied by zero fallacy proof
I have one problem with that proof. 1/0 does not equal 1. 2/0 does not equal 2. They both equal infinity. So it leads to the conclusion 0/0+infinity=0/0+infinity, which explains nothing since its true. —Preceding unsigned comment added by 96.255.72.164 (talk) 18:34, 18 October 2009 (UTC)


 * 1/0 obviously doesn't equal 1, but it doesn't equal ∞ either, as that would lead to 0 × ∞ equalling 1. If there's something you don’t understand about division by zero please read the article, don't use this talk page as a forum. Robo37 (talk) 18:47, 18 October 2009 (UTC)


 * I was unable to find a 1/0 in the fallacy proof section never mind an assertion that it was equal to 1. You'll need to be more specific about exactly where you find a problem in an article. Dmcq (talk) 18:58, 18 October 2009 (UTC)

I would like to make known the part where the article talks about 10/0. if one were to set 10/0=x, then x times 0 would have to equal 10. the article states that x*0=0, and not 10, but if you put 10/0 in for x, you'd get (10/0)*0=10. you'd then move the 0 in the denominator, resulting in 10*(0/0)=10. now, any number divided by itself equals 1, and 0 divided by any number equals 0, so 0/0 would equal 1 or 0. this would give you 2 possibilities, the first being 10*1=10, and the second being 10*0=10. the second would obviously be extraneous, and therefore x*0=10 if x=10/0 Didlybob123 (talk) 03:08, 28 February 2013 (UTC)

Kaplan example error?
There seems to be a basic flaw in the "Fallacies based on division by zero" example.

It says

"The following must be true:


 * $$0\times 1 = 0\times 2.\,$$"

and then

"Dividing by zero gives:


 * $$\textstyle \frac{0}{0}\times 1 = \frac{0}{0}\times 2.$$"

But that is NOT correct by basic algebra as you divide the entire equation by the value. So the equation should be


 * $$\textstyle \frac{0}{0}\times \frac{1}{0} = \frac{0}{0}\times \frac{2}{0}.$$

So either something is missing from the explanation or something else is wrong.--BruceGrubb (talk) 03:23, 1 November 2009 (UTC)


 * You've divided by zero twice. I'm not sure what you're up to. Perhaps you could divide the following by 5 instead of 0 in the same way just to show me what you're up to:


 * $$2 \times 3 = 6.\,$$


 * Thanks Dmcq (talk) 08:29, 1 November 2009 (UTC)


 * Sorry miswrote the above because I'm not used to doing formulas in wiki.


 * $$\frac{2 \times 3 = 6}{5}.\,$$ or $$\frac{2 \times 3}{5} = \frac{6}{5}.\,$$


 * So the above should be


 * $$\textstyle \frac{0\times 1}{0} = \frac{0\times 2}{0}.$$"


 * Yes that explains matters thanks. It should be


 * $$\frac{2}{5} \times 3 = \frac{6}{5}.\,$$
 * or
 * $$2 \times \frac{3}{5} = \frac{6}{5}.\,$$

or as you said the two multiplied together and then divided by 5.


 * These work out as
 * 0.4 × 3 = 1.2 for the first one
 * 2 × 0.6 = 1.2 for the second one
 * 6 ÷ 5 = 1,2
 * Your version
 * $$\frac{2}{5} \times \frac{3}{5} = \frac{6}{5}.\,$$
 * comes out as
 * 0.4 × 0.6 which is 0.24 whereas the right hand side is 1.2
 * Dividing everything on the left hand side would be correct if it was an addition or subtraction but not for a multiplication or division. For instance the following is quite correct
 * $$2 + 3 = 5 \,$$
 * so
 * $$\frac{2}{5} + \frac{3}{5} = \frac{5}{5} \,$$
 * or evaluating the bits
 * 0,4 + 0.6 = 1
 * If you hit yourself on the forehead and say oh yes! then that's fine otherwise you'll need a bit of practice with this sort of thing to make it all clearer. The article Fraction (mathematics) goes into all this but it is an encyclopaedia article rather than a tutorial. Dmcq (talk) 23:21, 1 November 2009 (UTC)


 * The problem is that $$\frac{0}{0}$$ is a special case as it is the one time the


 * if $$ {a \times b} = {c}$$ then $$ {a} = \frac {c}{b}$$ doesn't produce nonsense if b is 0 because any number times b equals 0. The problem is that c can be any number and you have two conflicting rules involved:


 * $$\frac{0}{b} = {0}$$ and $$\frac{a}{a} = {1}$$


 * If you go with the $$\frac{0}{b} = {0}$$ position then the equation makes sense again.--BruceGrubb (talk) 23:42, 1 November 2009 (UTC)

This just proves that $$\frac{a}{a} = {1}$$ is wrong. Can anyone honestly tell me that $$\frac{0}{0} = 1$$? This just makes no sense. The answer could either be undefined or 0. Just think about learning devision: I have 0 objects and I put them in 0 groups. How many objects are in a (nonexistent) group? If you say 1, something is wrong with you. 74.192.52.24 (talk) 20:14, 23 November 2010 (UTC)

Can anyone verify whether the "proof" is present in the Kaplan source? If it is, we don't really need the above discussion, and more importantly, we can undo this useless edit and bring the section back in line with the source. DVdm (talk) 20:29, 23 November 2010 (UTC)

In Elementary Arithmetic
Here, we mention an example of how many people satisfied with nothing can be satisfied with one apple. But this question should really be worded like this: "how many people that do not want anything can we satisfy with one apple, if it is mandatory to give everyone an apple?" In this case, each person would get an infinitely small part of the apple, so the number of people must be infinity. It is not any number - only zero over zero has an undefined, ambiguous quotient. Majopius (talk) 21:19, 16 December 2009 (UTC) contribs) 21:09, 16 December 2009 (UTC)

0/0
I think the current explanation is a bit confusing. If I may cite an "expert", this explanation by "Dr. Math" seems good. While the lim(x->0) x/x = 1, lim(x->0) 7x/x. (OR: Actually lim(x->0) ax/x = a, where a is any member of the set of complex numbers.)

Unsigned infinity
This article seems to be using the convention that ∞ and +∞ represent different concepts. This isn't wrong but it's not the only convention being used, as I mentioned recently at Wikipedia talk:WikiProject Mathematics. There should probably be some mention in the article to specify which notation is being used to avoid confusion.--RDBury (talk) 20:13, 20 January 2010 (UTC)

Featured Page
Can this get nominated? The author of this page is a genius. It's so easy to comprehend it's ridiculous. --204.49.80.2 (talk) 17:32, 18 February 2010 (UTC)

Vandalism
Should we protect this page? There has been lot's of vandalism lately. Somebody500 (talk) 00:18, 22 February 2010 (UTC)
 * You can file a request at WP:Requests for page protection. The current level is a bit low, so I'm not sure they will do it now. It's a very simple procedure. Go ahead and try. DVdm (talk) 08:20, 4 March 2010 (UTC)
 * ✅. I went ahead with this edit. Page is protected for two weeks. DVdm (talk) 21:49, 4 March 2010 (UTC)

removed chuck norris can divide by zero16:10, 13 May 2010 (UTC) —Preceding unsigned comment added by 69.214.14.30 (talk)

Preventing linebreaks
Does anyone know how to prevent a linebreak in the middle of the expression
 * a/0

? I tried putting &amp;nbsp; between the a and the slash, and the slash and the zero, but it didn't work. --Trovatore (talk) 00:12, 4 March 2010 (UTC)
 * I have replaced it with inline math: $$a/0$$ . DVdm (talk) 08:13, 4 March 2010 (UTC)
 * That's also problematic. If you force it to create a PNG, you get problems with line spacing; if you don't, the a comes out roman, which is not desired.  --Trovatore (talk) 08:28, 4 March 2010 (UTC)
 * Hm, indeed. I undid my mod. It seems to work now. There's no linewrapping anymore from where I'm sitting. Strange. - DVdm (talk) 09:46, 4 March 2010 (UTC)

