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I would say $\infty - \infty=0$,because even though $\infty$ is an undetermined number,$\infty = \infty$.so $\infty-\infty=0$.

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    $\begingroup$ Very relevant: en.wikipedia.org/wiki/Indeterminate_form. Hopefully, someone will write a nice answer. $\endgroup$ – Srivatsan Aug 30 '11 at 18:19
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    $\begingroup$ @Sri I see no "indeterminate forms" in the question. $\endgroup$ – Bill Dubuque Aug 30 '11 at 18:34
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    $\begingroup$ @Sri "Infinity" has many different meanings in mathematics. As such, the question is ill-posed as it stands. $\endgroup$ – Bill Dubuque Aug 30 '11 at 18:48
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    $\begingroup$ With natural numbers the result of taking $m$ items away from a set of $n$ items, $m\le n$, will result in a set with a unique number of items in it (no matter which subset of $m$ items were taken away), hence it makes sense to label the result as a specific number, namely $n-m$. With an infinite ($\aleph_0$) set of items, taking away an infinite subset doesn't uniquely determine the cardinality of the resulting set, hence $\infty-\infty$ is ill-defined in this context. $\endgroup$ – anon Aug 30 '11 at 19:28
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    $\begingroup$ Relevant reading: math.stackexchange.com/questions/36289/is-infinity-a-number $\endgroup$ – Qiaochu Yuan Aug 30 '11 at 22:39

13 Answers 13

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From a layman's perspective, imagine that I have an infinite number of hotel rooms, each numbered 1, 2, 3, 4, ...

Then I give you all of them. I would have none left, so $\infty - \infty = 0$

On the other hand, if I give you all of the odd-numbered ones, then I still have an infinite number left. So $\infty - \infty = \infty$.

Now suppose that I give you all of them except for the first seven. Then $\infty - \infty = 7$.
While this doesn't explain why this is indeterminate, hopefully you can agree that it is indeterminate!

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    $\begingroup$ I don't quite understand that last comment, but I'm almost certain that the answer is "no". $\endgroup$ – Altar Ego Aug 30 '11 at 19:19
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    $\begingroup$ @Altar: This is a well-known paradox. Google "Hilbert's hotel" for more information. $\endgroup$ – Fredrik Meyer Aug 30 '11 at 19:27
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    $\begingroup$ @Altar Ego: Congratulations on getting away from the usual limit examples. It would spoil the clarity of what you say, but let $A$ is a (finite) set of hotel rooms, say $a$ of them, and if $B$ is the set of occupied hotel rooms, and there are $b$ such, then the number of unoccupied rooms is $a-b$. So subtraction does give the answer in the finite case. $\endgroup$ – André Nicolas Aug 30 '11 at 19:42
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    $\begingroup$ @trony: I had "there are as many even numbers as there are natural numbers" in mind... $\endgroup$ – J. M. is a poor mathematician Aug 30 '11 at 23:50
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    $\begingroup$ I remember asking my math instructor a similar question, with a wide wide grin on my face. I honestly didn't understand the idea but I felt smart for challenging my instructor with a simple question. I remember him looking at me with a condescending look, shaking his head slowly, not saying a word, and then moving on with his lecture. I hated him that moment like no one else, for not teaching a wondering (wandering) student when he really, really should have. When that's what he was paid for, at the very least. $\endgroup$ – wilhelmtell Aug 31 '11 at 1:53
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Perhaps an uninteresting way of speaking about infinity, but one you surely will understand, the first one I was taught and which is at the level of Introductory Calculus. If you consider the sequence

$$1,4,9,16,25,36,\ldots ,n^{2},\ldots \tag{1}$$

and the sequence

$$1,2,3,4,5,6,\ldots ,n,\ldots \tag{2}$$

both go to infinity as $n$ tends to infinity (I will add if you deem necessary the meaning of "going to infinity"). The sequence obtained by subtracting $(2)$ from $(1)$ goes to infinity too

$$0,2,6,12,20,30,\ldots ,n^{2}-n,\ldots \tag{3}$$

If you consider now the sequence

$$\frac{1}{1},\frac{3}{2},\frac{8}{3},\frac{15}{4},\frac{24}{5},\frac{35}{6}% ,\ldots ,\frac{n^{2}-1}{n},\ldots \tag{4}$$

which goes to infinity too and subtract it from $(2)$ you get a sequence which tends to $0$

$$0,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},\ldots ,\frac{% 1}{n},\ldots \tag{5}$$

If you take sequence $(2)$ and subtract from it the sequence

$$0,1,2,3,4,5,\ldots ,n-1,\ldots \tag{6}$$

you get the constant sequence

$$1,1,1,1,1,1,\ldots \tag{7}$$

So the difference of two sequences both going to infinity may be a sequence which tends to $\pm\infty $, $0$ or a finite number.

