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0^0

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

ONE WORD, 00 = ?, /sci/ over

Name: Anonymous 2007-09-13 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

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?

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