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Does an equation containing infinity which is not equal to 0 or infinity exist? My math education stopped at poorly understanding trig so don't kill me please.

OK so the question I meant to ask was if any 'operation'? using infinity as a 'factor'? will yield 0 or +- infinity. I was looking at it as if infinity is a black hole which gobbles up everything on its side of the equals, then the equals it self as well, then the other side provided its never x0. The only thing capable of rendering not infinity is to multiply (or divide right?) by 0 or minus infinity. From what I understand now the symbol for infinity represents the possibility of an ever increasing number that is not finite...not necessarily "infinity" or infinite. So... can it be said that no function\equation\operation\expression even contains an infinite number as it would brick said action?

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$\operatorname{arctan}$$(\hspace{-0.02 in}\scriptsize+\normalsize \infty \hspace{-0.03 in}) \: = \: \frac{\pi}2 \;\;\;\;$ – Ricky Demer Mar 13 '14 at 6:28
This is rather badly formulated. Equations don't equal something (they may be true or false), but expressions may do so. But in the most common kind of expressions like artitmetic ones, $\infty$ cannot occur because it is not in the domain of the operations; though some would say things like $\frac1\infty=0$, it is really an abuse of notation for expressing a limit statement. But of course, if one allows this, there is no good reason to forbid an expression like $\frac1\infty+1$ either. – Marc van Leeuwen Mar 13 '14 at 9:39
Marc's comment should be the answer. answer: 'not applicable' I should of said 'operation' not 'equation'. – Mazura Mar 13 '14 at 16:23
$min(1, \infty) = 1$ – Thomas Eding Mar 13 '14 at 19:33
$\vert[0,1] \cap [2,\infty)\vert=0$ does not involve a limit disguised as $\infty$. – Loki Clock Mar 13 '14 at 23:24
up vote 28 down vote accepted

Before I answer your question, let me first clear up what I think is a point of confusion. In formal mathematics, $\infty$ is not a number. The reason that mathematicians do not treat $\infty$ as a number is that if we did, we would reach some conclusions that are clearly wrong.

For instance, one property numbers have is that you can subtract the same number from both sides of an equation and the equation will still be true. For example, I can subtract $1$ from both sides of the equation $x+1=4$ to get $x=3$. On the other hand, if I treat $\infty$ like a regular number and I subtract $\infty$ from both sides of the "equation" $\infty + 1 = \infty$, I end up with $1=0$, which is clearly false.

Instead, mathematicians think of $\infty$ as a limit. Roughly, this means that if you want to "plug in" $\infty$ into a function, you plug in bigger and bigger numbers and see what happens in the long term. For example, we write $$\lim_{x\to\infty}\frac{1}{x}=0$$ to mean that "as you plug bigger and bigger numbers into the function $f(x)=1/x$, the function becomes arbitrarily close to zero." You should convince yourself that this particular limit is right. In some cases the limit is infinite; all this means is that, as you plug in bigger and bigger numbers into the function, the function becomes arbitrarily large. For example,

  • $\lim_{x\to\infty} x = \infty$.
  • $\lim_{x\to\infty} x^2 = \infty$.

To answer your question, pretty much anything can happen when $\infty$ is involved. Let's look at the two examples I just gave. Even though both functions $f(x) = x$ and $g(x) = x^2$ go to infinity as $x$ goes to infinity, the second one grows a lot faster. Case in point: $f(100)=100$ and $g(100)=10\,000$. In fact, $g(x)$ grows so much faster that the difference $g(x) - f(x)$ (remember that this is just $x^2-x$) also goes to infinity as $x$ goes to infinity. You can convince yourself of this by plugging in values. In symbols, $$ \lim_{x\to\infty} (x^2 - x) = \infty.$$ So informally speaking, it is possible that $\infty-\infty=\infty$!

If this result seems counter-intuitive to you, it is because you are thinking of the two infinities on the left hand side of the equation $\infty-\infty=\infty$ as the same $\infty$: in fact, they are different. The first $\infty$ comes from the function $g(x)=x^2$, and in some sense it is bigger than the $\infty$ from the function $f(x)=x$ since $x^2$ gets bigger a lot faster than $x$ does.

In any case, you can come up with other functions (that is to say, you can approach $\infty$ at different speeds) that make the following statements true:

  • $\infty-\infty$ can equal anything between $-\infty$ and $+\infty$.
  • $\infty/\infty$ can equal anything between $-\infty$ and $+\infty$.
  • $\infty^0$ can equal anything between $0$ and $+\infty$.

Finally, there can be cases where plugging in $\infty$ doesn't give you any answer at all. If you took trigonometry you're probably familiar with the sine function, whose graph oscillates back and forth, like a wave, between $-1$ and $+1$. (I tried to put a picture of the graph of sine here, but I couldn't get it to work since I'm new to this site. Just search "graph of sine" on Google images and you'll see what I mean.) If you plug in larger and larger numbers into $\sin(x)$, you won't approach any fixed number. So $\sin\infty$ does not exist.

