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My brother was teaching me the basics of mathematics and we had some confusion about the positive and negative behavior of Zero. After reading a few post on this we came to know that it depends on the context of its use.

Why do we take 1/0 as positive infinity rather than negative infinity (we come close to zero from negative axis)?

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    We do not take $1/0$ as $\infty$, or as $-\infty$. Division by zero is undefined. – Arturo Magidin Apr 02 '12 at 20:19
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    @ArturoMagidin Of course by "we" you mean people doing mathematics in the real number system :) – Michael McGowan Apr 02 '12 at 20:28
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    @Michael: The only place I recall division by zero is well defined is the trivial ring, ${0=1}$. In this ring $1/0=0/0=1/1=1=0$. – Asaf Karagila Apr 02 '12 at 21:42
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    @Asaf That's true for rings, but there are plently of other algebraic structures where 1/0 is defined, e.g. see the answers below. – Bill Dubuque Apr 02 '12 at 22:00
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    You "learnt" it wrong. –  Apr 03 '12 at 00:24
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    I used to upset my 6th grade math teacher in the early 60s by insisting that 1/0 is ∞ – jacknad Apr 03 '12 at 10:34
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    @JackManey: why the scare quotes around learnt? It's unusual in American English, but more common elsewhere. – DSM Apr 03 '12 at 14:10
  • You cannot divide anything by zero, it's meaningless. Divide something by zero means you're not dividing by anything, so there is no operation, no output, no answer. Take a cake, serves 3 people equally @ 1/3 each, or 2 at 1/2 each, or 1 at 1 each, but if there are 0 people to serve, the cake remains uneaten, it's a no-op. – Lee Kowalkowski Apr 03 '12 at 21:40
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    ...Your brother doesn't happen to be a JavaScript programmer, does he? (Where 1/0 actually is Infinity *facepalm*) – Peter C Apr 04 '12 at 02:41
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    alpha123: That's normal for any system using IEEE 754 floating-point arithmetic, i.e. nearly everywhere that isn't a CAS. Javascript isn't alone in that regard and floating-point arithmetic is a mere approximation of math and thus adheres to other rules (you complain about $x+1=x$ for large $x$, too?) – Joey Apr 04 '12 at 07:01
  • @DSM I assume it wasn't a comment on the spelling, but suggesting that he didn't learn it. That is, you can't "learn" something that's wrong. – Andrew Grimm Apr 04 '12 at 07:50
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    @everyone thanku for explaining the misconception of mine, now i can figure out why 1/0 is undefined. why 1/x as x approaches to zero is also undefined. how 1/x makes infinity or -infinity as x approaches to zero from right or left respectively. – Rishav Mehta Apr 04 '12 at 11:31
  • @Lee: Wouldn't dividing the cake by 1 be a no-op? – LarsH Apr 13 '12 at 03:18
  • @LarsH: Dunno, in terms of performing a scientific experiment, if you do something but nothing happens, that's still a result. You can divide a cake by 1 (especially a small one) it is eaten, that's not a no-op. Dividing by zero is like not even performing the experiment to start with, or not eating the cake, if nobody eats it, the division has not taken place. – Lee Kowalkowski Apr 13 '12 at 14:56
  • @Lee: Sure, eating the cake is not a no-op. But that's separate from dividing the cake. The question isn't about dividing 1 by 0 and eating it. – LarsH Apr 13 '12 at 16:01
  • @LarsH: No, in my analogy, the cake is not divided unless it's consumed (sharing a cake between differing numbers of people, or the number of servings), you may have taken it too literally and let the analogy distract you. It's not about dividing the cake and subsequently eating it, the focus is the consumers of the cake, the reason for dividing. (We always divide for a reason, don't we?) Your question, "is division by 1 a no-op" was answered as "no, just because you did something and the answer was as before doesn't mean you did nothing". – Lee Kowalkowski Apr 15 '12 at 20:34
  • Some related posts: http://math.stackexchange.com/questions/1382013/is-frac00-infty-and-what-is-frac10-are-they-same-does-it-hold-true-for and http://math.stackexchange.com/questions/144526/is-it-wrong-to-tell-children-that-1-0-nan-is-incorrect-and-should-be-%E2%88%9E (You can find probably a few other similar posts.) – Martin Sleziak May 27 '16 at 07:41

13 Answers13

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The other comments are correct: $\frac{1}{0}$ is undefined. Similarly, the limit of $\frac{1}{x}$ as $x$ approaches $0$ is also undefined. However, if you take the limit of $\frac{1}{x}$ as $x$ approaches zero from the left or from the right, you get negative and positive infinity respectively.

