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0^0

Name: Anonymous 2007-09-08 23:37 ID:DQ/MKm+7

ONE WORD, 00 = ?, /sci/ over

Name: Anonymous 2007-09-09 0:02 ID:Heaven

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

Name: RedCream 2007-09-09 0:15 ID:SbZOeRVK

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

Name: Anonymous 2007-09-09 2:37 ID:pm+SizGU

0^0 is undefined.

Name: Anonymous 2007-09-09 4:19 ID:mNA1pyiu

00 = 1

Name: Anonymous 2007-09-09 5:07 ID:Heaven

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

Name: Anonymous 2007-09-09 5:26 ID:ifu2abK3

>>4

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

Name: Anonymous 2007-09-09 9:31 ID:9N03+9Gj

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

Name: Anonymous 2007-09-09 11:20 ID:ElBosyop

>>5

Fail.

Name: Anonymous 2007-09-09 11:59 ID:Ym7h/DUC

>>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.

Name: Anonymous 2007-09-09 14:30 ID:Cq2YfRC+

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.

Name: RedCream 2007-09-09 16:44 ID:QlN8U8zB

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.

Name: Anonymous 2007-09-09 16:59 ID:Heaven

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

Name: Anonymous 2007-09-09 17:06 ID:ejL9R1rN

>>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

Name: Anonymous 2007-09-09 18:24 ID:Cq2YfRC+

>>12

<<"how many zeros in zero">>

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

Name: Anonymous 2007-09-09 20:20 ID:i8iEItrg

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

Name: Anonymous 2007-09-09 20:21 ID:eSErl7ei

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.

Name: Anonymous 2007-09-09 22:15 ID:9hj5lYHj

>>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.

Name: Anonymous 2007-09-09 22:18 ID:9hj5lYHj

>>18 form = approach.

Name: Anonymous 2007-09-09 22:25 ID:Heaven

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

IDENTITIES

Name: RedCream 2007-09-10 6:02 ID:nx3Y3AX0

>>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.

Name: 4tran 2007-09-10 6:46 ID:Heaven

>>21
Please read >>15

Name: !8SqUCOTuGQ 2007-09-10 7:34 ID:zRRF1RBR

>>22

I was about to say so myself.

Name: Anonymous 2007-09-10 15:21 ID:HD3vKr9F

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.

Name: Anonymous 2007-09-10 15:27 ID:kZO/cjZR

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.

Name: Anonymous 2007-09-10 18:45 ID:Heaven

>>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.

Name: Anonymous 2007-09-10 20:02 ID:ciEDoOcI

>>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.

Name: RedCream 2007-09-11 3:57 ID:EhMU+V3q

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

Name: Anonymous 2007-09-11 11:12 ID:Heaven

>>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!

Name: Anonymous 2007-09-11 18:13 ID:UXdyeKkm

I fucking hate 0.

Name: Anonymous 2007-09-11 23:41 ID:QrqYFhSR

>>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.

Name: Anonymous 2007-09-11 23:44 ID:Heaven

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

Name: Anonymous 2007-09-12 10:24 ID:ct/YQpDz

>>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!!!

Name: Anonymous 2007-09-12 10:26 ID:ct/YQpDz

>>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.

Name: Anonymous 2007-09-12 10:44 ID:ct/YQpDz

>>32

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

Name: Anonymous 2007-09-13 1:24 ID:1q6N7lfg

>>29 wins

Name: Anonymous 2007-09-13 1:46 ID:Heaven

>>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.

Name: Anonymous 2007-09-13 1:59 ID:Heaven

>>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.

Name: Anonymous 2007-09-13 11:17 ID:PYFgqIap

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

Name: Anonymous 2007-09-13 11:42 ID:Heaven

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

Name: Anonymous 2007-09-13 11:50 ID:CYPhArhr

>>40
booyah, you told him, Vinny

Name: Anonymous 2007-09-13 11:56 ID:1X2+qmvX

>>37

Frege and Russell took 0 to be { {} }, but von Neumann took 0 to be {} (and n+1 to be n U {n}), and this is quite standard in many contexts.  Your assertion about "almost every contemporary text" is just plain wrong; you obviously need to read more contemporary texts.

See
http://en.wikipedia.org/w/index.php?title=Ordinal_number
for more about the von Neumann 0 = {} thing.


Name: Anonymous 2007-09-13 12:00 ID:1X2+qmvX

>>38

You blew it: you forgot to consider that K could be {}, which satisfies your requirement. So the set of such K might be { {} }, and so have one element.

And in fact when S is empty, the set of functions from {} -> S is { {} } and there is this one function.

See:

http://en.wikipedia.org/w/index.php?title=Empty_function

Name: Anonymous 2007-09-13 12:10 ID:Heaven

>>42
As long as you're going to drag the conversation down to the realm of citing wikipedia, http://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers

Name: Anonymous 2007-09-13 12:29 ID:R9BgNa/N

>>37

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.

I can't say it any better than wikipedia does "This definition works in naive set theory, type theory, and in set theories that grew out of type theory, such as New Foundations and related systems. But it does not work in the axiomatic set theory ZFC and related systems, because in such systems the equivalence classes under equinumerosity are "too large" to be sets. For that matter, there is no universal set V in ZFC, under pain of the Russell paradox."