In algebra
This appears in the section "In algebra": "Although most educated people would probably recognize the above "proof" as fallacious, the same argument can be presented in a way that makes it harder to spot the error." I imagine that this is a pretty spurious claim and would be difficult to support. We should probably just remove "Although most educated people would probably recognize the above "proof" as fallacious," and the sentence will still work. Setitup (talk) 23:15, 10 March 2010 (UTC)


 * I agree that such a claim doesn't really belong in an encyclopedia. How's "Although it is possible to recognize the above 'proof' as fallacious,"? --Zarel (talk) 05:30, 11 March 2010 (UTC)
 * Even better: "Although the proof is recognizably fallacious, ..." — Anonymous Dissident  Talk 06:40, 11 March 2010 (UTC)
 * "Recognizably" still sounds vaguely weasely. What's wrong with "Although the proof is fallacious"? --Zarel (talk) 20:53, 13 May 2010 (UTC)
 * Perfect. DVdm (talk) 21:14, 13 May 2010 (UTC)

dividend?
The page refers to a in a/0 as a "dividend". I only heard of dividends on the stock market. Is this standard usage? Tkuvho (talk) 16:10, 1 July 2010 (UTC)
 * Yes, see dividend. The dividend as far as a company is concerned is the amount to be divided before it is shared out amongst the shareholders. Dmcq (talk) 18:33, 1 July 2010 (UTC)

0/0 is called "indeterminate," not "undefined."
The Math Forum at Drexel webpage has a good explanation of the difference between undefined expressions such as a/0 and indeterminate expressions such as 0/0. Indeterminate expressions could have many different answers, and it can't be "determined" which one is appropriate for a given case. Undefined expressions have no answer because the operation on the numbers is nonsensical. —Preceding unsigned comment added by 72.83.187.227 (talk) 01:31, 11 August 2010 (UTC)
 * 0/0 is one of the classic indeterminate forms, yes. But as a value it's undefined.  That is, the function (technically partial function) that takes two real numbers x and y and returns x/y does not have the point (0,0) in its domain. --Trovatore (talk) 02:04, 11 August 2010 (UTC)


 * Almost anything can be called "undefined", as long as you don't define it (i.e. if the context is a set it's outside of). See Defined and undefined. 2&minus;5 is undefined in the positive numbers, for instance, and 4/5 is undefined in the integers. 0/0 can be defined to be an indeterminate form. 1/0 can be defined to be an indeterminate infinity. So you could argue that they're both undefined and both indeterminate. --Zarel (talk&sdot;c) 06:07, 11 August 2010 (UTC)

I agree that "0/0" is both an indetermination and undefined. But note that saying something is an indeterminate form does not define it; quite the contrary. And, unlike 2&minus;5, which may be undefined in some mathematical structures but has a meaning in others, there is no useful mathematical structure that assigns a meaning to "0/0". FilipeS (talk) 13:15, 22 March 2012 (UTC)

Complex Infinity?
Mathematica and Wolfram Alpha (and apparently Microsoft Math) return "ComplexInfinity" when asked to evaluate 1/0.

Wolfram MathWorld mentions them here:


 * http://mathworld.wolfram.com/ComplexInfinity.html

I think it deserves more of a mention than in the "in computer arithmetic" section. To me, "in computer arithmetic" should talk about how computers handle problems they're not intended to be able to solve. On the other hand, Mathematica is, as far as I know, returning what Wolfram Research believes to be the correct answer.

Thoughts? --Zarel (talk&sdot;c) 02:40, 12 August 2010 (UTC)
 * It's in the "Riemann sphere" section. --Trovatore (talk) 02:42, 12 August 2010 (UTC)


 * Not entirely. The MathWorld "Complex Infinity" article describes it as a number in the complex plane, not a Riemann sphere, and the words "complex infinity" are not used in the Riemann sphere section of our article. --Zarel  (talk&sdot;c) 03:13, 12 August 2010 (UTC)
 * OK, here's a tip &mdash; never rely on MathWorld for terminology. It's math is usually right as far as I can tell, but sometimes it just makes up words more or less.
 * As far as the substance goes, no, it is not a number in the complex plane. By definition the complex plane is a plane; it has no point at infinity.  When you add the point at infinity you don't have a plane anymore; you have a sphere.  --Trovatore (talk) 04:41, 12 August 2010 (UTC)

question about internet culture popularity
Should we make a reference to dividing by zero becomeing an internet meme as it is quite popular among message boards in demotavationals? --anon


 * We wait till things appear in WP:Reliable sources. It just needs an newspaper column for instance to mention its use on the web. Dmcq (talk) 08:50, 21 March 2011 (UTC)

What about Chuck?
Everyone knows that Chuck Norris candivide by zero? —Preceding unsigned comment added by 86.140.155.239 (talk) 23:43, 6 May 2011 (UTC)


 * He also served in the United States Air Force, that doesn't mean the article on the United States Air Force need mention him. One would only be interested in that if one was going to read about Chuuck Norris anyway, It has zilch interest of a reader just reading this article. Dmcq (talk) 10:01, 7 May 2011 (UTC)

The Word Problem is Not Helpful
They (you know who they are, don't you?) always phrase the word problem something like this: "So for dividing by zero – what is the number of apples that each person receives when 10 apples are evenly distributed amongst 0 people?"

But I prefer to phrase it like this: "Given 10 apples, how many piles of apples can you make if you put 0 apples in each pile?" (or, to keep it in terms of people instead of piles: "Given 10 apples, how many people can you give an apple to if you give 0 apples to each person?)

Clearly, here, the answer is an infinite number of piles or people.

I've been "trained" in all the right stuff, and believe that for the good of mathematicians everywhere, division by 0 must be undefined. I've even seen what I believe is the deep seated reason behind it (which I didn't look for in the article, and have to admit I didn't even read the entire article).

It's just that I think using that word problem as an example of why division by zero is undefined is not useful, and, in fact, counterproductive. With some thought, people (perhaps like me) can look at it the way I do, and realize that the word problem is just a sham--it does not do anything to prove that division by zero is undefined.

If you want to educate people, or be an encyclopedia, provide the real explanation. Rhkramer (talk) 00:27, 15 June 2011 (UTC)


 * If you put no apples into each pile, then they are not piles of apples. They are piles of air. You do not end up with infinite piles of apples. Your question is undefined. Fred775 (talk) 01:25, 12 January 2015 (UTC)

Pop Culture
There ought to be a section explaining the cultural impact made by the idea of division by zero. This is evident whenever one searches divide by zero, a series of playful images result showing how the modern culture has made somewhat of a "play" on the idea. — Preceding unsigned comment added by Ghost9420 (talk • contribs) 15:16, 16 June 2011 (UTC)


 * If you can find a WP:reliable source that discusses this then that could form the basis for such section. I've seen sources discussing all sorts of things so somebody may have thought of something like this. Without that it would really have to be something that actually used the concept in some meaningful way and was noted by someone, see WP:TRIVIA and WP:handling trivia. Dmcq (talk) 15:25, 16 June 2011 (UTC)