S.C.B.
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    Generally the only reason one sees 1/0 as infinity is because some systems (incorrectly) output infinity when given dividing by zero. – Ben Brocka Apr 03 '12 at 14:25
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    Why incorrectly? A limit tending to unsigned infinity can be well defined. – Anixx Apr 03 '12 at 20:09
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    @Anixx I think that Ben is referring to things such as the IEEE floating point standard, in which division of a nonzero float number by 0 is well defined even if you're not doing limits. – Agos Apr 04 '12 at 09:38
  • IEEE-754 doesn't have to give infinity as the result of 1/0; it has a type "NaN" (Not a Number) for these cases. – stevenvh Dec 19 '12 at 17:48
  • Isn't it better to say that the limit of 1/x as x approaches 0 is either plus or minus infinity, rather than saying that it is undefined? – Andrew Sep 18 '17 at 21:52
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    @Andrew I believe that's what my answer says. If you just say "the limit", you are implicitly saying "from the left AND the right." Therefore, if you simply say "the limit", it either converges to the same value from the left and the right, or its undefined. If its undefined because it has different values from the left and the right, you would say "the limit from the left" and "the limit from the right". – Ethan Brown Sep 18 '17 at 23:08
  • Would it not be more correct to say that $\frac{1}{0}$ = ∞ because $\frac{1}{-0}$ = -∞? Because $\frac{1}{0}$ ≠ $\frac{1}{-0}$ and given that 0 ≠ -0, it would be implied that $\frac{1}{0}$ is equal to positive infinity, and $\frac{1}{-0}$ is negative infinity. Right? – Swivel Oct 04 '18 at 01:06
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    @Swivel But 0 does equal -0. Even under IEEE-754. The only reason IEEE-754 makes a distinction between +0 and -0 at all is because of underflow, and for +/- ∞, overflow. The intention is if you have a number whose magnitude is so small it underflows the exponent, you have no choice but to call the magnitude zero, but you can still salvage the sign. In other words, -0 is meant for numbers just barely less than zero, ∞ for positive numbers too large to handle, and -∞ negative numbers to small to handle. Of course, these issues don't even come into play for "regular math." – Dominick Pastore Oct 15 '20 at 19:42
  • @DominickPastore I feel like that underscores the significance of 0 ≠ -0. There can be, for instance, theoretical and statistical significance to the distinction, especially since exact zero is seldom encountered. Additionally, in low level CS, it's important to consider the sign during comparisons because they are not equal. Numerically, you could argue they're equivalent in most applications, but the sign does changes the inherent value. They possess different meanings and serve different purposes. But in areas that a -0 isn't important, like traditional arithmetic, it's redundant. – Swivel Oct 16 '20 at 17:43
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    @Swivel Not really. You're conflating math and CS concepts. In a perfect world, floating point numbers would be a perfect analogue for real numbers, but that's not possible. Instead, we have IEEE-754, an abstraction trying to emulate them as usefully as possible, within X bits. Of course, there are compromises and differences. IEEE-754 also rounds almost every calculation, which "real math" doesn't do. For comparison, look at a program like Mathematica, and you'll find it neither rounds nor has a separate concept of -0: its representation of numbers is more expressive, so it doesn't need them. – Dominick Pastore Oct 17 '20 at 19:42
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    @Swivel Point is, -0 and 0 in IEEE-754 don't behave quite like 0 in real math because they aren't 0. If you want to be technical, they represent intervals $(-2^{-1075}, 0]$ and $[0, 2^{-1075})$, respectively. In pure mathematics, 0 is 0, no ifs, ands, or buts. Any abstract algebra book will explain that $0 = -0$ is one of the basic axioms upon which virtually all math is built. (And this is, after all, Mathematics Stack Exchange.) – Dominick Pastore Oct 17 '20 at 19:43
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$1/x$ does tend to $-\infty$ as you approach zero from the left, and $\infty$ as you approach from the right:

f(x)=1/x

That these limits are not equal is why $1/0$ is undefined.

