# Math Qestion (Zero, Infinity, etc.)
**URL:** https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645
**Category:** Factual Questions
**Created:** [June 22, 2003, 10:57pm UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645 "2003-06-22T22:57:04Z")
**Posts on this page:** 19
**Page:** 1
### Author:  [@Shiva418](https://boards.straightdope.com/u/Shiva418)
#### Post date: [June 22, 2003, 10:57pm UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/1 "2003-06-22T22:57:04Z")
I want to ask any mathematicians out there if there are any problems with this equation below. I know that 0^0 is usually considered nonsense, and that dividing by zere is a big no-no, but aside from those little points, is the algebra ok?
0^0 = 0^(n-n) = (0^n) / (0^n) = [(0^n) / 1] x [1 / (0^n)]
all of this then = 0 x oo [infinity symbol].
In case you’re wondering, this equation comes from an ontological essay by Aleister Crowley which can be found here: [http://www.hermetic.com/crowley/mwt/mwt\_05.html](http://www.hermetic.com/crowley/mwt/mwt_05.html)
The formula, which Crowley calls “The 0=2 Equation,” is supposed to be a mathematical explanation of a particular type of tantric theology/cosmology in which nirvana (which equals samsara), represented by 0^0, is seen as the interplay of the infinite and the infinitly small, as shown by the final 0 X oo.
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### Author:  [@quelquechose](https://boards.straightdope.com/u/quelquechose)
#### Post date: [June 22, 2003, 11:25pm UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/2 "2003-06-22T23:25:33Z")
First of all, 0^0 is defined, and it equals 1, just like n^0 where n is any real number. The same thing happens with 0! (0 factorial).
In your last step you equate 1/(0^n) with infinity. 1/0 is undefined, although lim n-\>0 of 1/n equals infinity. There are a number of proofs that 1=0 etc., and they are interesting to look at, but there’s always an error somewhere.
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### Author:  [@John\_Kentzel-Griffin](https://boards.straightdope.com/u/John_Kentzel-Griffin)
#### Post date: [June 22, 2003, 11:48pm UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/3 "2003-06-22T23:48:42Z")
0^0 does not equal 1. For non-zero n, n^0 equals 1, while for non-zero m, 0^m equals 0. 0^0 is not defined. 0! is well-defined and it equals 1.
Crowley is manipulating symbols for which no meaning is assigned. Using “0^0” in a proof is like using “+^/”; it has no meaning.
Aside from the fact that the derivation is nonsense from the very begining, is the algebra OK? I don’t really know how to answer that. Maybe the best answer to your question is: Mu.
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### Author:  [@monica](https://boards.straightdope.com/u/monica)
#### Post date: [June 23, 2003, 1:35am UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/4 "2003-06-23T01:35:56Z")
> [@](#):
>
> [1 / (0^n)]
Here’s your problem. The top part of the equation, obviously, will be 1. The bottom part will be 0 times many 0s. This, of course, will be zero. Therefore, you will have 1/0. Any number divided by zero is “undefined.”
Oh wow! I might have actually learned something in math class! :eek:
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### Author:  [@quelquechose](https://boards.straightdope.com/u/quelquechose)
#### Post date: [June 23, 2003, 2:00am UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/5 "2003-06-23T02:00:00Z")
> [@](#):
>
> 0^0 does not equal 1. For non-zero n, n^0 equals 1, while for non-zero m, 0^m equals 0. 0^0 is not defined.
I was always taught that it did equal 1. Windows calc, my TI-89 calculator, and all the math books I have seen it mentioned in agree that it is 1. Apparently (based on Google search) there is some debate as to whether or not it should be undefined, but I believe it is technically equal to one.
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### Author:  [@quelquechose](https://boards.straightdope.com/u/quelquechose)
#### Post date: [June 23, 2003, 2:30am UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/6 "2003-06-23T02:30:04Z")
I found a few old math books. I didn’t look very hard but I did find one that said 0^0 was undefined. :smack:
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### Author:  [@Exapno\_Mapcase](https://boards.straightdope.com/u/Exapno_Mapcase)
#### Post date: [June 23, 2003, 2:51am UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/7 "2003-06-23T02:51:12Z")
**shiva** all 418 of you should remember that in math or formal logic, if you start with a false (or nonsense) assumption, you can literally obtain **any conceivable answer** you desire.
If the original false assumption is made knowingly, then it is a deliberate attempt to deceive, nothing more.
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### Author:  [@Shiva418](https://boards.straightdope.com/u/Shiva418)
#### Post date: [June 23, 2003, 2:58am UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/8 "2003-06-23T02:58:10Z")
How’s about this. Don’t look at 0^0 as an integer. Look at it as pure zero. Why? Any number implies an index of 1 (1=1^1). The index can be understood to represent the number of spatial dimensions the variable is extended in. Thus, 0^0 COULD be used to represent zero extended in zero dimensions. I understand that from a strictly methematical p.o.v. there is a definite problem with beginning with 0^0, but if the above is taken into account, it kind of makes sense.
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### Author:  [@kunoichi](https://boards.straightdope.com/u/kunoichi)
#### Post date: [June 23, 2003, 2:58am UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/9 "2003-06-23T02:58:16Z")
[http://mathforum.org/library/drmath/view/55764.html](http://mathforum.org/library/drmath/view/55764.html)
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### Author:  [@mic84](https://boards.straightdope.com/u/mic84)
#### Post date: [June 23, 2003, 7:01am UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/10 "2003-06-23T07:01:14Z")
> [@](#):
>
> 0^0 = 0^(n-n) = (0^n) / (0^n) = [(0^n) / 1] x [1 / (0^n)]
>
> all of this then = 0 x oo [infinity symbol].
>
> …is the algebra ok?
No. 1/0 is -infinity in the same sense that it’s infinity.