Mahavira, and the number remaining unchanged
To divide zero times is to not divide. So to divide 7 by zero would be not to divide it, and let it remain 7. Because zero is the subject not an object (if zero were an object it would be one as a subject, but that's a philosophic question and not a mathematical). Now dividing "zero" seven times is another issue, because such an operation posits zero as an object by definition. Mathematics should do well to consider numbers as objects or subjects. For example, 1 as an object is singular even amidst multiplicity, 1 as a subject is unity or the whole considered at the whole without multiplicity. Nagelfar (talk) 02:12, 14 September 2011 (UTC)
 * Why did you write that? In what way are you hoping to improve the article? Dmcq (talk) 02:29, 14 September 2011 (UTC)
 * Maieutic improvement through dialectical discussion. For example, why should dividing by zero be undefined, but subtracting 50% from zero is still zero? Shouldn't it be -50% of undefined? I wouldn't add to article to of course avoid original research, but posting here as looking to those who may know sources of opinions of mathematicians based on notoriety of said opinions for improvement of said article. Nagelfar (talk) 04:09, 14 September 2011 (UTC)
 * If you want a discussion about things rather than improving the article then Reference desk/Mathematics would be better though it really is just for answering questions. There are no forums as such on WIkipedia. I can say though the whole basis of the way you're going about it strikes no chord with modern mathematics. Discussion of an individual instance of division and trying to give a value on its own would normally be seen as a fruitless idea. What is wanted is general ideas so one wants a general definition of division and have the result come out of that. And the general idea of division is that it is an inverse operation to multiplication. The result of a/b is c if c is the unique value such that a×c is b. If there is no such value one might be able to extend the system as described for the projective real line in the article where ≈ is used - but then one has problems with comparisons as it may be positive or negative and 0/0 still isn't defined and there's no good reason to define it as a particular value when any value can satisfy 0×c=0. Dmcq (talk) 09:00, 14 September 2011 (UTC)
 * "What is wanted is general ideas so one wants a general definition of division and have the result come out of that. <...> and 0/0 still isn't defined and there's no good reason to define it as a particular value <...>" Why not explain it in the article? The tradition is to put in separate places statements and explanations, but this clearly doesn't work well for an encyclopedia article. For example, purposes of defining things are to be explained in-place, so that possible and quite logical misconceptions be ruled out immediately; this is more reader-friendly. I'd do the work myself, but I'm afraid to go too far from sources or make a mess with my English, so I'm out. - 92.100.173.55 (talk) 15:15, 7 April 2013 (UTC)

Sanskrit mathematics & Division by zero as infinity
According to "The Universal History of Number" by Georges Ifrah, page 476: "''Khahâra. Sanskrit word for infinity. Literally "division by zero". Notably used by *Bhâskarâchârya. See Khachheda. ... Khachheda. Sanskrit term used to denote infinity. Literally "divided by zero" (from *kha, "space" as a symbol for zero, and chheda, "the act of breaking into many parts", "division"). Thus it is the quantity whose denominator is zero". The term is used notably in this sense by *Brahmagupta in his Brahmasphutasiddânta (628 CE)."'' Nagelfar (talk) 00:24, 19 September 2011 (UTC)


 * That's already in the article in the section 'Early attempts' Dmcq (talk) 07:55, 19 September 2011 (UTC)


 * No it's not. There's no mention of Bhâskarâchârya whatsoever, nor does it say it is equated with infinity, but rather zero. Nagelfar (talk) 03:55, 20 September 2011 (UTC)
 * From that section: "Bhaskara II (1114–1185) tried to solve the problem by defining (in modern notation) $$\textstyle\frac{n}{0}=\infty$$.[1] This definition makes some sense, as discussed below, but can lead to paradoxes if not treated carefully. These paradoxes were not treated until modern times." Dmcq (talk) 07:53, 20 September 2011 (UTC)
 * Yes I read that, this post was in response to that very section obviously. It neglects to say the same of Brahmagupta's interpretation (it says zero). There's no mention whatsoever of Khahâra. Nagelfar (talk) 13:37, 21 September 2011 (UTC)
 * You should have said what you meant. This is the English wikipedia, why should it stick in Indian words rather than just say infinity? Is there something of note about that word in this context? That section does talk about Brahmaguptra and you haven't said anything extra that's not there, so how should more be put in? Are you saying that the text is wrong about what he said or something? Please be clear in exactly what you are saying should be there. Dmcq (talk) 15:20, 21 September 2011 (UTC)
 * Let me stop my same quotation before the ellipses "According to "The Universal History of Number" by Georges Ifrah, page 476: "Khahâra. Sanskrit word for infinity. Literally "division by zero". Notably used by *Bhâskarâchârya. See Khachheda." now, all that information is new. I'm sorry my second quotation from that work was distracting. The point is, the word used for infinity (Khahâra) meant "division by zero" literally, now I'd say that is notable for any language, and it was used by Bhâskarâchârya, one not mentioned in the article. The article says Brahmaguptra claimed that division by zero equals zero, my quotation for Brahmaguptra says he used it as infinity, that is something from my second quotation that is "not there". Is that clearer? It's nothing that wasn't in my first posting. Nagelfar (talk) 20:04, 21 September 2011 (UTC)
 * Bhaskara II is Bhâskarâchârya. Zero divided by zero was zero according to Brahmaguptra, not just any division by zero. He represented a number other than zero divided by zero by the number over zero, he did not say it was a particular value but he did make a mess of things saying for instance that 5/0 × 0 was 5. You don't get that by replacing 5/0 by infinity, he did not change 5/0 into something else. Dmcq (talk) 21:28, 21 September 2011 (UTC)
 * In the very least, I think that there being words for infinity in Sanskrit which mean literally "division by zero" is of note to the history portion. Nagelfar (talk) 22:37, 21 September 2011 (UTC)
 * The Sanskrit for infinity is ananta for without end. People later on interpreted n/0 as infinity but he didn't call it infinity, he called it divided by zero which is exactly what it is. Dmcq (talk) 11:25, 23 September 2011 (UTC)

The simple parts.
The division being undefined because of the multiplication doesn't make sense. A number times zero should equal the number, n x 0 = n. N is a number which represents real value. So if I have 8 Coke bottles and I multiply them zero times, the 8 Cokes do not magically disappear, they are still there. In reverse of that, a number divided by zero would be would also equal the number n/0 = n. — Preceding unsigned comment added by Gummifaustus (talk • contribs) 00:45, 19 September 2011 (UTC)
 * n x 0 = n and n x 1 = n would give n x 1 − n x 0 = 0 or n x (1 − 0) = 0 which is  n x 1 = 0. Nothing magical happens when you multiply. If you get no deliveries of a six pack of coke you don't have any coke. If you get one delivery of a six pack of coke you have a six pack of coke. Dmcq (talk) 08:01, 19 September 2011 (UTC)

x/y=0
If dividing by 0 was possible,then there could be 2 numbers:x and y to satisfy x/y (or y/x)=0 Since I don't know any numbers which fulfill this,then dividing by 0 is impossible.Oh,and x and y are not 0 — Preceding unsigned comment added by 86.104.51.117 (talk) 16:14, 21 May 2012 (UTC)


 * I am not sure why you stuck this here. This page is about improving the article and we cannot use people's own arguments, the stuff has to be from books or papers or a very reliable web source. As to the argument itself no reason was given for the conditions and your not being able to think of some numbers doesn't constitute a proof. Dmcq (talk) 09:05, 23 May 2012 (UTC)

Zero divided by zero
Any number times zero equals zero,so zero divided by zero equals any number you want — Preceding unsigned comment added by 188.25.224.223 (talk) 18:14, 12 June 2012 (UTC)


 * That's why this thing is said to be undefined. The symbol 0/0 has no meaning. See the article. - DVdm (talk) 18:57, 12 June 2012 (UTC)


 * I think the article probably should have a separate section about 0/0 as it has a special status as an indeterminate and has been discussed in detail itself. Dmcq (talk) 09:25, 28 July 2012 (UTC)

1/0 > Infinity
I read all explanations in this article, but I am not sure if this was mentioned. The logic is simple. If we devide a number (let's say 1) by an infinitesimal number, we'll have an infinity as result. The zero is still smaller than the infinitesimal number so deviding a number (1 in our case) by zero, will result as a bigger number than infinity. But this is absurd since by deffinition there is nothing bigger than infinity. So division by zero is impossible. Such a number can not exist (neither exist the zero by the way, but we have some abstract understanding for it as an absence of something). — Preceding unsigned comment added by Stan3u (talk • contribs) 22:32, 11 February 2013 (UTC)

The cases where the subject matter of the article makes sense should be treated prominently in the lead
For some time, I'm not sure how long, you've been able to read the lead section without any explicit mention of the fact that division by zero (note that this is the title of the article, therefore what the article is supposed to be about) actually sometimes does make sense.