Emre
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    Hence the rationale for having $1/0=\infty$ in standard floating-point arithmetic, and also the less common $1/-0=-\infty$. – lhf Apr 03 '12 at 10:51
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    Even if the limits were equal 1/0 wouldn't be defined. – millimoose Apr 03 '12 at 14:38
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    the limit of 1/0 is not equal to 1/0. One requires a continuous number system etc, the other does not. Your argument - that there would be a definition were there to be a consistent number 'approached' in th e – Cor_Blimey Apr 03 '12 at 20:11
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    (whoops - I pressed enter by mistake): the limit of 1/0 is not equal to 1/0. The limit of 4/(x -> 2) is not 2; it only (ever) approaches 2: in the limit x is infinitely close to 2, but is still infinitesimally not 2: just as the limit can never be reached so can x never reach being an identity of 2 (in my book ;P). Notwithstanding my point of view on that aspect, I agree that you are right :) – Cor_Blimey Apr 03 '12 at 20:17
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    @Cor_Blimey: no, the limit of $4/x$ as $x\to2$ is $2$, it doesn't approach $2$. $2$ is the limit because it's the number which is approached. – DSM Apr 03 '12 at 21:53
  • @millimoose - Not sure that's correct. For instance for 1/x^2 both the left limit and right limit for x -> 0 are +infinity. Then the limit is +infinity, so defined. – stevenvh Dec 19 '12 at 17:51
  • @stevenvh I phrased that too vaguely and you've misunderstood. I meant that even though in your example the limit of 1/x^2 as x->0 is defined, the value of 1/0^2 still isn't. Therefore, this answer isn't the correct explanation of why 1/0 is undefined. (And has disgustingly many upvotes for how patently wrong it is.) "Has a limit for x->x0" is not equivalent to "is defined at x0". Neither does one imply the other because it's trivially easy to define functions for which one holds true and the other doesn't. – millimoose Dec 19 '12 at 18:39
  • @millimoose - Sorry, I still don't agree. lim(1/x^2) for x->0 is defined: it's +infinity. That may not be a number in R, but it has a defined symbol, for which a limited number of operations are defined. Like infinity + infinity = infinity, but infinity - infinity is not defined. lim(1/x) on the other hand has no defined value, not in R, nor + or -infinity. That's undefined, which is different from + or -infinity. – stevenvh Dec 21 '12 at 12:44
  • @stevenvh You're conflating two ideas: "the value of a function at $x$" ($f(x)$); and "the limit of a function as $x$ approaches $x0$" ($\lim_{x \to x_0} f(x)$). These are two completely distinct concepts. $\infty + \infty$ does not equal $\infty$; nor does it make sense if you're being strict. (Infinity isn't a number and you can't really do arithmethics on it.) It's just a shorthand used when learning how to do limits. It'd be more verbosely (and thus precisely) phrased as "the sum of two terms that approach $\infty$ itself approaches $\infty$". – millimoose Dec 21 '12 at 23:55
  • @stevenvh You're also still misunderstanding what I was originally trying to say. The facts presented in this answer are, in and of themselves, correct. The conclusion drawn from them is not. (The erroneous conclusion being that whether $f(x_0)$ is defined somehow follows from whether $lim_{x \to x_0} f(x)$ exists. It does not, because it's possible to have functions be defined at a point where they do not have a both-sided limit; and it's possible to have functions have a limit at a point where they're not defined.) – millimoose Dec 22 '12 at 00:01
  • @millimoose - The conclusion is correct inasmuch the limit can't be defined if left and right limit are different. That does not imply that it would be defined if they are equal. And $\infty$ + $\infty$ is $\infty$, by definition. $\infty$ is not a number, but that doesn't mean you can't define a ring for it. And operations with reals, like $\infty$ + r = $\infty$. The fact that $\infty$ + $\infty$ = $\infty$ is the reason why $\infty$ - $\infty$ is not defined; you can't find a result which is consistent with the definition for addition. – stevenvh Dec 22 '12 at 06:26
  • @stevenvh Okay, one last try. The OP's question doesn't mention limits. Therefore he's asking whether the value of the expression $1/0$ is defined. (Which it isn't.) The last sentence of this answer is "That these limits are not equal is why $1/0$ is undefined." – once again, the value of the expression, not the limit of a rational function where the denominator approaches 0. It is this conclusion, which is an actual (if incorrect) answer to the actual question, that I'm talking about. You're arguing about an entirely different statement that while correct isn't an answer to this question. – millimoose Dec 22 '12 at 14:39
  • @stevenvh Also, an algebraic structure that includes $\infty$ and defines $\infty + r$ is not the same algebraic structure as the common real numbers. You can't really make a mathematical argument based on a set of axioms you just made up for the sake of that argument alone. The set of real numbers does not contain $\infty$. No operation over two real numbers (including, but not limited, to the operation of division, and the numbers $1$ and $0$) can have a result of $\infty$; and $\infty + \infty = \infty$ is a statement that has no meaning whatsoever when talking about the real numbers. – millimoose Dec 22 '12 at 14:53
  • @millimoose - No, $\infty$ isn't a real number, I never said it was. But you can extend the reals to include $\infty$, that's the hyperreals, and then you can define operations which include reals and $\infty$. (Probably not a ring like I claimed earlier; none of the 4 basic operation has an inverse for $\infty$.) – stevenvh Dec 23 '12 at 07:40
  • @stevenvh Is this question a question about arithmetics on hyperreals? No? Then it doesn't matter as to what you can or cannot do them, and everything you said is completely off-topic, talking past me to argue for the sake of arguing. This is supposed to be a discussion of the above answer, not free-association rambling about mathemathical concepts that come into mind when reading it. – millimoose Dec 23 '12 at 14:23
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The place where you typically see $1 {\color{red}/} 0 = \infty$ is when doing arithmetic in the projective line. (I've added color to ${\color{red}/}$ to better distinguish it from the ordinary division operation on the real numbers) The binary operation ${\color{red}/}$ is defined for every pair of projective real numbers except $(0,0)$ and $(\infty, \infty)$:

  • $ x {\color{red}/} y = x/y$ when $y \neq 0$
  • $ x {\color{red}/} 0 = \infty$
  • $ x {\color{red}/} \infty = 0$
  • $\infty {\color{red}/} x = \infty$

where $x,y$ denote ordinary real numbers. (one can define the other arithmetic operations too)

The projective line has only one infinite element. In the projective line, the same number $\infty$ is at both "ends" of the ordinary line. There is another common number system -- the extended real numbers -- that has two infinite elements: $+\infty$ and $-\infty$. Make particular note that $1 {\color{cyan}/} 0$ is undefined for the arithmetic of extended real numbers. (where again I've added color to distinguish)

Unfortunately, people often use $\infty$ instead of $+\infty$. So, when someone writes $\infty$, it can be unclear whether or not they are doing arithmetic in the projective real line, or in the extended real line.

And, just to be clear, $1/0$ is undefined as well.

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    And what about $x {\color{red}/} 0$? – Raphael Apr 03 '12 at 09:16
  • Whoops! Corrected. –  Apr 03 '12 at 19:13
  • @Cor_Blimey: By the conventions of this answer, $\infty / \infty$ doesn't even make sense, and $\infty {\color{red} / }\infty$ is undefined. In the projective numbers, the unusefulness of extending ${\color{red} / }$ to include a value for $\infty {\color{red} / }\infty$ is exactly the same as the unusefulness of extending to $0 {\color{red} / } 0$. Honestly, if you really believe that you need an arithmetic structure where $0/0$ or $\infty/\infty$ is defined, then you should probably look at a http://en.wikipedia.org/wiki/Wheel_theory –  Apr 03 '12 at 21:42
  • @Cor_Blimey : When you want to get a meaningful value for ∞/∞, you need to do the same as you'd do in the case of 1/∞ or 1/0 : look at the limit for x → ∞ in your specific context. For x/x (= lim x → ∞ (x/x)), the limit value is equal to 1. – John Slegers Jan 06 '16 at 16:22
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Since you are having this confusion, I think it helps to consider the concepts of zero, infinity and "undefined".