Or to prarphrase: it's informal; it's deprecated; it doesn't really work.

I would like you to name a couple of contemporary texts which use this definition and state that it is formal, because I don't believe you.

If you don't believe wikipedia and its citations then I direct you to http://mathworld.wolfram.com/OrdinalNumber.html .

Name: Anonymous 2007-09-13 12:35 ID:R9BgNa/N

>>38

You're logic breaks down the the case where S=E (ie. functions from the empty set to the empty set).

In this case the set ExS=E. There is one subset then of ExS, which is E. So let's set K=E and see if it is a map from E to S.

the condition you give is that "if (a,b) appears in K, then no other element of K is (a,c) for any c not equal to b" or in other words that "it is NOT TRUE that (there is some (a,b) and (a,c) both in K where b is not equal to c)".

Well, as K is empty, the bracketed statement is false, and so the entire statement is true; hence, by your own definition (which is indeed correct), the empty set IS a function from E to E.

Which is what I said in the first place.

Actually generally, by your definition again, there is ALWAYS exactly one function from the empty set to another set (the empty function). QED.

Name: Anonymous 2007-09-13 12:37 ID:R9BgNa/N

>>44

You're a dickhead. Everything you say is wrong. And you're sage bombing this thread because you can't handle it. GTFO and die.

Name: Anonymous 2007-09-13 14:58 ID:Heaven

>>47
Well gosh, I can't argue with that kind of logic!

Name: Anonymous 2007-09-13 15:34 ID:XIxmaZVw

X^0 is defined mathematically as X^y/X^y = X/X. Thus 0^0 = 0/0, but you cannot divide by 0, so 0^0 is undefined.

Name: Anonymous 2007-09-15 0:36 ID:dNZPpenp

let g(0)=f(0)=0
lim(x->0) f(x)^g(x)=
lim(x->0) e^(ln(f(x)^g(x)))=
e^(lim(x->0) g(x)*ln(f(x))=
e^(lim(x->0) g(x)/(1/ln(f(x)))

L'hopital can take it from there.

Name: Anonymous 2007-09-16 8:24 ID:viGiB6Vx

>>49

No, indeterminate, not undefined.

Name: Anonymous 2007-09-16 9:14 ID:bbXnekRD


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Name: Anonymous 2007-09-16 14:11 ID:xUqXlg3a

isnt anythign to the 0 power automatically 1?

Name: Anonymous 2007-09-16 15:11 ID:jFmKVW87

Who honestly cares how you define the naturals.  All you need is one of these babies:
http://en.wikipedia.org/wiki/Natural_number_object

In a NNO 0^0 = 1

Name: Anonymous 2007-09-16 23:38 ID:Heaven

00=0

Name: 4tran 2007-09-17 4:00 ID:Heaven

>>50 wins.

Name: Anonymous 2007-09-17 16:19 ID:+dNOhqg+

I have a science project. I need to take a coffee can, and make it so that if the can is pushed forward it will roll forward and then roll backward to the starting point. what  do you suggest?

Name: Skordocott 2007-09-17 19:21 ID:z4NUvuX+

Tape a weight to the inside wall of the can.

Name: Anonymous 2007-09-17 21:01 ID:Heaven

i suggest pushing it up a hill.

Name: Anonymous 2007-09-19 10:36 ID:QsaH9aRG

ITT people with no real advanced maths education mix up really basic shit.

Name: Anonymous 2007-09-19 11:12 ID:FGCA/x60

GET INSIDE THE CAN

Name: Anonymous 2007-09-19 11:16 ID:FGCA/x60

SET YOUR HAIR ON FIRE

Name: Anonymous 2007-09-19 11:16 ID:FGCA/x60

IT WON`T MAKE THE CAN MOVE, BUT IT WILL BE REALLY COOL

Name: Anonymous 2007-09-19 11:24 ID:FGCA/x60

FIREWORKS, GASOLINE AND BLOODFLIES.
PUT THEM IN THE CAN.

Name: Anonymous 2007-09-20 10:28 ID:La6n9QaK

ITT we divide by zero

Name: Anonymous 2007-09-20 14:26 ID:VclrQsH6

By using Carls therm of convinces you will get 0^0=1

Name: anon 2007-09-20 19:15 ID:4VM8EAVq

WTF????

Name: Anonymous 2007-09-22 8:10 ID:/Ck9Aicf

>>28
Massive fail. See >>29.

Name: Anonymous 2007-09-22 21:56 ID:56spn8T2

Google says so. It must be true.

http://www.google.no/search?q=0%5E0&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

Name: Anonymous 2007-09-22 22:04 ID:rBihE97x

can you apply L'Hospital's rule here or is this not one of the indeterminate forms?

Name: 4tran 2007-09-23 1:12 ID:Heaven

>>70
see >>50

Name: Anonymous 2007-09-25 2:08 ID:vx8PYMVr

>>50

Fail.

>>56

Fail by proxy.

Name: 4tran 2007-09-28 20:27 ID:Heaven

>>72
How does it fail?
Don't change these.
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