MvH's recent edit (this one) removed the only obscure allusion to that fact in the lead. I don't think that's the right way to go. There needs to be an explicit mention, not in a note but in visible text. I am not claiming my text is necessarily the optimal way of achieving that; any suggestions for making it less technical are welcomed. --Trovatore (talk) 22:31, 4 March 2014 (UTC)

Calculator image
Do we really need two calculator division-by-zero screenshots, one saying "ERROR 02 DIV BY ZERO" and the other saying "$$\infty$$"? I assume - the latter being an Android calculator on a phone - they're functionally identical and that you can't perform further operations on the infinity value. --McGeddon (talk) 17:36, 6 June 2014 (UTC)


 * You may be right that they are functionally identical.
 * But they look very different.
 * In systems that use IEEE floating point, you *can* perform further operations on the infinity value (sometimes spelled "Number.POSITIVE_INFINITY", "+inf", etc.). --DavidCary (talk) 19:10, 15 November 2014 (UTC)

Meadows
Perhaps, in the interest of completeness, a small note about a meadow of numbers could be included? For those of you unaware a meadow is a commutative ring with a total inverse operator satisfying two equations which imply 0^−1 = 0. It's also known as a zero-totalized field. The paper I copied the definition from: arxiv.org/pdf/0901.0823. I managed to find one at springer from 2007 http://link.springer.com/article/10.1007/s00224-007-9035-4. Unfortunately, I'm not nearly knowledgeable enough in number theory to make this understandable to the common man.

The search I ran at springer to find the papers: http://link.springer.com/search?query=Zero+totalised+field&facet-discipline=%22Mathematics%22 — Preceding unsigned comment added by Tuvok302 (talk • contribs) 21:06, 2 March 2016 (UTC)

The algebra section is wrong fix or remove it.
EmpCarnivore (talk) 23:01, 17 April 2016 (UTC)
 * Re: It says 0/0=x can be found threw the unknown value of 0*x=0. That isn't true. Those aren't the same x. Here is proof.


 * 0*x=0 If what they say is true then I could just divide by zero and get 0/0=x. So to show you that doesn't work I will do that.
 * 0/0*x=0/0 Now that I have done that they aren't equal. We have a extra 0/0. To say 0/0=x Then you are removing the 0/0 by the x. Your saying 0/0=1, but we don't know that. The reason why we are using the equation 0*x=0 is to figure out the answer. Also 1*x doesn't equal 1, so 0/0 can't equal 1. It can equal to 0, infinity, or a none existing number.
 * What they said 0/0=x What algebra says. 0/0=0/0*x.
 * They did that or this
 * (0/0*x) /x-0/0+x = 0/0
 * x=0/0 simplified
 * You need to do the same thing to both sides
 * With numbers: x=3 0/0=0 Where they are both variables
 * 3=0
 * x=0/0 simplified
 * You need to do the same thing to both sides
 * With numbers: x=3 0/0=0 Where they are both variables
 * 3=0


 * What algebra says:
 * (0/0*x) /x-0/0+x=0/0 /x-0/0+x
 * x=0/0/x-0/0+x simplified
 * With numbers: x=3 0/0=0 Where they are both variables
 * 3=3


 * See, the problem is, no one is going to read that. It is completely unclear what your objection is.  It might be possible to figure it out by sufficiently careful attention to that wall of text you've put up, but you haven't given anyone a reason to want to put that effort in.
 * If you can express your objection clearly and concisely, then someone will be willing to respond to it. Who knows; you might even convince us.  But right now there's just no actionable content. --Trovatore (talk) 21:26, 17 April 2016 (UTC)
 * I second Trovatore's suggestion. Take it in baby steps, one or at most a few lines at the time, and propose a change to the existing text. Don't assume that we are all illiterate Neanderthals, some of us may have a math degree so might be able to understand a mistake when it is clearly pointed out. Adrian two (talk) 22:47, 17 April 2016 (UTC)
 * I didn't assume you were illiterate. I am sorry for making you feel this way. I was just repeating the same explanation in different ways here I put a better shorter explanation by the Re: EmpCarnivore (talk) 23:02, 17 April 2016 (UTC)
 * Please do not edit previous comments in the Talk page, makes it very hard to follow. Add new thoughts or replies to the bottom of the page. Also, please identify clearly which part from the existing article you're challenging or trying to improve. Thank you. Adrian two (talk) 23:17, 17 April 2016 (UTC)
 * Once again, please do not edit existing comments (even when they are yours). Add stuff to the bottom of the page. Quote clearly the lines from the text you're challenging or trying to improve. Something like, for the sake of an example: "since 2 is the value for which the unknown quantity" is incorrect, and should be stated as "since 20 is the value for which the unknown number" Adrian two (talk) 01:08, 18 April 2016 (UTC)
 * I'm sorry I didn't see your comment when editing it I was just trying to help. I am still learning about wikipedia. The section that needs to be removed is division as the inverse of multiplication literally all of it is either wrong or pointless
 * 6/0=x assumes 0/0=1
 * 0/0=x the proof here is just wrong I can do the same thing to prove 6/6=is all numbers. 6/6*0=0*x Because all numbers times 0=0 6/6 is equal to all numbers. — Preceding unsigned comment added by EmpCarnivore (talk • contribs) 01:24, 18 April 2016 (UTC)
 * Take a deep breath, and slowly read the section in the article. Check the example first
 * $$\frac{6}{3}=2$$
 * since 2 is the value for which the unknown quantity in
 * $$?\times 3=6$$
 * is true.
 * Any objections so far?
 * Now move to the next part:
 * But the expression
 * $$\frac{6}{0}=\,x$$
 * requires a value to be found for the unknown quantity in
 * $$x\times 0=6.$$
 * Any objections so far?
 * But any number multiplied by 0 is 0 and so there is no number that solves the equation.
 * So far, so good? Adrian two (talk) 01:40, 18 April 2016 (UTC)

EmpCarnivore (talk) 01:57, 18 April 2016 (UTC)Can you change 6/0=x into 0*x=6 or the other way around. Let us try. 6/0*0=x*0 Well we have zero division. How about 0*x=6 Well 0/0*x=6/0 zero division. My point is that these equations aren't the same and no proof has been provided to let people now that. 6=2*x is different from these examples. 6/2=2/2*x We know that 2/2=1. I can prove that 0/0 doesn't. If it does than 1=2 Both of those are wrong.
 * First, please sign using 4 tilde's after your post.

Next, I'll try to reply to your points, quoting them first:


 * Can you change 6/0=x into 0*x=6 or the other way around.
 * Yes, the two above are equivalent: if one is true, the other one must be true. If one is impossible to solve, the other one is also impossible to solve.
 * Let us try. 6/0*0=x*0 Well we have zero division.
 * This is your way of converting. It is not required to do it that way.
 * $$\frac{6}{3}=2$$
 * since 2 is the value for which the unknown quantity in
 * $$?\times 3=6$$
 * is true.
 * This is just stating that Division is the inverse of multiplication, just as subtraction is the inverse of addition.
 * How about 0*x=6 Well 0/0*x=6/0 zero division. My point is that these equations aren't the same and no proof has been provided to let people now that.
 * Again, see above. You are using an additional divisor on both sides, that is not required.
 * But the expression
 * $$\frac{6}{0}=\,x$$
 * requires a value to be found for the unknown quantity in
 * $$x\times 0=6.$$
 * The above is just using the fact that Division is the inverse of multiplication, no proof required for that. — Preceding unsigned comment added by Adrian two (talk • contribs) 02:18, 18 April 2016 (UTC)

I know that division is the inverse of multiplication. But your still using 0/0 as 1. If you can make one into the other using proofs and algebra then do it correctly and visual and I will walk away right now. If you can't then I will continue to be right. Also you added nothing what was the point.EmpCarnivore (talk) 02:29, 18 April 2016 (UTC)
 * Baby step #1


 * $$\frac{6}{3}=2$$
 * since 2 is the value for which the unknown quantity in
 * $$?\times 3=6$$
 * is true.