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  • $\begingroup$ Or negative infinity! $\endgroup$ – orlp Oct 18 '11 at 10:08
  • $\begingroup$ @nightcracker: That's right. $\endgroup$ – Américo Tavares Oct 18 '11 at 11:11
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When trying to invent new number systems with numbers we name, for example, "infinity", we must define the rules of operation. If you decide to adjoin the symbol $\infty$ to, say, the real numbers, then you must decide which properties you want the symbol to have.

For example, do you want (which is reasonable) $x+\infty=\infty$ for every real number $x$? If so, your new system, $\mathbb{R} \cup \{\infty\}$ can no longer be a ring, so you lose some of the important properties of the system.

Usually, the symbol $\infty$ is only used to indicate that limits "grow beyond any number", for example. In this case, $\infty-\infty$ depends on what limits are in question.

In other situations, $\infty$ is used to formally "complete" a topological space, say $\mathbb{C}$. All new students of topology learn that the sphere and $\mathbb{C} \cup \{\infty \}$ are homeomorphic spaces, that is, essentially the same.

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The expression $\infty - \infty$ is called indeterminate because $\infty - \infty$ could be anything in the set $[-\infty, \infty] = \mathbb{R} \cup \{\pm \infty\}$. Consider the limit

$$ \lim_{x \to \infty} (f(x) - g(x)). $$ If $f(x)$ and $g(x)$ are polynomials (or arbitrary functions which tend to infinity), the limit is of the form $\infty - \infty$, but we can concoct examples where the limit can be any number. For example, let $\alpha$ be an arbitrary real number, and define $f(x) = x+ \alpha$ and $g(x) = x$. Then the limit is $\alpha$. Furthermore, if $f(x) = x^2$ and $g(x) = x$ (or vice versa), then the limit becomes $+ \infty$ (or $-\infty$). Therefore there is no reasonable way to define $\infty - \infty$.

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    $\begingroup$ I will take $\infty$ to mean a quantity which grows without bound. There are many such quantities, but the distinction is usually made after one is introduced to calculus, and since my answer is based in elementary calculus, I will not worry about the distinction here. $\endgroup$ – JavaMan Aug 30 '11 at 19:07
  • $\begingroup$ It is "indeterminate", not "indeterminant". Also I would say indeterminate form or expression, because being indeterminate here means exactly that no definite value (quantity) can be unambiguously associated to the expression. $\endgroup$ – Marc van Leeuwen Aug 21 '13 at 10:25
  • $\begingroup$ @JavaMan Be careful about notation. $$[a,b]=\{x \in \mathbb{R} : a \leq x \leq b\}.$$ Since, $\infty \notin \mathbb{R}$, $$[\infty,\infty]$$ doesn't really make sense in interval notation. $\endgroup$ – beep-boop Sep 11 '14 at 22:05
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    $\begingroup$ @alexqwx Actually, notations like $[0,\infty]$ are actually used quite commonly in fields for which it would be useful, for example measure theory. And I can think of at least one formal system where writing $[-\infty,\infty]=\Bbb R^*$ is valid, because the interval is defined not with respect to $\Bbb R$ but $\Bbb R^*$ where $\Bbb R^*=\Bbb R\cup\{-\infty,\infty\}$ and $-\infty,\infty$ are arbitrary sets not in $\Bbb R$. $\endgroup$ – Mario Carneiro Feb 10 '15 at 19:30
  • $\begingroup$ This argument assumes that ∞−∞ should be defined as a limit, which is not obvious. $\endgroup$ – PJTraill Nov 14 '18 at 20:51
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There is a simple way to show how we can define such problems. .Simply write the equation and try to find $x$.

$\infty - \infty =x$

$\infty =x+ \infty$

If you try to put some values of $x$, such as 1,2,3,.... you will notice that all numbers will be correct because if you add any number to infinity,it will be again infinity or if you select any negative number , infinity will not loss anything from infinity, it will be again infinity. If so what is $x$? x can be any number, therefore we say such x as indeterminate.

If you try the same method to other indeterminate expression such as $$\frac{0}{0}$$

$$\frac{0}{0}=x$$ what is $x$?

$$\frac{0}{0}=\frac{x}{1}$$

$$x.0=0.1$$

$$x.0=0$$ $x$ can be any number because if we multiple any number with zero result will be zero. Again similiar result as we got for $\infty - \infty$.