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I was assuming ∞ = [1,+1,+1,+1...] then ∞+1=∞ would make sense. I was unaware of limits which I believe cheat the spirit of my question because of my lack of knowledge to word it correctly. Another way of what I meant is once you reach the 'limit' and have gone on to infinity you can no longer calculate an answer with out infinity or 0 coming out the other side of the equals. Or nothing can remove infinity from an equation except infinity or 0. So far as I understand it real infinity would break any equation making it unsolvable as it is an un-calculable quantity. – Mazura Mar 13 '14 at 5:13
Don't feel bad for not knowing about limits: it took people a few hundred years to come up with them even after calculus had been invented. I think you're right that as soon as you add infinity to an equation you can't really talk about solving it. – user134824 Mar 13 '14 at 5:25
@Mazura: But if you don't (want to) know about limits, you should not use $\infty$ in expressions at all; it makes no sense. Without notion of limit, you can only use "infinite" sensibly in the meaning of "not finite"; for instance it makes sense to say there are infinitely many prime numbers (the set of prime numbers is not finite); this does not involve any limit. But expressions, no! – Marc van Leeuwen Mar 13 '14 at 9:47
$\infty^0$ can’t possibly be anything but $1$, could it? In other words, no matter how quickly your function approaches infinity, raising the result of that function to the zeroth will always turn it into $1$, so no matter what you are using $\infty$ as shorthand for, $\infty^0$ is still going to be $1$, right? Is that a typo or am I missing something? – KRyan Mar 14 '14 at 1:05
@KRyan no, you can produce a limit that shows $\infty^0$ to be anything you want; for example, $$ \lim_{x\to\infty} x^{1/ln x} = \infty^0 = e$$ This is why such expressions are called "indeterminate forms"; there's not enough information in $\infty^0$ to determine a single value for it. – KutuluMike Mar 14 '14 at 1:19

Yes. There are lots of limits involving infinity where the answer isn't $\infty$ or $0$. For example:

$$\lim_{x\to\infty} \frac{2x-1}{x+3}=2$$

$$\lim_{x\to\infty} \left(1+\frac1x\right)^x=e$$

Do these address the question you were asking?

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If you want to be boring, you could also say $\lim \limits_{x \to \infty} \frac{\alpha x}{x} = \alpha$ – Soke Mar 13 '14 at 2:27
If it does I don't understand it. What is lim and the arrow mean. – Mazura Mar 13 '14 at 2:28
Yeah, the trouble is that infinity isn't a number. If you write something like "$1/\infty=0$", it doesn't technically make sense. The usual way of making sense of it is to take the expression $1/x$, then let $x$ get larger and larger, and see whether the expression gets closer to some particular value. In the case of $1/x$, it gets closer and closer to $0$, so we say that the "limit" of $1/x$ as "$x$ goes to $\infty$" is equal to $0$. – G Tony Jacobs Mar 13 '14 at 2:37
Operations performed with infinity itself are meaningless, because infinity is not a number. Mathematicians don't try to incorporate infinity into arithmetic operations, because it's not meaningful, unless you make it meaningful by talking about limits. – G Tony Jacobs Mar 13 '14 at 7:30
If you're working with limits, and doing actual math, then $0\cdot\infty$ can actually equal any real number, depending on how you're getting to $0$ and to $\infty$. – G Tony Jacobs Mar 13 '14 at 7:31

Certainly an equation like $1/\infty+r$ will equal $r$, so one can get well-defined finite values other than zero as well. But more commonly such an equation does not have a single well defined value at all: $\infty-\infty$ can be any real number or none.

For instance, we can run into an expression of the form $\infty-\infty$ in trying to evaluate $\lim_{n,k\to\infty} n-k$. So depending on how $n$ and $k$ grow relative to each other this "$\infty-\infty$" may behave in many ways: for instance if $k=n$ then we get the natural case $\infty-\infty=0$. On the other hand if $k=n+m$ then we get $\infty-\infty=1-m$! That's one way of saying that the general limit doesn't actually exist, which shows equations involving $\infty$ don't necessarily have a specific value. Another way of looking at this: if you think $\infty+1=\infty,$ then we should have $\infty-\infty=1$! So we really can't just do arithmetic with this symbol.

There are alternative number systems like hyperreal and surreal numbers in which something very similar to the symbol $\infty$ is actually a number, i.e. extensions of the real numbers with infinitely small nonzero and infinitely large elements. The simplest one has an element called $\varepsilon$ which is smaller than all positive reals but greater than $0$. Then $1/\varepsilon$ is greater than all the real numbers, and so is something like your $\infty$. But then to make these new symbols behave like numbers we must also admit things like $2\varepsilon$ and $-\varepsilon^2$, and so there's still no unique notion corresponding to $\infty$. Anyway, in such a system there are lots of equations involving $\infty$ that evaluate to finite nonzero numbers: for instance we no longer have $\infty+1=\infty,$ because the infinitely large numbers still obey the usual rules of arithmetic.