 * The above does not require any division (or any other arithmetic / algebra) to 'prove' that it is true. It is just using the fact that Division is the inverse of multiplication. Agree or not? Adrian two (talk) 03:18, 18 April 2016 (UTC)

Are you serious. Stop talking like I am stupid.x*3=6 x*3/3=6/3 x=2. Make 6/0=x into 6=x*0 and show you work. Don't repeat yourself and don't make me repeat myself.EmpCarnivore (talk) 03:26, 18 April 2016 (UTC)


 * Sigh. Please read: "The concept that explains division in algebra is that it is the inverse of multiplication." Also, please click on division and slowly read:
 * For instance,
 * 6 ÷ 3 = 2
 * since
 * 3 × 2 = 6.
 * I do not need to make one into the other. Adrian two (talk) 03:42, 18 April 2016 (UTC)

You just said "sigh", repeated yourself, and now I have to explain this again. I am not stupid. That web page shows division as the inverse of multiplication. I already know this. I use it all the time, but something you don't know or at least any more is that it is skipping steps. It is skipping steps. It is skipping steps. Got that. It is skipping y/x*x=1y, because everyone with a 1st grade education should know that x/x=1. But with x as zero that doesn't apply, because 0/0 isn't 1. Got that. To assume that I am wrong without actually trying it yourself is wrong. 6/3*3=2*3 6*1=2*3 (With the steps not skipped). 6=2*3 (With the steps skipped), 6/0*0=x*0 0/0=x*0 we know that x=1/0 This is just a simple everything is equal to itself equation. (Steps fully included). 6=x*0 Here we seem to be missing 0/0 (Steps left out) If they are both the same and yours should apply to everything then why do we get different answers. Because you assume 0/0=1 when using division as inverse multiplication. 0/0 isn't 1 EmpCarnivore (talk) 04:10, 18 April 2016 (UTC)


 * Let me try to break this impasse. First consider the equation $a⁄b = c$. This is the same as writing $a ⋅ b^{-1} = c$, where $b^{-1}$ is the multiplicative inverse of $b$, that is, the solution of the equation $x ⋅ b = 1$ (this is just the fancy way of saying that division is the inverse of multiplication). We solve the equation for $a$ using these steps
 * There are two assumptions that must be valid in order for this manipulation to be correct. One of these is that the associative law must hold (step 2 to step 3) and more to the point for us, the multiplicative inverse of $a ⋅ b^{-1} = c$ must exist (step 3 to step 4). If either assumption is false, this computation is not valid. That does not say that the result is false, just that this way of obtaining it doesn't work. In a quasigroup (which does not satisfy the associative law), the result here is valid but you can not prove it with this computation. Now to get back to the problem at hand, what happens when $(a ⋅ b^{-1}) ⋅ b = c ⋅ b$? EmpCarnivore is arguing that in going from the first to the last step, you must use this computation and therefore 0/0 = 1 (which can then be shown to be false). Adrian two is arguing that the first and last step are equivalent by definition, so this computation is moot and no statement about 0/0 is being made. I don't completely agree with Adrian two (this is an oversimplification) but he is essentially right. The problem with EmpCarnivore's argument is that in order for this computation to work, you must assume that 0 has an inverse. With that assumption it is no surprise that you conclude that 0/0 = 1, since that is what it means for 0 to have an inverse. In other words, the argument is circular (but subtle). Bill Cherowitzo (talk) 04:31, 18 April 2016 (UTC)
 * There are two assumptions that must be valid in order for this manipulation to be correct. One of these is that the associative law must hold (step 2 to step 3) and more to the point for us, the multiplicative inverse of $a ⋅ (b^{-1} ⋅ b) = c ⋅ b$ must exist (step 3 to step 4). If either assumption is false, this computation is not valid. That does not say that the result is false, just that this way of obtaining it doesn't work. In a quasigroup (which does not satisfy the associative law), the result here is valid but you can not prove it with this computation. Now to get back to the problem at hand, what happens when $a ⋅ 1 = c ⋅ b$? EmpCarnivore is arguing that in going from the first to the last step, you must use this computation and therefore 0/0 = 1 (which can then be shown to be false). Adrian two is arguing that the first and last step are equivalent by definition, so this computation is moot and no statement about 0/0 is being made. I don't completely agree with Adrian two (this is an oversimplification) but he is essentially right. The problem with EmpCarnivore's argument is that in order for this computation to work, you must assume that 0 has an inverse. With that assumption it is no surprise that you conclude that 0/0 = 1, since that is what it means for 0 to have an inverse. In other words, the argument is circular (but subtle). Bill Cherowitzo (talk) 04:31, 18 April 2016 (UTC)
 * There are two assumptions that must be valid in order for this manipulation to be correct. One of these is that the associative law must hold (step 2 to step 3) and more to the point for us, the multiplicative inverse of $a = c ⋅ b$ must exist (step 3 to step 4). If either assumption is false, this computation is not valid. That does not say that the result is false, just that this way of obtaining it doesn't work. In a quasigroup (which does not satisfy the associative law), the result here is valid but you can not prove it with this computation. Now to get back to the problem at hand, what happens when $b$? EmpCarnivore is arguing that in going from the first to the last step, you must use this computation and therefore 0/0 = 1 (which can then be shown to be false). Adrian two is arguing that the first and last step are equivalent by definition, so this computation is moot and no statement about 0/0 is being made. I don't completely agree with Adrian two (this is an oversimplification) but he is essentially right. The problem with EmpCarnivore's argument is that in order for this computation to work, you must assume that 0 has an inverse. With that assumption it is no surprise that you conclude that 0/0 = 1, since that is what it means for 0 to have an inverse. In other words, the argument is circular (but subtle). Bill Cherowitzo (talk) 04:31, 18 April 2016 (UTC)
 * There are two assumptions that must be valid in order for this manipulation to be correct. One of these is that the associative law must hold (step 2 to step 3) and more to the point for us, the multiplicative inverse of $b = 0$ must exist (step 3 to step 4). If either assumption is false, this computation is not valid. That does not say that the result is false, just that this way of obtaining it doesn't work. In a quasigroup (which does not satisfy the associative law), the result here is valid but you can not prove it with this computation. Now to get back to the problem at hand, what happens when ᙭᙭᙭? EmpCarnivore is arguing that in going from the first to the last step, you must use this computation and therefore 0/0 = 1 (which can then be shown to be false). Adrian two is arguing that the first and last step are equivalent by definition, so this computation is moot and no statement about 0/0 is being made. I don't completely agree with Adrian two (this is an oversimplification) but he is essentially right. The problem with EmpCarnivore's argument is that in order for this computation to work, you must assume that 0 has an inverse. With that assumption it is no surprise that you conclude that 0/0 = 1, since that is what it means for 0 to have an inverse. In other words, the argument is circular (but subtle). Bill Cherowitzo (talk) 04:31, 18 April 2016 (UTC)
 * There are two assumptions that must be valid in order for this manipulation to be correct. One of these is that the associative law must hold (step 2 to step 3) and more to the point for us, the multiplicative inverse of ᙭᙭᙭ must exist (step 3 to step 4). If either assumption is false, this computation is not valid. That does not say that the result is false, just that this way of obtaining it doesn't work. In a quasigroup (which does not satisfy the associative law), the result here is valid but you can not prove it with this computation. Now to get back to the problem at hand, what happens when ᙭᙭᙭? EmpCarnivore is arguing that in going from the first to the last step, you must use this computation and therefore 0/0 = 1 (which can then be shown to be false). Adrian two is arguing that the first and last step are equivalent by definition, so this computation is moot and no statement about 0/0 is being made. I don't completely agree with Adrian two (this is an oversimplification) but he is essentially right. The problem with EmpCarnivore's argument is that in order for this computation to work, you must assume that 0 has an inverse. With that assumption it is no surprise that you conclude that 0/0 = 1, since that is what it means for 0 to have an inverse. In other words, the argument is circular (but subtle). Bill Cherowitzo (talk) 04:31, 18 April 2016 (UTC)
 * There are two assumptions that must be valid in order for this manipulation to be correct. One of these is that the associative law must hold (step 2 to step 3) and more to the point for us, the multiplicative inverse of ᙭᙭᙭ must exist (step 3 to step 4). If either assumption is false, this computation is not valid. That does not say that the result is false, just that this way of obtaining it doesn't work. In a quasigroup (which does not satisfy the associative law), the result here is valid but you can not prove it with this computation. Now to get back to the problem at hand, what happens when ᙭᙭᙭? EmpCarnivore is arguing that in going from the first to the last step, you must use this computation and therefore 0/0 = 1 (which can then be shown to be false). Adrian two is arguing that the first and last step are equivalent by definition, so this computation is moot and no statement about 0/0 is being made. I don't completely agree with Adrian two (this is an oversimplification) but he is essentially right. The problem with EmpCarnivore's argument is that in order for this computation to work, you must assume that 0 has an inverse. With that assumption it is no surprise that you conclude that 0/0 = 1, since that is what it means for 0 to have an inverse. In other words, the argument is circular (but subtle). Bill Cherowitzo (talk) 04:31, 18 April 2016 (UTC)