Another important property that all indeterminate expressions can transform to another form. please see examples below.

Example 1: $$0.\infty=$$

$$=0.\frac{1}{0}=\frac{0}{0}$$


Example 2: $$\frac{\infty}{\infty}=$$

$$ =\frac{\frac{1}{0}} {\frac{1}{0}} =\frac{0}{0}$$


Example 3: $$\infty - \infty=$$

$$ =\frac{1}{0}-\frac{1}{0}= \frac{1-1}{0}=\frac{0}{0}$$


Example 4: $$1^\infty=x$$

$$ln(x) = ln(1^\infty)= \infty.ln(1)=\infty.0=\frac{1}{0}.0= \frac{0}{0}$$

If $ln(x)$ is indeterminate , thus $x$ is also indeterminate

Similiar way can be shown for $\infty ^ 0$ as shown in example 4. These examples show that indeterminate expressions can convert to other type. It is very important to find limit values in mathematics.

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    $\begingroup$ I can't comment on the merit of the bottom half of this answer, but I can say that the top is respectable. As such, I'll act on good faith and +1. Feel free to modify the bottom--I do not understand it enough to fully help you clarify it. $\endgroup$ – 000 Mar 17 '12 at 21:00
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$$ \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2 + y^2)^2} \, dy\,dx \text{ is actually different from }\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2 + y^2)^2} \, dx\,dy. $$ One of these is $\pi/4$; the other is $-\pi/4$.

That's infinity minus infinity. I.e. if you integrate over the part of the square where $x^2-y^2$ is positive, you get $\infty$, and if you integrate over the part that's negative, you get $-\infty$.

If you have $\infty$ minus a finite positive number, then the integral would be $\infty$ either way; if you have $-\infty$ plus a finite positive number, then it's $-\infty$ either way. And if both the positive and negative parts are finite, then you get the same number either way. This sort of bad behavior of integrals, where changing the order of integration can change the number you get as the bottom line, can happen only when the positive and negative parts are both infinite.

The same thing happens with infinite series, as you'll see if you google: conditional convergence Riemann rearrange

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If we interpret $\infty$ to mean some infinite surreal number such as $\omega = \{ \mathbb{N} \mathrel{|} \emptyset \}$, then yes, $\infty - \infty = 0$, because all surreal numbers have additive inverses.

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    $\begingroup$ Note that under this interpretation, you have $\infty+1>\infty$ and $2\infty>\infty$, unlike most of the other answers here, so having $\infty-\infty=0$ is consistent since $\infty$ is not actually the biggest number in the surreals. $\endgroup$ – Mario Carneiro Feb 10 '15 at 20:07
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The result is Indeterminate.

The reason being because you could never come to a concise answer:

Is the 1st infinity larger? Then answer would be +infinity.

Is the 2nd infinity larger? Then answer would be -infinity.

Are the same size? Then answer could be 0.

Since the size of infinity is unknown, we cannot determine any of these situations and therefore the answer is Indeterminate.

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    $\begingroup$ Well: Take this example $\lim_{x \to \infty} (x+2)-(x)=2$. This is an expression of the form $\infty-\infty$, but "the result" is 2, even though one "infinity is larger" than the other. $\endgroup$ – Fredrik Meyer Aug 30 '11 at 18:55
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    $\begingroup$ But you've given an answer to a different thing from to the question: you determined the expression, but in the question it is indetermined (underdetermined,indeterminate) $\endgroup$ – Gottfried Helms Aug 31 '11 at 2:57
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One reason the answer is indeterminate is because you can find sequences $x_n,y_n$ of real numbers such that $x_n,y_n \to \infty$ and $(x_n-y_n)$ can converge to any real value or $\pm \infty$.

For example pick $a \in \Bbb{R}$ and $x_n=n+a,y_n=n$. Then $x_n-y_n=a,\ \forall n$ and therefore $x_n-y_n \to a$.

For $x_n=2n,y_n=n$ you get $\infty$. Switch the order and get $-\infty$.

Indeterminations are usually taught in analysis courses, and the main reason they are called this way is that you cannot say what is the value of the limit from the start. Other examples are $\infty \cdot 0, 1^\infty, \infty^0, \frac{\infty}{\infty},\frac{0}{0}$...

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In any number system where one class of numbers (the infinite) is bigger than all the numbers in another (the finite) there will be should usually be more than 1 "infinity". For example, in the hyper-reals, if $H$ is infinite, $2H, \frac H2, H+1,$ and $H-1$ are all infinite.