In the most basic sense, division is the opposite of multiplication. Thus, the fact that 2 x 3 = 6 implies that 6 / 3 = 2.

1 x 0 = 0. Applying the above logic, 0 / 0 = 1. However, 2 x 0 = 0, so 0 / 0 must also be 2. In fact, it looks as though 0 / 0 could be any number! This obviously makes no sense - we say that 0 / 0 is "undefined" because there isn't really an answer.

Likewise, 1 / 0 is not really infinity. Infinity isn't actually a number, it's more of a concept. If you think about how division is often described in schools, say, number of sweets shared between number of people, you see the confusion. If I go around some people giving them 0 sweets each, how many people do I need to go around until I have given away my 1 sweet? An infinite number? Kind of, because I can keep going around infinitely. However, I never actually give away that sweet. This is why people say that 1 / 0 "tends to" infinity - we can't really use infinity as a number, we can only imagine what we are getting closer to as we move in the direction of infinity. However, in this case, the number of sweets I have is never changing, so I'm not really getting closer to anywhere. Even this logic doesn't really work.

The long and short of it is that 1 / 0 doesn't really make sense as a calculation. When we do use the notion of infinity we tend to use positive infinity where it doesn't matter purely by convention. However, if you think about it too hard you start to get into philosophy and stuff, like "what actually is infinity?" and "wait, what is a number"?

The things people are talking about where it does are different ways of using numbers so they don't really count. For example, in the trivial ring, there is only one number, which works like a 0 (add it to anything and you get that thing) and a 1 (multiply it by anything and you get the same thing again) and makes sense because you can only add it to or multiply it by itself to get itself. It's pretty boring actually, but in that case this one number - let's call it x - is both 0 and 1, so 1 / 0 = x / x = x because everything equals x. As you can see, this is a bit of a cheat because we don't even have enough numbers to have a notion of 1 / 0 in the way you're thinking of it.

Fred
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    +1 for "infinity isn't actually a number". That's a very important point, many beginners confuse that, probably due to the ubiquitous abuse of notation. – Frank Apr 04 '12 at 08:28
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    +1 because "0 / 0 could be any number" - didn't think about that. – Camilo Martin Apr 13 '12 at 01:17
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In Calculus sometimes we write

$$\frac{1}{0_+}=+\infty \,,$$ $$\frac{1}{0_-}=-\infty \,,$$

This notation makes $\frac{1}{0}$ sometimes $+\infty$ and sometimes$-\infty$, but you have to be careful what it means. It doesn't literally mean $\frac{1}{0}$, it's actual meaning in Calculus is

a limit of the type $\frac{f(x)}{g(x)}$, where $\lim f(x) =1$ and $\lim g(x)=0$.

While writing $\frac{1}{0}$ is much more convenient than writing that expression, it has the downside that many people take it to mean literary $1$ divided by $0$.

Sometimes in calculus $\frac{1}{0}$ is $+\infty$, sometimes it is $-\infty$ and sometimes it doesn't exists. But while the meaning is somehow closed to the idea of $\frac{1}{0}$, in calculus $\frac{1}{0}$ means something else that division.

After reading few post on this we came to know that it depends on the context of its use.

Actually what happens is more subtle. In two different areas of mathematics that notation is used to denote two different things. When you say $\frac{1}{0}$, if you mean standard division of real numbers, then that is not defined, no matter what the context is...

There are actually many situations when the unfortunate use of the same notation/definition for different things creates confusion....