In other words 6/0=x isn't 6=x*0 and the entire division as inverse needs to be put in the Fallacies section. With the ending being all of these are wrong because they assume dividing by zero has the same properties as diving by any other number. btw If 0/0=x*0 and 0/0*x=0/0, then the only real number that 0/0 can be is 0. In reality people just assumed things were being messed up because it involved the inverse of 0, but in reality it is just human error. Programmers and mathematicians need to learn the difference between such things.EmpCarnivore (talk) 04:42, 18 April 2016 (UTC)


 * Note to : please indent your talk page messages as outlined in wp:THREAD and wp:INDENT. Thanks. - DVdm (talk) 08:02, 18 April 2016 (UTC)

Note to all involved. This entire discussion can (and probably should) be stopped by adding proper sources to the intro of section Division by zero, and to the example and conclusion of subsection Division by zero. Where did the example come from? Perhaps everything is indeed neatly sourced in the Division by zero section, but perhaps it is not. Who can tell without proper inline citations with references to actual pages? If nobody knows, we probably should replace it with an example that can be backed by a good source, perferably one with a page that is accessible by Google Books, so everyone can single-click-verify it. Otherwise anyone can challenge it as wp:original research beyond wp:CALC. A source would immediately and effectively moot future discussions like this, that, as we all know, tend to go nowhere. - DVdm (talk) 08:02, 18 April 2016 (UTC)
 * Actually, It has gone somewhere. I said that multiplicative inverse section is wrong because for that section to be right as it currently is 0/0 needs to equal to 1. It doesn't. https://en.wikipedia.org/wiki/Division_(mathematics)   In abstract algebra it says basically what I said. It is wrong. Also if the webpage is right then why, do I get a different answer when doing the same thing. Well (this version of) division as the inverse multiplication is skipping steps that include 0/0 or 0^0. They are both the same and can't be given the value 1. Which is required for the webpage to be true.  — Preceding unsigned comment added by EmpCarnivore (talk • contribs) 13:06, 18 April 2016 (UTC)


 * please indent your talk page messages as outlined in wp:THREAD and wp:INDENT and sign all your talk page messages with four tildes ( ~ ). What you said about why "multiplicative inverse section is wrong", is, without a concrete sourced alternative, of no importance, and off-topic, so no further comment to it is appropriate here. Also note that Wikipedia is not a reliable source for itself—see wp:CIRCULAR. Other similar comments will be deleted and you will get another warning on your talk page.
 * waiting for response from other contributors about proper inline sources for current text in said section and subsection (now tagged for inline citations). - DVdm (talk) 13:25, 18 April 2016 (UTC)


 * https://www.khanacademy.org/math/pre-algebra/order-of-operations/arithmetic_properties/v/inverse-property-of-multiplication "5*1/5=5/5=1" here 6*b*1/b=x*b if b=0 then 0*1/0=0/0=undefined http://m.wolframalpha.com/input/?i=0%5E0&x=0&y=0   so     6*1≠6*undefined in the equations 6*0/0=x*0 (with all steps) 6=x*0(skipping to the last step). I am not wrong and I am sorry if you think I am. It is on topic, it is useful, and I am only to help. EmpCarnivore (talk) 14:36, 18 April 2016 (UTC)


 * Here are some better sources that agree with it on a whole. Some more help. http://math2.org/math/algebra/basicidens.htm https://www.mathsisfun.com/numbers/inverse.html http://mathworld.wolfram.com/DivisionbyZero.html If you need it.EmpCarnivore (talk) 20:06, 18 April 2016 (UTC)

Riemann Sphere
The Riemann sphere is a compact complex manifold where 1/0 is defined to be the "point at infinity"

This model of the complex plane shows up in RF engineering. Complex values like admittance and impedance can be represented with this sort of model of the complex numbers. And division by zero naturally occurs in this domain; an open circuit is conveniently thought of as having zero admittance. Since admittance is the reciprocal of impedance, such a circuit naturally has infinite impedance.

Maybe its not worth mentioning, however. I cannot find a primary source for Riemann ever defining arithmetic as the wikipedia page describes. — Preceding unsigned comment added by 2601:184:4100:1480:B460:E479:9E82:CA52 (talk) 14:08, 18 April 2016 (UTC)
 * Hmm, interesting. Not only do I not know of any direct writings by Riemann describing this arithmetic, I don't even recall ever hearing anyone claim explicitly that Riemann ever had anything to do with the Riemann sphere.  But that doesn't prove much as I've never really looked into Riemann's history much.
 * Anyway, I wish you luck tracking it down. If you find anything out about the history of the name "Riemann sphere", though, I would think it would belong more at the Riemann sphere article than at this one. --Trovatore (talk) 04:56, 19 April 2016 (UTC)

Looks like it was invented by Carl G. Neumann, independently by Felix Klein; Riemann had nothing to do with it: http://math.stackexchange.com/a/895572/36677 At any rate wouldn't mentioning the Riemann sphere be a good idea here? Since it's a rare example of a structure where 1/0 is well behaved. — Preceding unsigned comment added by 2601:184:4100:1480:5C59:A21A:AD4C:C099 (talk) 13:29, 25 April 2016 (UTC)
 * It's mentioned in the second paragraph of the lead section, and elaborated a little bit in the third subsection of the "Calculus" section. I would not be opposed to making it more prominent. --Trovatore (talk) 19:00, 25 April 2016 (UTC)

OR
The section without references (section 2.1), maybe we can use section templates.  333-blue  23:07, 29 September 2016 (UTC)


 * That's not OR, it's a spelled-out explanation of the sort you would find in a middle-school math textbook. It deserves a "citation needed" at best. Elwoz (talk) 23:33, 29 September 2016 (UTC)


 * Agree, and the subsection was already tagged. It is referenced now, and the tag is removed. - DVdm (talk) 08:37, 30 September 2016 (UTC)

why?
In the article, it mentions that the reason why 0/0 is undefined is because theres nothing you can divide to get a assuming a isnt 0. what does that mean? Also, if a/0=undefined, does infinity times = something? UB Blacephalon (talk) 20:51, 2 March 2020 (UTC)
 * Per the wp:talk page guidelines, we cannot discuss the subject here, just the article. Questions about the subject should be asked at our wp:reference desk/Mathematics. Good luck. - DVdm (talk) 22:47, 2 March 2020 (UTC)
 * OK thanks!UB Blacephalon (talk) 03:06, 3 March 2020 (UTC)

How many noughts in a row are necessary to make their sum add up to one whole?
Extracted from Junior News (Al Nisr, Dubai), 5th March 1987:

To understand why 1 ÷ 0 cannot be evaluated, consider what happens to 1 ÷ y when y is made smaller and smaller.

First, think about 1 ÷ ¼. In particular, ask how many "quarters" are required to add up to one whole. We need four: ¼ + ¼ + ¼ + ¼ = 1, so 1 ÷ ¼ = 4. (This illustration will prove useful later when we try to investigate 1 ÷ 0).

Similarly, 1 ÷ 1/1000 (or 1 ÷ 0.001) = 1000, because 1000 "thousandths" in a row will add up to 1.0

As y steadily decreases, approaching closer and closer to zero, 1 ÷ y apparently just keeps growing without any upper limit. No matter how enormous a number you care to write down, we can always find a value of y such that 1/y will equal or exceed it.

Now comes the most important step - make y so small that it becomes equal to zero. Before looking for the answer, consider the following question:

How many noughts in a row are necessary to make their sum add up to one whole?

That is easy - no matter how many noughts we have, we will never reach a value equal to 1. So within the domain of ordinary numbers, "one divided by zero" does not have an answer. Notice how we are adopting exactly the same approach as we did with 1 ÷ ¼, and with 1 ÷ 0.001.

