0^0

ONE WORD, 0⁰ = ?, /sci/ over

[u][i]?[i] = 1[/u]

0⁰ = 0^1-1 = 0¹x0^-1 = 0¹/0¹ = 0/0 = 1

0^0 is undefined.

0⁰ = 1

Oh oh paradox, I guess we have to scrap our whole system of mathematics and start over, and this time get rid of that fucking repeating nines bullshit

>>4

Either that, or it's a representation of an indeterminate form, depending on the context it's used in.

>>3
RedCream, I like it when you're straight and to the point, and don't sound like a pompous asshole.  Good job.

>>5

Fail.

>>9
Not entirely. Apparently in computer science, 0⁰ = 1 by convention.
Everything on my computer says so, e.g. http://www.google.com/search?q=0%5E0
My two calculators refuse to play ball, though.

0^0 = 1 makes lots of maths a lot easier, so it's taken for granted, and included in the idea that any real number taken to the power of zero is 1.

The thing is, that at the basis, X⁰ asks this question:

"How many X are in X?"

Ref. X/X.  The answer is ONE.  So, 0/0 is "how many zeros in zero", to which the answer is ONE.

>>12
That is a really fucking stupid attempt to explain it.

>>12

No, no it doesn't. Not at all.

Powers are firstly defined for integers and rationals trivially, except that X^0 = 1 is X|=0. 0^0 is defined as indeterminate.

If we were to treat 0 as a real number, then X^0 = e^ln(X)*0  That's how real powers are generally defined. But ln(0) is undefined, so once again we get an indeterminate form

>>12

<<"how many zeros in zero">>

The answer is also 0, and 5, and over 9000.

How come RedCream is allowed to divide by zero in his "proof"?

As stated by >>3, 0^0 = 0/0

x = 0/0
0*(x) = (0/0)*0
0 = 0
Therefore 0^0 is all real numbers.

>>12 is an intuitive form, and doesn't fucking work at all. try explaining how exactly you can have sqrt2 apples from sqrt2(1 apple + 2 apples).

0^0 is indeterminate form but is usually taken to mean 1, since it makes binomial expansions nice, for one of many many other things.

>>17 is a nice non-rigorous demonstration of why 0/0 can be any number at all, since 0x = 0, for all real x.

>>18 form = approach.

0 added 0 times = 0
0 multiplied 0 times = 1

IDENTITIES

>>16
I didn't divide by zero in the OH NOES NEVAR DOO DAT!!11!1! sense.  By saying (0/0), I'm actually saying:

"How many zeros are in zero?"

The answer is obviously 1.

If instead we had (1/0) to examine, we'd find ourselves saying:

"How many zeros are in one?"

... to which we have no answer.  How many zeros are there in any real number?  Zero?  One?  Infinitely many?  Fuck!  That's why we leave it undefined.

But to ask about the special case of how many zeros are in zero?  Well, that's resolvable.  The answer is 1.

>>21
Please read >>15

>>22

I was about to say so myself.

There's some lee-way with 0^0.  It's considered 1 in some systems and undefined in others.  If you type 0^0 in Mathematica for instance, you'll get "undefined".  If you type 0^0 into Maple, you'll get 1.

>>11
>0^0 = 1 makes lots of maths a lot easier, so it's taken for granted, and included in the idea that any real number taken to the power of zero is 1.

It also makes a lot of math more difficult as well.  It's a double-edged sword.  That's why it's sometimes considered to be 1 and sometimes considered undefined.

Definitive Answer:

It depends which "0" you are talking about. If "0" is the Natural Number then 0:={} and 0^0:=#{all functions from 0 to 0}.

As 0={}, there is only one map from 0 to itself, ie. {} (ie. the function with dom = {} and range = {}). So in this case the answer is 0^0 = #{{}}=1 (because we want the size of the set of functions and there is only 1 function).

However, if we get to the real numbers, 0 is not defined in this way. In fact it's definition is not important. What is important is that 0^0:=exp(0 ln 0) but as ln 0 is not defined, 0^0 is, by definition, not defined. So 0^0 when 0 is a real number has no definition. (This definition stands up because exp x is not simply e^x; exp x is the series by definition.)

