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0^0

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Re: 0^0

#201

Post by doetoe »

mathgrad wrote: post #114
rickhanlonii wrote: post #86

It's very important to note here that 0^0=1 is a shorthand and not a truth . Mathematicians are absolutely not stating that they have proven, or that it is true, that 0^0=1. It is a definition, not a claim of equality. They're not saying "0^0 is 1" in the sense that they say "1+1 is 2" or "0.999... is 1". They're saying "we define 0^0 to be 1". The difference is more than just pedantry, it strikes at the core of why…

This is not true. A very natural way to define a^b is as the number of functions for a set with "b" elements to a set with "a" elements. In this case there is exactly one function from the empty set to the empty set, and we have proven 0^0 = 1. This is no different than defining addition and then proving 1+1 = 2.

I agree that your definition is a very natural one, but I think your conclusion is mistaken. Defining a function in the same, set-theoretical terms (http://en.wikipedia.org/wiki/Implementation_of_mathematics_i...), a function is a subset of the cartesian product, and the Cartesian product of the empty set with itself is empty.

EDIT: I realize that you are right. The Cartesian product of the empty set with itself is empty, but it does have a subset, namely the empty set, which happens to be a function.

I'm not so sure this settles anything though. There is exactly one function 0 -> a for every non-empty set a and there is no function b -> 0 for any non-empty set b. Now you have to decide how to define a function to treat the case 0 -> 0. I'm no expert in set theory, but I would think this can be defined to be either without getting into any kind of trouble.

Re: 0^0

#202

Post by Perseids »

pavelrub wrote: post #197
Perseids wrote: post #192

Earlier quoted context omitted.

It depends on where you are coming from. If your context are ordinal numbers then you are right and a typical definition of 0 is {} and of s(x) is x∪{x}. But if you working with finite fields for example then you only have an addition operation. "Successor" does not make much sense there, since 1+1+…+1=0 for the right amount of additions (you are calculating modulo a prime). Since 0+1:=1 is trivial you usually start…

You don't usually define 2 at all when talking about abstract structures (it makes little sense to call the polynomial 2 as being "the 2" of the field of rational functions, for example). 2 is something that exists only in N, and talking about it in other structures makes sense only when you are referring to a ring homomorphism Z->F or something similar.

That definitely is a more clean way to look at the matter, yes. Nonetheless I've seen the definition I stated above a few times and the merit is that you do not have to take an implicit indirection every time you state something like "2≠0". And as a ring homomorphism ℤ->R for any ring R with identity element is already completely defined and in effect identical to the definition "2:=1+1…" for every positive whole number, it is really just a different way of formulating the same idea.

Re: 0^0

#203

Post by pavelrub »

tomp wrote: post #198
pavelrub wrote: post #193

Earlier quoted context omitted.

This is simply wrong. 1+1=2 is the same kind of statement as 2^1=2 - it follows from the definition of the operation and from a general rule (it would be absurd here to say that 2^1 is the definition of 2, or that 2^1=2 is part of the definition of ^). 0^0 is just a meaningless symbol unless given explicit definition. There is no popular set of axioms where 1+1=2 is taken as an axiom - it is always proven. The symbol…

Maybe your parent meant that "1 + 1 = 2" is the definition of "2", in the sense that we write the symbol "2" to denote the Peano successor of "1", the Peano successor of "0", which is not a successor.

Maybe. Formally this is inaccurate as the successor function is not the addition function, and one has to prove that S(S(0))=S(0)+S(0).

Edit: it should also be noted that this equality shouldn't seem "obvious": the name "successor" here is misleading, as it automatically links to our everyday understanding of a successor as being "the next number", in which case it is indeed obvious. But formally it's just a meaningless symbol that can be changed to other meaningless symbols by certain meaningless rules. Our everyday life association of (apple) -> S(0), (apple,apple)->S(S(0)) is then a kind of a model for those meaningless things, but from within the formal system - they have no meaning.

In general I also do not like the invocation of formal arguments here, I think it misses the point. The bigger issue is that for us humans there is an obvious difference in kind between 1+1=2 and 0^0=1 - this should be obvious. Trying to hide this difference by appealing to different kinds of language trickery and formalism is an indication of doing something wrong, not of discovering anything new. The difference is still there, and we should understand why it's there, not force it to disappear (compare to the claim that every human act is egoistic in nature - what information does it really give us about the human nature?).

Re: 0^0

#204

Post by pavelrub »

thomasahle wrote: post #200
pavelrub wrote: post #193

Earlier quoted context omitted.

This is simply wrong. 1+1=2 is the same kind of statement as 2^1=2 - it follows from the definition of the operation and from a general rule (it would be absurd here to say that 2^1 is the definition of 2, or that 2^1=2 is part of the definition of ^). 0^0 is just a meaningless symbol unless given explicit definition. There is no popular set of axioms where 1+1=2 is taken as an axiom - it is always proven. The symbol…

What do you think the definition of ^ is then? The ^ operator is defined as: 0^0 = 1 x^y = exp(ylog(x)) if x != 0 0^y = 0 if y != 0 Or similarly set theoretically.