However, that is not quite the end of the story. In mathematics, it is convenient to invent a special entity, named "infinity" and written like an 8 lying on its side. It is really a "hypervalue", defined as "the quantity which is greater than absolutely any ordinary number".

Infinity is not a normal number. It does not obey the laws of arithmetic. For example, what is the value of "infinity plus one"? Once again, there is no answer; (calculations like that do not have a place in mathematics). We could of course try and invent another hypervalue, but eventually we would need an infinitely large supply of them.

It is better to confine ourselves to only one "hypernumber", so that adding to (or even doubling or squaring) infinity does not change its nature - or its "value". Thus, if you really insist on some kind of answer, infinity plus one equals infinity.

--DLMcN (talk) 09:26, 31 August 2020 (UTC)

0/0 = x, fact
I know I've said this before but I think I might of thought of some proof that can end all of this division by zero shit forever. For those who don’t know what logarithms are, log(a) b is the same as saying "With $$a^x$$, what does x have to be to get b?". Lets look at log(2) 256 as an example, with $$2^x$$, what does x have to be to get 256? The answer is 8, because $$2^8$$ = 256. Now lets look at log(1) 1, with $$1^x$$, what does x have to be to get 1? x can of course be any number, so I will just keep the answer as x. Because log(a) b = log b/log a (with log being log(10), it still works the same with any other positive logarithm) we get

x = log(1) 1 = log 1/log 1 = 0/0

We can use a similar method to prove that 1/0 doesn't have an answer, with $$1^x$$, what does x have to be to get 10?

log(1) 10 = log 10/log 1 = 1/0

If you don’t understand the above proof you can get the same answers by asking "If you are travailing at 0mph, how long does it take you to travel 0 meters?" or simply "How many 0's go into 0?" for 0/0 and "If you are travailing at 0mph, how long does it take you to travel 1 meter?" or "How many 0's go into 1?" for 1/0. Well okay maybe 'x' isn't a suitable answer for the first one, but you have to at least accept that the answer can be any number, and that it deserves at least a small mention in this article. Robo37 (talk) 15:32, 12 August 2009 (UTC)


 * I fail to see how this differs from what you wrote above. The answer is the same: we want 0 divided by 0 to be a number, a single number. If we can't define it that way, it's not very useful to define it at all. You are definitely right that the correct way to define it if we thought it useful would be as an entity that assumes all possible values at once. The logarithms do not make that argument less valid (although they are completely unnecessary here), but they also do not make the definition any more useful. — JAO • T • C 18:42, 12 August 2009 (UTC)


 * I'm not asking for you to tell me if I'm right or not, I'm asking for something to be said about it in this article. I can't see a single bit of text in it that says that 0/0 can be any number, and since the entire article is about division by zero it would make sense to mention it don’t you think? I fail to see why it isn't important and if we all went by your "if it doesn’t produce a single number it doesn’t produce any" rule then it would be impossible for a number to have a square root. Robo37 (talk) 20:50, 12 August 2009 (UTC)


 * Yes, you have a point here in that there are lots of useful multivalued functions, of which the complex logarithm, and therefore the unrestricted square root function, are prime examples. Many even assume an infinite number of values in each point. I still don't see the usefulness of something that simultaneously assumes all values in its codomain, though. But if you're looking to add something like "without the usual requirement that a/b be a unique number, 0/0 could instead have been defined as an entity assuming all values at once" to the article, I wouldn't be opposed to that. — JAO • T • C 05:36, 13 August 2009 (UTC)


 * Please add a citation to a journal where such a way of dealing with it is used otherwise it would be original research. Please see that link for Wikipedias policy as regards that. Dmcq (talk) 10:25, 13 August 2009 (UTC)


 * Something like that would be fine Jao. How is that original research? Robo37 (talk) 16:55, 13 August 2009 (UTC)


 * For something like this the appropriate mathematical term is 0/0 is an indeterminate form. Saying 0/0=x would be removed in a flash because it is not any sort of standard mathematical notation. You thought of it yourself and therefore it is original research in wikipedia terms. The article about original research is quite specific about this, it is one of the core policies of wikipedia. Actually I do know of a reference for stuff like this, see James Anderson (computer scientist). Really I don't advise following his path as it goes nowhere. Dmcq (talk) 17:17, 13 August 2009 (UTC)


 * Simply saying "0/0=x" or anything like that would be stupid, I'm asking for something like "without the usual requirement that a/b be a unique number, 0/0 could instead have been defined as an entity assuming all values at once" as Jao said above or "arguably, 0/0 can, in fact, have any number as it's answer, but representing this under a single value is where the difficulty lies”. If it’s a fact (and it seems pretty clear that it is), then why shouldn’t it be mentioned? Robo37 (talk) 17:39, 13 August 2009 (UTC)


 * 0/0 could be defined as anything you want, even 1. The problem is that any definition would not follow the usual rules. In fact 00 is actually defined as 1 in many circumstances even though in others it is best left undefined and is also an indeterminate form. There just is no point in your definition that I can see and there isn't a literature saying that it has a point. What would be your reason for doing so? What would be the difference between what you are saying and what James Anderson (computer scientist) wrote and why would what you wrote be any better? Dmcq (talk) 20:09, 13 August 2009 (UTC)


 * As an aside, I think there is a basic misconception about the possibility or impossibility of dividing by zero. If someone says "we can't divide by zero", someone else will invariably hear "nobody has yet managed to divide by zero". And since it's so extremely easy to come up with a way of dividing by zero, they think they have seen something original. (I'm not saying this is what Robo37 does here, but it certainly is the way the BBC story on Anderson's nullity was worded.) The fact, as you point out and as the article maybe needs to be clearer on, is that 0/0 can be defined to be something (and that it already has been done, several times!), it just can't be defined to be something more useful than "undefined". A useful division by zero should let us, for instance, completely solve the equation x2 = x by dividing both sides by x. As far as I know, not even wheel theory accomplishes that, because it's just not accomplishable. Now, there's a gauntlet for all presumptive zero divisors (pun intended) to pick up. — JAO • T • C 05:14, 14 August 2009 (UTC)
 * There are a few bits missing, it links to defined and undefined which is a badly written article and it misses any mention of an indeterminate form in the calculus section. I think the article could be cleaned up a bit too. However it does mention that the value is not defined except in some special circumstances and it isn't too bad an article. It is also read by a lot of people, so overall there is a higher bar to aim at when editing it than many other articles, but of course anyone who's willing to put in the effort to try is welcome to give it a go. Dmcq (talk) 10:15, 14 August 2009 (UTC)


 * Why is there so much original research here? If you think you are correct, publish a paper, have it peer reviewed, accepted, and published in a journal, and then we can link to it. Or at least bring it up with the Math help desk. Talk pages are supposed to be for improving articles, not introducing original research. --Zarel (talk) 22:31, 14 August 2009 (UTC)


 * All I'm doing is suggesting an improvement to this article. Robo37 (talk) 23:00, 14 August 2009 (UTC)
 * As far as I recall you titled this section "0/0 = x, fact". That is not an accepted part of mathematics. Jao has tried to extract some sort of idea of what the problem is from what you said seeing that you haven't been able to read the article without difficulty. The precise ideas you have had though are simply not suitable for inclusion though and what Zarel said about them is correct. Dmcq (talk) 23:17, 14 August 2009 (UTC)


 * We have the No Original Research rule for a good reason - if we didn't, Wikipedia talk pages would be full of debate over whether or not your original research was correct or not. For instance, to me, it is obvious that your "fact" of "0/0 = x" is incorrect and an abuse of notation, but article talk pages are not a good place for the discussion of why. The "No original research" rule ensures that if you disagree, you take it up with professional mathematicians instead of us, and you can come back to us when professional mathematicians agree with you. --Zarel (talk) 10:51, 2 December 2009 (UTC)

Hmmm, now if you had said that 0/0 ∈ ℝ, you might have gotten some consensus. When you consider that division is the process of repeated subtraction, asking "how many times can you subtract the divisor (the denominator, here zero) from the dividend (the numerator, the other zero) before you have a remainder of or less than the divisor (zero)?" (the literal meaning of division) then clearly any real number answers correctly for the quotient, and all answers you can comprehend although indeterminate are members of ℝ. If you do the long division of 0 goes into 0, you'll see that no matter what you choose for your quotient, immediately upon back-multiplying and subtracting that product of your number from ℝ times zero, you have zero and thus are done. In that sense, your wanting to use "x" is understandable, but you need to precisely state x ∈ ℝ. (Remembering the definition of division is always helpful in resolving these issues, when you try an operation such as 1/0, you'll immediately see you'll be subtracting zero from 1 without bound, hence making it really clear why 1/0 is usually considered to be boundless and expressed 1/0 ∈ ±∞ .) —Preceding unsigned comment added by 130.111.163.179 (talk) 12:59, 30 September 2010 (UTC)

Grade seven revisited
Long ago, way back when a four-function calculator was consider high tech, I got into a sparring match with my grade seven math teacher, which did not end well so far as I was concerned.