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@SanathDevalapurkar Not so. I'm using $\infty$ to symbolize a unbounded sequence of real numbers. Mazura, I've added an example of when $\infty-\infty\neq 0$. The basic issue is that $\infty$ is not a number, so one can't always treat it like one. – Kevin Carlson Mar 13 '14 at 2:24
@SanathDevalapurkar: Not true, at least in the extended reals. One has $1/\infty = 1/(-\infty) = 0$ but $1/0$ cannot be consistently defined (is it $\infty$ or $-\infty$?). Your comment is only correct in $\mathbb C_{\infty}$, which I doubt the OP intended. – MPW Mar 13 '14 at 2:50
@MPW Right - I was talking about $\mathbb{C}_\infty$. Thank you for clarifying that that was probably not what the OP intended. – Sanath K. Devalapurkar Mar 13 '14 at 2:51
Infinity is simply not a number, and you can't do arithmetic with it unless you define very carefully what everything is supposed to mean. If you try to work intuitively with infinity, it leads to contradictions easily. – G Tony Jacobs Mar 13 '14 at 7:36
@Mazura That's like saying that 1/5=5, because when you divide it into 5 parts, you have 5 pieces. Remember that division is the reverse of multiplication, so if you conclude that 1/∞=x, then it follows that x*∞=1. – Brilliand Mar 13 '14 at 22:17

The function that you are thinking about is a function with asymptotes. For example, $f(x)=\dfrac{1}{x}$ never reaches $x=0$. Similarly, other functions only exist within certain domains and ranges,depending on the function itself. So, to answer your question, yes, functions which neither pass through 0 or approach $\infty$ do exist. In fact, they are often in common use in math.

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You are expressing a common misunderstanding about the extended real number $+\infty$. It isn't really different from any other number in the sense that it simply represents a single quantity, not some mysterious concept of an object that is ever increasing without bound.

The only real difference between an ordinary real number and $+\infty$ is that $+\infty$ is not a point on the number line. Although we can consider a larger geometric object -- the extended real number line, which adds two endpoints to the ordinary number line.

The arithmetic of the extended real numbers isn't very complicated: it is mostly common sense, and the only trickiness is that all of the arithmetic operations pick up the same sorts of subtleties you see with division. e.g. $(+\infty) + (-\infty)$ is left undefined for similar reasons to the fact that $0/0$ is left undefined.

As an example of this trickiness, think about solving $x = 2x-1$. Normally, you would subtract $x$ from both sides to simplify the equation: but that's only allowed if subtraction is actually defined. And if you don't know that $x \neq \pm \infty$ then you don't know if subtraction is allowed! You need to split the problem into three cases: in case 1, we assume $x = +\infty$, in case 2 we assume $x = -\infty$, and in case 3, we assume $-\infty < x < +\infty$ and solve in the normal way. It turns out that $x = 2x-1$ actually has three solutions: $x = 1/2$, $x = +\infty$, and $x = -\infty$

A list of various arithmetic facts includes

Some examples of

  • $x \leq +\infty$ for all $x$
  • $x \geq -\infty$ for all $x$
  • $ x + (+\infty) = +\infty$ whenever $x \neq -\infty$
  • $ x + (-\infty) = -\infty$ whenever $x \neq +\infty$
  • $(-\infty) + (+\infty)$ is not defined
  • $x+y = y+x$ for all $x,y$, when defined
  • $(x+y)+z = x+(y+z) $ for all $x,y,z$, when defined
  • $ x \cdot (+\infty) = +\infty$ whenever $x > 0 $
  • $ x \cdot (+\infty) = -\infty$ whenever $x < 0 $
  • $ 0 \cdot (+\infty)$ is not defined
  • $x \cdot y = y \cdot x$ for all $x,y$, when defined
  • $(xy)z = x(yz) $ for all $x,y,z$, when defined
  • $x / (+\infty) = 0$ if $x \neq \pm \infty$
  • $(+\infty) / x = +\infty$ if $0 < x < +\infty$
  • $x/0$ is undefined

and so forth. You can compute the arithmetic of exended real numbers through limits if you like: e.g. if you failed to remember the addition tables that would say $(+\infty) + (+\infty) = +\infty$, then you could figure it out by computing

$$ (+\infty) + (+\infty) = \lim_{\substack{x \to +\infty \\ y \to +\infty}} x + y = (+\infty)$$

Of course, doing things the other way around -- using extended real numbers to help compute limits -- is somewhat more useful.

Don't make the mistake of confusing $+\infty$ with the concept of a variable that increases without bound -- instead, you should think of $+\infty$ as being the point on the number line that such a variable approaches.

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