Confusion arises when people start talking about taking the limits of stuff like x/x as x->0. However, a limit of a function at a certain point is not necessarily the value of the function at that point. In fact for this to be true it is necessary and sufficient that the function be continuous. And by strict definition, if you cannot simply calculate the value of a function as it is written then it has no value, even if it may have a limit. For example x/x has no value at x=0 because 0/0 is not defined (this is because x/y is defined as the unique number z such that zy =x and it is obvious that no such number exists for x=y=0) whereas x/x as x->0 has limit 1.

>>25
Wow. First of all, the set-theoretical definition of the natural numbers identifies 0 as { {} }, not {}. 1 is identified with { {}, { {} } }. Your definition of exponentiation would give 0^1 = #{all functions from 0 to 1} = 2.

>>21

God, RedCream, do you ever stop spouting crap?

>The answer is obviously 1.

If it's so obvious, why is this thread here? The answer is that it's indeterminate. Cope with it.

>>27
How many zeros are in zero?  That answer is 1.  The question itself may be localized, but the local answer is still correct.

>>28
how many 4s are in 48?
4+4+4+4+4+4+4+4+4+4+4+4 = 48
that's twelve 4s!

how many 0s are in 0?
0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0 = 0
thats twenty-four 0s!

I fucking hate 0.

>>28

intuitive approaches will get you so far, and then they'll shit all over your face.

a^b, a > 0, as b --> 0, --> 1
0^c, as c --> 0, --> 0

enjoy, assfuck.

>>31 That should be a >= 0. ITT we discuss stupid bullshit, further trivialising /sci/.

>>26

25 here, everything you just said was wrong.

(quote) Wow. First of all, the set-theoretical definition of the natural numbers identifies 0 as { {} }, not {}. 1 is identified with { {}, { {} } }.

Answer: Firstly, 0 is defined as {} not {{}}. The point is that "0" (the cardinal number) is equivalent (ie. has the same "number of elements", intuitively speaking) as a set that we can say has 0 elements. So if what you said was true, the set {x}~{{}} (because the bijection x ~> {} allows this equivalence) so #{x}=0, which is nonsense. Look this up anywhere and it will tell you that 0 is defined as {}.

>(quote) Your definition of exponentiation would give 0^1 = #{all functions from 0 to 1} = 2.

Secondly, it is not _MY_ defintion of exponentiation: it is one the THE standard formal definition of exponentiation for natural numbers (and cardinal numbers in general).

I will clarify it. In this definition x^y = #{all functions from x to y}. Hence 0^1 = #{all functions from {{}} to {}} = #{} = 0 (there are no functions from {{}} to {}). In general, in fact, by this definition 0^y = 0 where y>0. As I showed before, when y=0, 0^y=1.

You swapped the variables (which is my fault for not clarifying the definition), so what you should have got is 1^0 = #{all functions from {} to {{}}.} = #{ the only one is the empty function, the function with domain and range = {}.} = 1.

You didn't get this answer because you had 0={{}} and 1={{}, {{}}.} but 1={{}} and 2={{}, {{}}.}. So what you actually were trying to work out is 2^1=2, which is right!!!

>>33

25 here again. "
In this definition x^y = #{all functions from x to y}." should be "In this definition x^y = #{all functions from y to x}."

Sorry about that.

>>32

you have no business complaining about other people trivialising /sci/ while posting something that is just completely wrong.

>>29 wins

>>33
Frege and Russell both defined 0 as { {} }. Furthermore, this is the definition used in almost every contemporary text. Please take set theory 101 before trying to have a conversation with the adults.

>>33
PS: There are no functions from the empty set to any set. If we denote the empty set as E and any other set as S, then the set of functions from E -> S is equivalent to the set { subsets K of ExS : if (a,b) appears in K, then no other element of K is (a,c) for any c}. In other words, the set of relations between E and S in which no element of E is related to more than one element of S. But ExS = E, and the set of subsets of E is E, and #(E) = 0, so obviously the number of subsets of ExS satisfying the necessary property is also 0.

>>38
combinatorics tells us when theres no way to do something theres only one way to do it.

>>39
Cardinality tells us that the set of functions from the empty set to any set has cardinality 0.