What is the result of $\infty - \infty$?

Question

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

Answer 1

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.

Answer 2

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 $\infty - \infty$.

Answer 3

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.

Answer 4

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!

Answer 5

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 $\infty \cdot 0, 1^\infty, \infty^0, \frac{\infty}{\infty},\frac{0}{0}$...

Answer 6

$$ \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2 + y^2)^2} \, dy\,dx \text{ is actually different from }\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2 + y^2)^2} \, dx\,dy. $$ 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

Answer 7

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.

Answer 8

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.

Answer 9

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: $$0.\infty=$$

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

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

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

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

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

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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.