Also, contrary to what the author of the essay thinks, 0x oo isn’t always an expression we can’t evaluate. For example (x) \* (1/x) when x goes to infinity can be written(well, not really, but following the logic of the essay) as oo x 0, yet it clearly equals 1.
(and, of course, 0^0, division by 0, etc. etc. are undefined anyway)
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### Author:  [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)
#### Post date: [June 23, 2003, 7:48am UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/11 "2003-06-23T07:48:59Z")
[Here’s](http://db.uwaterloo.ca/~alopez-o/math-faq/node40.html#SECTION00530000000000000000) a discussion from the sci.math FAQ. The author writes
> [@](#):
>
> Some people feel that giving a value to a function with an essential discontinuity at a point, such as _x_[sup]_y_[/sup] at (0,0), is an inelegant patch and should not be done. Others point out correctly that in mathematics, usefulness and consistency are very important, and that under these parameters 0[sup]0[/sup] = 1 is the natural choice.
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### Author:  [@Achernar](https://boards.straightdope.com/u/Achernar)
#### Post date: [June 23, 2003, 10:17am UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/12 "2003-06-23T10:17:34Z")
That link appears to have a mistake. It says that (according to Kahan) if lim f(x) = 0 and lim g(x) = 0, then lim f(x)[sup]g(x)[/sup] = 1. This is clearly wrong. In fact, the link goes on to say that Möbius made the same mistake. Am I missing something?
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### Author:  [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)
#### Post date: [June 23, 2003, 10:32am UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/13 "2003-06-23T10:32:28Z")
Kahan states that
> [@](#):
>
> 0.0[sup]0.0[/sup] should be 1, because if _f_(_x_),_g_(_x_) → 0 as x approaches some limit, and _f_(_x_) and _g_(_x_) are analytic functions, then _f_(_x_)[sup]_g_(_x_)[/sup] → 1.
The condition that _f_(_x_) and _g_(_x_) be analytic functions saves Kahan from error. (“counterexamples” like _f_(_x_) = _e_[sup]-1/_x_[/sup] and _g_(_x_) = _x_ don’t count because _e_[sup]-1/_x_[/sup] isn’t analytic at zero.)
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### Author:  [@Achernar](https://boards.straightdope.com/u/Achernar)
#### Post date: [June 23, 2003, 10:38am UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/14 "2003-06-23T10:38:44Z")
What about f(x) = 0 and g(x) = x?
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### Author:  [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)
#### Post date: [June 23, 2003, 11:35am UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/15 "2003-06-23T11:35:01Z")
You’re absolutely right, of course. The author of the FAQ has mad an omission.
Along with Kahan, the FAQ article also cites
Louis M. Rotando and Henry Korn.The Indeterminate Form 0[sup]0[/sup]. Mathematics Magazine,Vol. 50, No. 1 (January 1977), pp. 41-42.
If you have access to JSTOR, you can read this [here](http://links.jstor.org/sici?sici=0025-570X%28197701%2950%3A1%3C41%3ATIF%3E2.0.CO%3B2-J).
The upshot is that the FAQ author forgot to mention that _f_(_x_) and _g_(_x_) both have to not be the zero function. For the proof to go though, their Taylor expansions have to be at least degree 1.
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### Author:  [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)
#### Post date: [June 23, 2003, 12:38pm UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/16 "2003-06-23T12:38:55Z")
First, 0^0 = 1. You can say it is undefined, but that’s an error. Since 0^x = 0, for x different from 0, it follows that the exponential function of two variables x^y is discontinuous at (0,0), but there are no laws about discontinuous functions. It does mean that you cannot make an argument going from non-zero values to 0 since such arguments are usually based on continuity. Any sources that state the contrary are simply wrong.
As for the original post, that is pure unadulterated bilgewater. Although you sometimes see 1/0 = oo, that is meaningless since you cannot carry out any arithmetic with oo. There is a consistent arithmetic with infinitesimals and their reciprocals, which are infinitely large, but even so you cannot divide by 0.
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### Author:  [@Exapno\_Mapcase](https://boards.straightdope.com/u/Exapno_Mapcase)
#### Post date: [June 23, 2003, 2:36pm UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/17 "2003-06-23T14:36:52Z")
> [@](#):
>
> \*Originally posted by Shiva418 \*
> \*\*How’s about this. Don’t look at 0^0 as an integer. Look at it as pure zero. Why? Any number implies an index of 1 (1=1^1). The index can be understood to represent the number of spatial dimensions the variable is extended in. Thus, 0^0 COULD be used to represent zero extended in zero dimensions. I understand that from a strictly methematical p.o.v. there is a definite problem with beginning with 0^0, but if the above is taken into account, it kind of makes sense. \*\*
No matter how thin you slice it, it’s still baloney.
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [June 23, 2003, 2:44pm UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/18 "2003-06-23T14:44:42Z")
The best reason I know of to define 0[sup]0[/sup] as 1 is because it keeps the binomial theorem elegant. Otherwise, you need a special case to handle (a + 0)[sup]1[/sup].
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### Author:  [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)
#### Post date: [June 23, 2003, 9:58pm UTC](https://boards.straightdope.com/t/math-qestion-zero-infinity-etc/183645/19 "2003-06-23T21:58:57Z")
> [@](#):
>
> _Originally posted by **Hari Seldon** :_
> First, 0^0 = 1. You can say it is undefined, but that’s an error.
No, it is not an error. 0[sup]0[/sup] has been defined to be 1, at least be some people. The people who choose to define it this way do not claim that they can _prove_ that 0[sup]0[/sup] = 1 (so it doesn’t matter that the function _x[sup]y[/sup]_ is not continuous). They just claim that it is the most convenient definition, particularly for the reason that **ultrafilter** gave.