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- October 5th 2012, 07:50 AM #1

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## 0^0

I've recently been involved in a discussion regarding the value of 0

^{0}in which I confidently asserted that 0^{0}is undefined, largely because x^{0}is always 1, while 0^{x}is always 0, leading to a contradiction at 0^{0}.

However, I've since seen discussions of the subject that say a consensus has "recently" been reached around 0^{0 }= 1, seemingly on utility grounds, with x^{0}being a much more important result than 0^{x}.

I'd be interested to hear some expert views on what the current thinking is. Thanks a lot!

- October 5th 2012, 08:44 AM #2
## Re: 0^0

Here is a webpage dedicated to that. Exponentiation - Wikipedia, the free encyclopedia

- October 5th 2012, 11:07 AM #3

- October 5th 2012, 06:14 PM #4
## Re: 0^0

It's not a good enough proof for me. The rule comes from the index law , with . However, . Wouldn't mean "How many 0's go into 0"? Surely there are an infinite number of them that can go into 0, since 0 + 0 + 0 + ... + 0 (infinity times) = 0. This gives THREE possible answers for what is!

So regardless of what the limit of is, it will NEVER be able to be defined at that point, and so it is definitely not enough to say that .

- October 5th 2012, 09:30 PM #5

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## Re: 0^0

in general, if and it may still be the case that:

takes on different values depending on what functions we choose for f and g.

for example, if f(t) = g(t) = t, then (the right hand limit doesn't exist, as h(t) is undefined, if h is to be real-valued).

however, if , and g(t) = at, then .

a lot depends on what one MEANS by a^{b}: is it a function of a, a function of b, or a function of both? what domain for a and b are we talking about, and how do we interpret what the expression a^{b}is?

for example, if one takes a^{b}to be one is faced with the quandry of what should be.

this is related to "what is infinity"? not all infinite quantities are "the same", we can't really treat ∞ as a real number without sacrificing some of the arithmetic structure of the real numbers.

that said, there are some reasons for treating 0^{0}as being 1.

one of the reasons i find most compelling is that:

for finite (non-empty sets) A and B, |B|^{|A|}represents the number of functions from A to B. if we count the empty set as a finite set, then 0^{0}should represent the number of functions from the empty set TO the empty set, and there is only one such function: the empty function (a "blank sheet of paper as the graph").

another compelling reason is that the power rule:

is not valid for n = 1 at x = 0, unless 0^{0}= 1.

**********

in many situations, "0 is a special case", and deciding how to deal with that special case is context-sensitive. it makes little sense to say "0^{0}= 1", without supplying more information about what kinds of things 0 and 1 are. things they might be:

natural numbers (as in combinatorics)

rational numbers

real numbers

complex numbers

polynomial functions

real-valued constant functions

values of other real-valued functions

bottom and top elements of a boolean lattice

people new to the subtleties of advanced mathematics often think mathematical ideas have some some of "eternal epistemological state", perfect and unchanging. "1" IS a certain something, and nothing else, forever and ever. the truth is, mathematical concepts are a bit more flexible than that: the same string of symbols may mean entirely two different things in different arenas. this is why mathematical definitions can be so intricate: we're trying to define the "scope of discussion" (and what rules apply).

- October 6th 2012, 12:02 AM #6

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- October 6th 2012, 02:20 AM #7

- October 6th 2012, 10:36 AM #8

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## Re: 0^0

Its a matter of definition and context... I don't think there's a real situation in math where the result of a calculation is gonna be 0^0, leading you to say "Well gee, thats undefined". There's gonna be a limit or something in which case there would be an answer.

BTW. Intuitively, I would still say if we HAD to choose a value for 0^0, it'd be 1... even the case of 0^x supports it. 0 raised to a positive power is 0, but remember 0 raised to a negative power is infinity.. so 0^0 is the bridge in between.

Still, saying 0^0 is equal to 1 would be very misleading in certain situations because the function 0^x would be defined but discontinuous at that point.

- October 6th 2012, 12:52 PM #9

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## Re: 0^0

to amplify:

there is NO choice of value for f(x,y) = x^{y}, that makes this function continuous at (0,0).

(interesting historical side-note: Augustin Cauchy (yes...THAT Cauchy) was of the mind, and had even published that 0^{0}= 1, until an anonymous submission (merely signed "S" i believe) exhibited a counterexample (two functions as in my earlier post), and he reversed his position. so this "consensus" the OP speaks of is neither "recent" nor actual. such musings border on the philosophical...there really isn't a "right" answer, but there are several wrong ones).