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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 2 '12 at 20:19
@ArturoMagidin Of course by "we" you mean people doing mathematics in the real number system :) – Michael McGowan Apr 2 '12 at 20:28
@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 2 '12 at 22:00
You "learnt" it wrong. – user5137 Apr 3 '12 at 0:24
@JackManey: why the scare quotes around learnt? It's unusual in American English, but more common elsewhere. – DSM Apr 3 '12 at 14:10

13 Answers 13

up vote 69 down vote accepted

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.

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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 3 '12 at 14:25
Why incorrectly? A limit tending to unsigned infinity can be well defined. – Anixx Apr 3 '12 at 20:09
@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 4 '12 at 9:38
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

$1/x$ does tend to $-\infty$ as you approach zero from the left, and $\infty$ as you approach from the right:


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

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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 3 '12 at 10:51
Even if the limits were equal 1/0 wouldn't be defined. – millimoose Apr 3 '12 at 14:38
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 3 '12 at 20:11
(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 3 '12 at 20:17
@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 3 '12 at 21:53

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 3 '12 at 9:16
Whoops! Corrected. – Hurkyl Apr 3 '12 at 19:13
or what would infinity / infinity be? (ignoring the perspective that there are gradients of infinity) – Cor_Blimey Apr 3 '12 at 20:23
@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 – Hurkyl Apr 3 '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 6 at 16:22

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.

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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 4 '12 at 8:28
+1 because "0 / 0 could be any number" - didn't think about that. – Camilo Martin Apr 13 '12 at 1:17

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....

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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 2 '12 at 21:57
@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 3 '12 at 2:21
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 3 '12 at 13:25
True, but in all those situations, division means something else. – N. S. Apr 3 '12 at 15:45
Why do you think that? – Bill Dubuque Apr 3 '12 at 16:12

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 $\infty$.

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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 $\mathbb{C} \cup \infty \approx \mathbb{P_1}(\mathbb{C})$.

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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.

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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 2 '12 at 21:42
@Asaf, We should throw them into the deep end of the pool and let them swim or drown. Gives them character. – Will Jagy Apr 2 '12 at 21:49
But the the case of the extended complex plane, there is no $+\infty$ or $-\infty$, just one $\infty$. – Robert Israel Apr 2 '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 3 '12 at 0:43

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.

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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 3 '12 at 10:49
Yes, and 1.0/-0.0 yields negative infinity. Srsly. – TonyK Apr 3 '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. – alpha123 Apr 4 '12 at 2:47
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 4 '12 at 7:10

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.

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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 $0$.

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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).

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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 2 '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 3 '12 at 7:48

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 $\frac{0}{0}$.

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