N. S.
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    It is not true that "1/0 is not defined, no matter what the context is". Indeed, there are already a couple answers showing that is is well-defined in certain contexts. – Bill Dubuque Apr 02 '12 at 21:57
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    @BillDubuque arithmetically, $1/0$ simply means the inverse of $0$ for multiplication. While you can define $1/0$ in other contexts, usually by compatcification of plane or line, it's meaning is different than the arithmetic meaning. No matter how you define it $\frac{1}{0} \cdot 0$ cannot be $1$ without breaking the properties of arithmetic.... That is really what I meant by that comment... – N. S. Apr 03 '12 at 02:21
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    Not true. The only ring where zero is invertible is the zero ring. But what you wrote was a much more general assertion not restricted to rings, viz. "1/0 is not defined no matter what the context". That is false. There are various algebraic structures generalizing rings where 1/0 has a meaning, both algebraically and geometrically. – Bill Dubuque Apr 03 '12 at 13:25
  • True, but in all those situations, division means something else. – N. S. Apr 03 '12 at 15:45
  • Why do you think that? – Bill Dubuque Apr 03 '12 at 16:12
  • Because the poster said that he is learning the basic of mathematics, which means arithmetic, basic algebra and calculus. By number he probably means a real number and by fraction he probably means division of real numbers.... – N. S. Apr 03 '12 at 16:41
  • Of course this is more a philosophical issue: if you extend the reals to a new structure, which is {\it not a ring}, is it the same operation or not? No matter how natural the extension is, in my opinion it is a new operation, and $1/0$ only makes sense in the extension.... – N. S. Apr 03 '12 at 16:55
  • If "context" in your "no matter what the context" is restricted to real number contexts, then it would help to explicitly mention that. Many of the extended number systems preserve all of the original ring structure, differing only in the extended part, where some operations may be partial vs. total, and where some laws may no longer hold universally. As such, the operations are natural extensions of the original operations, at least by every definition of "natural extension" that I've ever encountered. – Bill Dubuque Apr 03 '12 at 18:13
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Works just fine if you are discussing Möbius transformations in $\mathbb C \cup \{\infty\},$ called the extended complex plane or the Riemann sphere. Otherwise, no.

azimut
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Will Jagy
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    The OP wrote that his brother is teaching him some mathematics, I think it's still a bit early to discuss Moebius transformations... :-) – Asaf Karagila Apr 02 '12 at 21:42
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    @Asaf, We should throw them into the deep end of the pool and let them swim or drown. Gives them character. – Will Jagy Apr 02 '12 at 21:49
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    But the the case of the extended complex plane, there is no $+\infty$ or $-\infty$, just one $\infty$. – Robert Israel Apr 02 '12 at 22:00
  • @Robert, I did notice mention of $\pm \infty.$ I just decided to ignore it. Not what I would have done if I had been giving the first answer, rather than the fifth or so. – Will Jagy Apr 03 '12 at 00:43
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It depends on context. When dealing with rational functions or trigonometric functions, it can make sense to deal with a line $\mathbb{R}\cup\{\infty\}$ in which a "neighborhood of $\infty$" is a set that looks like $(-\infty, a)\cup(b,\infty)$, so the line becomes topologically a circle; the two ends meet at $\infty$. A similar point of view is heavily relied on when studying functions of a complex variable, and you're appending just one $\infty$ to the plane $\mathbb{C}$. It is in that sort of context that it makes sense to say that $1/0$, or $5/0$, etc., is $\infty$.

5

As far as I am aware, when doing floating point math on most computers in most computer languages, 1.0/0.0 will yield positive infinity (-1.0/0.0 yielding negative infinity). I think this question is extremely dependent on context and usage.

Mark
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    Hence the rationale for having both $0=1/\infty$ and $-0=1/-\infty$, distinct values that compare as equal. – lhf Apr 03 '12 at 10:49
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    Yes, and 1.0/-0.0 yields negative infinity. Srsly. – TonyK Apr 03 '12 at 11:27
  • As a programmer, I can say that most programming languages either generate a compiler error or throw a runtime exception on division by zero, the notable exception being JavaScript, in which 1/0 actually is positive infinity and -1/0 (and 1/-0 :P) are negative infinity. – Peter C Apr 04 '12 at 02:47
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    alpha123, in many programming languages (most notable those descending syntax- or semantic-wise from C) 1/0 is interpreted as an integer division where this is indeed an error. If you force floating-point numbers you get the same result you attribute only to Javascript. E.g. 1/0.0 will usually yield positive infinity in those languages. – Joey Apr 04 '12 at 07:10
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Since I didn't see it in the comments : in the context of complex numbers, you can have $1/0 = \infty$ without this problem, with the convention that $\infty = - \infty ( = \lambda \infty$ for all $\lambda \neq 0$).