He made completely valid points, but he still seemed to be missing something essential about how the problem appears to the novice as yet unencumbered with mathematical convention. After our interaction, I remained as inarticulate about what my math teacher had skipped over as I was before, so I found his arguments correct, but nevertheless unsatisfying.

And so I suddenly get the urge to see how this is old chestnut plays out on Wikipedia, and "whoa! time machine". Is that you Mr W? Are you editing Wikipedia from your retirement cabin on the lake? Because this is the same damn thing.

A compelling reason for not allowing division by zero is that, if it were allowed, many absurd results (i.e., fallacies) would arise.

The fallacy here is the assumption that dividing 0 by 0 is a legitimate operation with the same properties as dividing by any other number.

What that passage actually demonstrates is that defining 0 as its own multiplicative inverse (0/0 = 1) instantly leads to the absurd. I quickly refreshed myself on the axioms of field theory, and the axioms pertaining to a/b exclude the cases where b = 0. Defining 0/0 to some quantity does not appear to violate the existing axions of field theory, although we quickly ruled out 0/0=1 as leading to absurdity. By the same argument we can quickly rule out 0/0=a for any non-zero a.

But what about defining 0/0=0?

This isn't inconsistent with the axioms of field theory based on my quick review. Maybe it's inconsistent with a theorem of field theory, such as x=x-1 having a single solution (er, a single non-negative solution) as proven from the existing axiom which excludes b=0 from any quantity a/b.

But that's kind of narrow-minded about axioms, and we ought to know better than to proceed on the basis of this kind of axiomatic narrow-mindedness because of non-Euclidean geometry.

So why don't we define 0/0=0 (and 0/x=0 and x/0=0 all around) and amend the theorem to x=x-1 having a singular solution (the multiplicative identity) for x strictly positive.

While we're at it, we can also thoroughly subordinate division under multiplication, so that we can't actually divide both sides of an equation by any quantity; whereas we can multiply both sides of an equation by any chosen reciprocal. In this formalism, you can't divide an equation by zero, you can only multiply by 1/0=0, as defined in the previous paragraph. And you really can't get into much fallacy trouble multiplying both sides of an equation by 0 (be careful with strict inequalities such as a < b, however).

So why don't we do this? Here are my two primary guesses:
 * 1) it's uniformly infelicitous to multiply both sides of an equation by 0 (wherever you were going, this doesn't help you get there, even if you salvage your proof later—leaving behind some foolish but non-fallacious deadwood).
 * 2) when progressing to infinitesimals, this peculiar pointwise discontinuity clutters your statements and proofs to no useful effect

It appears to me on superficial review that defining 0/a=0 and a/0=0 for all a is not precluded by fallacy, nor by necessary rather than whimsical field theory axiomatization, but by pure infelicity: it simply doesn't buy you anything you can finally wield to your advantage; refusing to extend the definition of division in this way votes a certain kind of deadwood off the island at first point of contact.

I have far more background in computer science, and in this realm it is surely not a win in many contexts to lose your shit over deadwood that isn't causing any real trouble; better to let the computation proceed uniformly and deal with any issues that arise at the other end. What's a NaN here or there between friends that probably disappears again at a later step anyway? Did we really need to raise a bat signal to the overlord of the galaxy over one piddly NaN (the Toyota model where any task station can stop the entire assembly line in a heartbeat with one tug of a nearly rope, for 10,000,000 parallel arithmetic units spread over a 1000 distinct systems)? Perhaps exception handling at scale is already hard enough without the hair trigger?

Back to math, what I don't want in response is people crawling all over me because I'm lacking some vital pinprick of sophistication here. I'm deliberately standing in for a clever 14-year-old who will not be even slightly impressed with an answer that amounts to this: after you have worked within field theory for another decade, it will make you sick to your stomach to even think about defining a/0=0, because so much older and so much wiser.

Worst. Possible. Answer. Abominably unacceptable to clever, independent-minded 14-year-old.

Can we make this more usefully explicit, or am I smoking a crack pipe here?

What Mr W. ought to have told me: the seemingly obvious 0/0=1 leads to an instant quagmire, the weirder 0/0=0 (augmented with a/0=0) leads to no particular quagmire, but it's completely infelicitous and doesn't buy you a a darn thing as a working mathematician (proceed directly to tautology, do not collect $200).

My other perspective comes from quantum mechanics where destroying information is regarded as pissing away useful energy. Multiplying by zero is special in that way: it's the only multiplication that destroys information.

This is why fields don't promise for all a: a*b*b-1=a when b=0 (b=0 is at least implicitly ruled out in conventional axiomatization because 0-1 is undefined).

In the infelicitous extension of field theory where 0/a=0, a*b*b-1=a except when b is an information-destroying black hole (b=0).

[*] See The Black Hole War (2008) concerning the controversial non-existence of information-destroying black holes in the real universe; only science book I've ever read where I got halfway through and said to myself "hmmm, better luck next time" as I glazed over into a perfect non-absorbent mirror. Coincidentally, next time is circa next week, because it's presently en route from my library for a second kick at the cat. &mdash; MaxEnt 00:49, 3 May 2021 (UTC)


 * This is clearly a subject that interests you. There are many books that will answer your questions. Essentially, we do not define 0/0 = 0 and abandon, for example, the other axioms of Field Theory which would make this "definition" fail to be well defined, because a great deal of interesting mathematics has already been written with the standard definitions. We could, of course, define something new, call it a quasi-Field, and try to come up with consistent axioms that would allow your definition. In fact, many people, ranging from the advanced mathematician to people almost totally ignorant of mathematics, have written extensively on this subject, and I know at least one grade school teacher who teaches your definition to her class, has for years, and isn't about to stop. Rick Norwood (talk) 11:43, 3 May 2021 (UTC)

Creative Numbers...
Creative numbers have purported to solve the problem of division by zero. I'm no mathematician but perhaps someone who understands the subject matter more critically could check out http://science.mistu.info/Math/Numbers/Creative_numbers_and_division_by_zero.html and see if it warrants expanding the article. 71.207.183.254 (talk) 18:22, 4 January 2013 (UTC)


 * Its original research, with no rigorous proof or peer review. It doesn't merit inclusion here. Mind  matrix  18:40, 4 January 2013 (UTC)


 * I guess you would know better than I would. But, it makes sense to me as a layman. 71.207.183.254 (talk) 19:31, 4 January 2013 (UTC)

I know how to properly divide by 0 and do 0 math in general, so I reviewed the reference for you No. Victor Kosko (talk) 19:03, 3 June 2018 (UTC)

Feel free to correct me, if I am wrong.
Wouldn't it be simpler to use examples such as Foucault's pendulum, as a real world example of dividing by zero to get an observable result? Sin 0 being at the equator. At the Equator, 0° latitude, a Foucault pendulum does not rotate. In the Southern Hemisphere, rotation is counterclockwise.

Or is this page simply concerned with the arithmetical concept? 49.185.200.59 (talk) 04:35, 18 May 2022 (UTC)
 * This is solely about the arithmetic concept, but it does address division by zero as a limit of a function. Your example is the limit where the period of precession approaches infinity because the precession rate approaches zero. –LaundryPizza03 ( d c̄ ) 04:45, 18 May 2022 (UTC)