$H-2H=-H$ which is negative infinite.

$H - \frac H 2 = H/2 $ which is positive infinite.

$H-(H+1)=-1$ which is negative finite.

$H-(H-1)=1$ which is positive finite.

$H-H=0$ which is zero.

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$\quad\quad\quad\quad\quad\quad\quad\quad\quad (\infty - \infty) = \infty(1 - 1) =\, $$\infty\cdot0$

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    $\begingroup$ This assumes that ∞−∞ is well-defined without stating what properties we require of it. $\endgroup$ – PJTraill Nov 14 '18 at 20:54
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Infinity does not lead to contradiction, but we can not conceptualize $\infty$ as a number. The issue is similar to, what is $ + - \times$, where $-$ is the operator. The answer is undefined, because $+$ and $\times$ are not the kind of mathematical objects that $-$ acts upon. (We could of course define something, but then we are no longer talking about the usual operations, we are changing the meaning of the symbols.)

We can not conceptualize $\infty$ as a number to a large part because $\infty - \infty$ does not have a good interpretations. In particular for any finite set, the size of the set, its cardinality, is the same as the place of its last member, its ordinal. So if I have a concrete set $\{a, 57, \text{apple}\}$, then one of the members is at the place $3$, because there are $3$ members of the set. That is no longer true for infinity.

This is shown by the Ross–Littlewood paradox, you receive $10$ coins at each time step, and then you give a single coin back. The question is, how many coins do you have after infinite many steps. The problem is called a paradox because it has a simple and intuitive answer, unfortunately not many people agree which answer is simple and intuitive.

The first answer is, you have $\infty$ coins, which is a notational shorthand for, the total number of coins in your possession growth without limit. Proponents of this answer parse the problem in terms of total value where individual coins loose their identity. One could interpret that as taking the ten coins to the bank, then wiring the value of a single coin back. The bank updates your balance and the operation at a time step $t$ is $b_{t+1}=b_{t} + 10 - 1$.

The other answer is $0$, because each coin only spends a finite time in your possession and so after an infinite amount of steps you gave each and every coin back. In this case, one puts the coin into a line, one by one, adds $10$ coins to the end of the line, and gives $1$ coin back from the front of the line. That means that the $n$-th coin spends the timesteps $n/10$ until $n$ in your possession, and you give it away in a finite amount of time. (You get the $65$-th coin at time step $6$ and give it away during the time step $65$.) In "the end," infinity does not have an end" you have given all coins away and so the balance is $0$.

The difference between the two scenarios is, in the second the order of the coins is preserved, in the first the coins are piled and one only looks at the total value of the coins. And if we modify the second scenario a little bit, we end up with different answers. If you give back always the last coin you received, then you keep coins $1,\dots,9$ from the first step, coins $11, \dots, 19$ from the second and so on and you get to keep an infinite amount. (I use the order of received here to number the coins, not the place the coins lie in at the end.) If you take the ten coins, put $1$ on a pile, and you put $9$ in a line, and you do that for the first $n$ steps and after that you get lazy and just put all the coins into the line, then you will give all the coins in the line away, but you end up with a pile of $n$ coins.

These different procedures could all be plausibly called $\infty - \infty$, and we just constructed ways to parse that expression in a way to get any answer between $0$ and $\infty$. This is the issue with infinity, we have to be precise how to conceptualize $\infty$ and we can not just treat $\infty$ like a kind of number.

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infinity minus infinity is not defined since infinity itself is not defined

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    $\begingroup$ Welcome to math.SE! I'm afraid this answer is incorrect; there are many valid ways of defining infinity. Look at any of the answers above for a correct explanation of the issue. Please improve your answer, or it may be deleted. $\endgroup$ – Zev Chonoles Aug 30 '11 at 21:51
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    $\begingroup$ Will it really be deleted just for being incorrect? $\endgroup$ – Bonanza Aug 30 '11 at 23:37
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    $\begingroup$ It's not even wrong.... $\endgroup$ – The Chaz 2.0 Aug 30 '11 at 23:53
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    $\begingroup$ "Infinite" as a modifier is defined in various contexts in various ways. I would be surprised to see bare "infinity" defined in a mathematical text. Maybe in a book on mathematical theology. One could argue that the above answer is the only right one. $\endgroup$ – André Nicolas Aug 30 '11 at 23:59
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    $\begingroup$ This should not be deleted. André has a very good point. $\endgroup$ – Jonas Meyer Aug 31 '11 at 2:14

protected by Zev Chonoles Aug 30 '11 at 23:16

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