Whatever it is, 2^1=2 is not part of it - it follows from it.

Re: 0^0

#205

Post by Aardwolf »

thomasahle wrote: post #200
pavelrub wrote: post #193

Earlier quoted context omitted.

This is simply wrong. 1+1=2 is the same kind of statement as 2^1=2 - it follows from the definition of the operation and from a general rule (it would be absurd here to say that 2^1 is the definition of 2, or that 2^1=2 is part of the definition of ^). 0^0 is just a meaningless symbol unless given explicit definition. There is no popular set of axioms where 1+1=2 is taken as an axiom - it is always proven. The symbol…

What do you think the definition of ^ is then? The ^ operator is defined as: 0^0 = 1 x^y = exp(ylog(x)) if x != 0 0^y = 0 if y != 0 Or similarly set theoretically.

Isn't that a circular definition, since, exp is e^x?

Re: 0^0

#206

Post by sireat »

dvanduzer wrote: post #109
baddox wrote: post #91

Earlier quoted context omitted.

I don't think the fifth axiom is considered particularly mysterious anymore. The traditional fifth axiom clearly isn't a logical result of the first four, since it can be replaced with other parallel postulates to yield non-Euclidean geometries which are themselves perfectly workable and consistent. In fact, that section of the Wikipedia article notes that Beltrami proved the independence of the parallel postulate.

> clearly It's mysterious because it is only clear in retrospect. The fact that you can negate the parallel postulate and get a system that is still self-consistent is incredibly mysterious. This is what sets mathematics apart. We "make it up" like other human accomplishments, and yet we can't actually just make it up. If we arbitrarily defined 0^0 = π, we'd just be speaking nonsense.

We are clearly into the http://en.wikipedia.org/wiki/The_Unreasonable_Effectiveness_... territory and why most mathematicians seem to be Platonists.

Re: 0^0

#207

Post by vog »

I find it a bit disappointing that the article considers only limits that would justify "0^0 = 0" and "0^0 = 1". In fact, the termin "0^0" as a limit can reach any value - the same way as "0/0" can reach any value.

Im stressing this because back in school, I really thought that although 0^0 is undetermined, it can only reach exactly 0 or 1, but nothing else. This is of course wrong, but no teacher was able to tell me why. For some time I even thought I found a new theorem and tried to prove it. Later, I told some math guru about this, he thought about a minute, and told me two functions f(x) and g(x) whose limit is each 0, but for which the limit of f(x)/g(x) is 2 (or any other value, if you adjust f(x) and g(x) accordingly).

Having said that, in most cases "0^0 = 1" is a useful convention, especially in a purely algebraic context when polynomials are involved.

Re: 0^0

#208

Post by pavelrub »

Perseids wrote: post #202
pavelrub wrote: post #197

Earlier quoted context omitted.

You don't usually define 2 at all when talking about abstract structures (it makes little sense to call the polynomial 2 as being "the 2" of the field of rational functions, for example). 2 is something that exists only in N, and talking about it in other structures makes sense only when you are referring to a ring homomorphism Z->F or something similar.

That definitely is a more clean way to look at the matter, yes. Nonetheless I've seen the definition I stated above a few times and the merit is that you do not have to take an implicit indirection every time you state something like "2≠0". And as a ring homomorphism ℤ->R for any ring R with identity element is already completely defined and in effect identical to the definition "2:=1+1…" for every positive whole num…

My point is that when talking about the definition of 2, it is enough to restrict ourselves to the natural numbers - since that is where 2 is coming from. If we then want to extend this symbol to other places - the meaning of such an extension will be given by a map, not by requiring a different definition. That is - there is only one 2, everything else is ψ(2) for the homomorphism ψ:Z->R.

Re: 0^0

#209

Post by stephencanon »

sold wrote: post #186
stephencanon wrote: post #78

The real problem here is that x^y is a single shorthand which refers to a few fundamentally different mathematical concepts (which happen to have significant overlap with each other). First, it refers to a function f:C x N --> C, defined in terms of repeated multiplication. f(x,0) is 1 for all x != 0, and so we adopt the convention that f(0,0) is also 1. But it also refers to a function g:C x C --> C, defined as g(x,…

I'd like to add x^y is also used other contexts - e.g. for cardinal numbers and ordinal numbers.

Absolutely; I certainly didn't intend that anyone take my list as exhaustive.

Re: 0^0

#210

Post by SamReidHughes »

bumbledraven wrote: post #176

Although standard mathematica notation glosses over the difference, it's important to distinguish between the following two versions of the power operation, which I'll call pow1 and pow2: pow1(x, y) = x^y where y is any integer pow2(x, y) = x^y where y is any real The value of 0^0 is 1 for (pow-a) and 0 for (pow-b).

But lim x->0 (x^x) = 1. Why should 0^0 be 0 then?
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