Here "getting close to $\infty$" means getting very large in modulus, no matter the direction (including, but not exclusively, both directions of the real axis). This is useful, for example, when considering rational fractions : they always admit a limit in $\infty$, and take values in $\mathbb{C} \cup \infty$.

Of course this is strongly linked to projective geometry as well, since $\mathbb{C} \cup \infty \approx \mathbb{P_1}(\mathbb{C})$.

Albert
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First off, in the conventional system of real numbers, the expression $\frac{1}{0}$ is considered as undefined, that is, it has no value. "Infinity" does not exist in the real numbers. $0$ is not in the domain of the function $f(x) = \frac{1}{x}$.

There is good reason why it is not defined: the function $f(x) = \frac{1}{x}$ is usually taken to mean "give me the multiplicative inverse of $x$", but $0$ lacks a multiplicative inverse. This is easy to see. The function $f(x) = 0x$ is totally non-injective -- there is no subset of $\mathbb{R}$ containing more than one point on which it is injective. We lose all information about what $x$ was when we put it in there. So how can we hope to have a value that lets us invert it?

That being said, there are contexts in which we may take $\frac{1}{x}$ to stand for something more general, and where $\frac{1}{0}$ may be meaningful. One such example, where we can say $\frac{1}{0} = \infty$, is the real projective line and its complex equivalent, the Riemann sphere. In this case, the infinity is unsigned, that is, $-\infty = +\infty$. If we choose to use a generalized system of "numbers" that includes $-\infty$ and $+\infty$ as separate elements, then we usually still don't define $\frac{1}{0}$ because $\frac{1}{x}$ approaches $-\infty$ as $x$ approaches $0$ from the left and $+\infty$ as it approaches $0$ from the right. We could take it as one or the other if we wanted to, but the choice would be arbitrary.

A big caveat of these systems is that the laws of arithmetic will be different in them. For example, in the above-mentioned RPL, caveats occur when dealing with expressions involving infinities.

2

Division by zero can also be made sense of in an algebraic structure known as a "wheel".

From the Wikipedia article:

Wheels discard the usual notion of division being a binary operator, replacing it with multiplication by a unary operator $/x$ similar (but not identical) to the reciprocal $x^{-1}$, such that $a/b$ becomes short-hand for $a\cdot /b=/b\cdot a$, and modifies the rules of algebra such that

  • $0x\neq0$ in the general case.
  • $x-x\neq0$ in the general case.
  • $x/x\neq 1$ in the general case, as $/x$ is not the same as the multiplicative inverse of $x$.

So as you can see, one has to discard a whole lot of the "usual" rules of algebra in order for division by zero to even make sense. As others have done an excellent job of explaining, in the standard real numbers, it does not make sense to divide by $0$.

Zev Chonoles
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Infinite numbers do exist in the hyperreal number system which properly extends the real number system, but then their reciprocals are infinitesimals rather than zero. Thus the idea of $\frac{1}{0}$ can be interpreted as saying that if $\epsilon$ is infinitesimal then $\frac{1}{\epsilon}$ is infinite. This resolves your problem because it shows that $\frac{1}{\epsilon}$ will be positive infinity or infinite infinity depending on the sign of the original infinitesimal, while division by zero is still undefined. This viewpoint helps account for all indeterminate forms as well, such as $\frac{0}{0}$.

Mikhail Katz
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$\frac{1}{0}$ is not defined...end of story. It is not infinity or minus infinity (these things are not numbers either).

fretty
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    That simply passes the buck to precisely defining "number". But there is no precise definition of number in general use. In fact some structures where such expressions are valid are called "number systems" by some authors. – Bill Dubuque Apr 02 '12 at 21:29
  • Well my reply was written to account for the fact that the original poster clearly has no knowledge of higher maths. Back then you accept what a "number" is intuitively. Technically I should say that infinity and minus infinity do not belong to the same "number system" that the OP is working in but that is just being pedantic for no reason. – fretty Apr 03 '12 at 07:48