# zero raised to itself

by Shinobiku
Tags: raised
 P: 5 I was asked in another forum what zero raised to itself was. Any other number raised to zero equals one, but zero raised to any number equals 0. But what about zero itself. I already tried explaining it through thinking of zero as a limit of getting really small, but now this question has got me thinking. Any suggestions on how to think through this?
 PF Patron Sci Advisor Emeritus P: 10,400 It is indeterminate. - Warren
 P: 1,210 Hi Shinobiku, I think that to answer to your question, we have to examine the logic system itself. The most known logic system is the Boolean logic, which is based on 0 XOR 1 connective. Fuzzy logic is based on fading between 0 XOR 1. A non-Boolean logic is based on 0 AND 1. In Boolean or Fuzzy logic 0^0 is unknown because any base value is some cardinal X >= 0. Any cardinal X>0 is based on some non-empty set {x}, where cardinal X=0 is based on the empty set {}. So, what we have here is |{x}|^0 and |{}|^0. By Boolean and Fuzzy logic |{x}|^0=1, but |{}|^0 is not well-defined because it is based on no set's content. So, I think the answer to your question is beyond the limitations of Boolean and Fuzzy logic. What do you think? Organic
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## zero raised to itself

Organic: Since you asked "what do you think", I think you chattered and chattered and finally said YOU were unable to answer the question.

The question is very simply answered: 0 to the 0 power is "undetermined" (that's an adjective: noun form is "indeterminant").

Notice that is NOT the same as saying "undefined". 1/0 is undefined because "1/0= x" is the same as "1= 0*x" which is untrue for all possible x. 0/0 is "undetermined" because "0/0= x" is the same as "0= 0*x" which is TRUE for all possible x. Another way of saying that is if we have f(x)->1, g(x)->0, then lim f(x)/g(x) does not exist while there are functions f(x)->0, g(x)->0 such that lim f(x)/g(x) exists and can be any given number. Similarly, given any number, we can find functions f(x)->0, g(x)->0 such that lim f(x)g(x) exist and is equal to that number.
 P: 1,210 Hi HallsofIvy, Thank you for your corrections about "1=0*x"(which is UNTRUE for all possible x) and "0=0*x"(which is TRUE for all possible x). What i mean is this: any nonempty set's content {x} which is some singleton, can be used as some base for x^0=1 where 1 means base value (some singleton) exists. In the case of the empty-set our content is NOTHING and by Boolean logic NOTHING^0 is gibberish. By non-Boolean logic we can define another set's content, wich is not {} ,not {.) (singleton) and not a collection of singletons {....}. This object has the ability to simultaneously be in two different states like > AND = ,0 AND 1, closed AND opened, and so on. Therefore it exists between any distinguished singletons as a non-localized object {.___.}. There are exactly 0 singletons in this object therefore its base value = 0, but because it exists ( unlike the empty set content, then __^0=1 or |{__}|^0=1, exactly as x^0=1 and |{x}|^0=1 in Boolean or Fuzzy logic. Organic
 P: 5 Alright, here's what i've come up w/ before i read anyones replies, which all had something to say, but left me w/ everyone saying it's indeterminate. Let y = f(x)^g(x), take the ln of both sides ln(y) = ln(f(x)^g(x)) ln(y) = g(x)*ln(f(x)), now exponentiate both sides e^ln(y) = e^g(x) * e^ln(f(x)) y = f(x)*e^g(x), then take the limit as f(x)->0 and g(x)->0 y = 0*1 = 0 Can anyone see any flaws? Although I don't see anything wrong, it doesn't feel right to me. I believe this can only work if both f(x) and g(x) are finite polynomials. Any thoughts? EDIT: I just caught my own mistake (30 min. later) go back to the third line: ln(y) = g(x)*ln(f(x)), now take the limit as both f(x) and g(x)->0 I believe that ln(f(x)) approaches -infinity faster than g(x) approaches 0. If so, then ln(y) = -inf. but since this is a limit, then as ln(y) -> -inf., means that y -> 0 But this depends on the rate at which g(x)->0 and ln(f(x))-> -inf. Any thoughts?
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 ln(y) = g(x)*ln(f(x)), now exponentiate both sides e^ln(y) = e^g(x) * e^ln(f(x))
No, that's wrong. If ln(y)= g(x)*ln(f(x)) then
eln(y)= eg(x)*ln(f(x)
= eln(f(x)^g(x))

or y= f(x)g(x) which is exactly what you started with.
 P: 5 Plz notice the edit
 PF Patron Sci Advisor Emeritus P: 10,400 This thread was pointless after the second post. - Warren
 P: 5 I don't really like the stuff i came up w/, i've asked some ppl at my college, as well as tested it in mathematica using different methods and this is what i've come up w/: I'm just going to post here what i just posted in my other forum. I've talked w/ a friend of mine who is a math major (senior), and he likes to think of it as any number to the 0 power is the same as that number divided by itself, so a^1/a^1 = a^(1-1) = a^0 but in the case of zero, 0^1/0^1 is already indeterminate. That's one way to look at it. And then i went ahead and plugged a few things into mathematica: In[2]:= 0^0 Power::indet : Indeterminate expression 0.^0. encountered. Out[2]= Indeterminate In[3]:= Limit[x^x, x->0] Out[3]= 1 In[4]:= Table[{n,n^n},{n,.1,0,-.01}]//TableForm Out[4]//TableForm= 0.10 0.794328 0.09 0.805159 0.08 0.817047 0.07 0.830151 0.06 0.844674 0.05 0.860892 0.04 0.879189 0.03 0.900147 0.02 0.924742 0.01 0.954993 0.00 Indeterminate So, from those 3 methods, it seems that it is apporoaching 1 as a solution
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 Originally posted by Shinobiku I don't really like the stuff i came up w/, i've asked some ppl at my college, as well as tested it in mathematica using different methods and this is what i've come up w/:
0^0 IS INDETERMINATE. PERIOD. THERE IS NO QUESTION, NOR ANY ROOM FOR DISCUSSION.

- Warren
 PF Patron Sci Advisor Thanks Emeritus P: 38,400 Well, you're no fun at all![:D]
P: 364
 Originally posted by Shinobiku So, from those 3 methods, it seems that it is apporoaching 1 as a solution
Although I'm quite afraid that chroot will kill me for continuing this thread, I will say that the fact that limx-->af(x) exists does not imply that f(a) exists.
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 Originally posted by StephenPrivitera Although I'm quite afraid that chroot will kill me for continuing this thread
*BANG*

- Warren
 P: 5 Indeterminate does not mean there is no answer, it just implies that the current method is not suitable. Just because it is indeterminate does not imply there can be no discussion. If you don't like the topic, then there is no need for you to read it. Stephen: From what i know, when something doesn't exist at a particular value, then for many purposes in mathematics, it is then appropriate to look at it's behavior in the limits that it approaches the value at which it does not exist. This said, the answer lies in what context the question is brought up, and what situation the solution would be used.
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