0^0 • Hacker News Board

Re: 0^0

#211

Post by stephencanon » Thu, Apr 03, 2014, 9:47 AM UTC

Sniffnoy wrote: post #87

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,…

But if you define 0^0=1 in general, it doesn't cause a problem here -- that definition never disagrees with x^y=exp(ylog(x)), it just defines it at the point 0^0, while the latter leaves it undefined. In other words, it's possible to make a common extension of the two; they don't actually give different values in any case. Of course, doing this makes exponentiation discontinuous at (0,0), but seeing as it already had…

Right, but when you're working with the function on C x C, it doesn't really add anything either to arbitrarily pick 0^0 = 1, except for consistency with that other function that happens to be written the same way.

I'm not really opposed to saying 0^0 = 1; it's the only reasonable choice if we're going to insist on using the same notation for these two functions, and I don't expect that's going to change.

Re: 0^0

#212

Post by stephencanon » Thu, Apr 03, 2014, 9:51 AM UTC

analog31 wrote: post #53

Perhaps a related question: How should it be defined in a math library for a programming language? Should it return 1, or throw an exception?

Math libraries (and language standards) should follow the guidance in IEEE-754 unless they have a very good reason not to do so.

"pow(x, +/-0) is 1 for any x (even a zero, quiet NaN, or infinity)."

"pown(x, 0) is 1 for any x (even a zero, quiet NaN, or infinity)."

- but -

"powr(+/-0, +/-0) signals the invalid operation [and returns NaN]."

"pown" refers to the function on R x N defined by repeated multiplication; "powr" is the function on R x R defined by exp(y log(x)). "pow" refers to the mental hodgepodge of the two that most people intend when they write x^y without really thinking about it.

Re: 0^0

#213

Post by taejo » Thu, Apr 03, 2014, 9:51 AM UTC

Aardwolf wrote: post #205

thomasahle wrote: post #200

Earlier quoted context omitted.

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?

If one defines x^y this way, one usually defines exp by a power series, and then defines e as the number such that e^x = exp(x).

Re: 0^0

#214

Post by arketyp » Thu, Apr 03, 2014, 10:20 AM UTC

pavelrub wrote: post #193

rntz wrote: post #105

Earlier quoted context omitted.

In a certain sense, "1+1 is 2" is also merely a definition, in the same sense that "0^0 is 1" is a definition. Addition can be formally defined in mathematics; we habitually omit this definition because it is tedious, and because addition is such an intuitive operation that we do not require a definition in order to reason about it. Much as the question "what if the parallel axiom didn't hold?" leads to alternative g…

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…

This is a good point. Your grandparent's original distinction was that 1+1=2, just as 2^1=2, follows from a definition while 0^0=1 is immediately defined. However, the parent's still valid point was clearly that even the plus operator is in the larger sense just as arbitrary as the definition of 0^0, concurring with the overall theme of this thread that mathematics ultimately gives no credibility to intuition.

Re: 0^0

#215

Post by thomasahle » Thu, Apr 03, 2014, 10:38 AM UTC

Aardwolf wrote: post #205

thomasahle wrote: post #200

Earlier quoted context omitted.

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?

We can just define exp as:

  exp(x) = 1 + x/1! + x*x/2! + x*x*x/3! + ...

Re: 0^0

#216

Post by thomasahle » Thu, Apr 03, 2014, 10:40 AM UTC

pavelrub wrote: post #204

thomasahle wrote: post #200

Earlier quoted context omitted.

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.

Uhm, anything that is part of a definition follows from that definition.

`A and B implies A' and so on.

In the end a function, in the mathematical sense, is just a long (possibly infinite) table of `from' and `to' values. We could argue that one definition of ^ is more or less complex than another, for some definition of complexity, but I'm not sure which would win, and if it would be useful.

Re: 0^0

#217

Post by philbarr » Thu, Apr 03, 2014, 10:41 AM UTC

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…

Can anyone recommend a really good book on mathematics that demonstrates the "beauty" of it? I'd love to really learn and understand about Fast Fourier Transforms and the like, but any book I have is from my old college days and is just so bloody tedious.

I used to really love maths when I was younger but it got beaten out of me by endless repetition and "now do Questions 1 - 50" tedium.

Now I get scared whenever I see the "summation" symbol.

Re: 0^0

#218

Post by ubercow13 » Thu, Apr 03, 2014, 11:37 AM UTC

einhverfr wrote: post #144

chacham15 wrote: post #121

Earlier quoted context omitted.

> If you do not accept it as true (an explicitly accept it as false), then you can prove all of Hyperbolic Geometry What is the equivalent in this analogy if you do not accept that 0^0=1 (i.e. accept that 0^0=0)?

I think you break algebra if you start treating y = x^2/x as distinct from y = x, and the both the former and x/x have the same problem with x=0. Edit: As has occurred to me, you might actually have a point, in that x/x may be a continuous function, but so is 0^x, which gives you a different limit. So it may actually determine whether in context it makes sense to treat 0^0 as x/x or whether it makes sense to treat it…

y=x^2/x and y=x aren't the same thing though. The first is undefined at x=0, the second is not. How does this break algebra?

Re: 0^0

#219

Post by Chattered » Thu, Apr 03, 2014, 11:42 AM UTC

j2kun wrote: post #151

mathgrad wrote: post #128

Earlier quoted context omitted.

There is a very natural definition of numbers as sets. We define 0 to be the empty set and we define the successor function by S(x) = {x} union x. Then the natural numbers are the smallest set containing 0 and closed under the successor operation. This is the standard way to define the natural numbers within ZFC set theory. This is admittedly very formal and not how the lay person thinks of natural numbers. However,…

Levying the "authority" of ZFC doesn't change the fact that your definition is still arbitrary. I'm certain you could pick a different definition of natural numbers within ZFC and get 0^0 = 0.

The given definition only tells you what natural numbers are; it doesn't directly tell you what 0^0 is. For that, the set theoretic definition of exponentiation was provided.

That definition is not arbitrary, but is an instance of the very general definition of exponentiation given in category theory. Accordingly, A^B is the set of maps from B to A, and for non-empty finite sets, the number of such maps is the number of elements in A raised to the number of elements in B. If you extend this to cover empty and infinite sets, you get the full definition of cardinal exponentiation, which has 0^0 = 1 and has other cute things like

2^(the cardinal of the natural numbers) = the cardinal of the reals.

There are a few obviously good properties about the given definition of natural numbers (the Von-Neumann ordinals), and one of them is that the size of each natural number is, well, that natural number.

    0 = {} and contains 0 elements.
    1 = {0} and contains 1 element.
    2 = {0,1} and contains 2 elements.
    3 = {0,1,2} and contains 3 elements.

And so on. For any definition of natural numbers which has this feature, you find that when you consider their category where morphisms are all the functions between them, then again, the exponential (in the category theoretic sense) is precisely the one expected in arithmetic, and 0^0 = 1.

Re: 0^0

#220

Post by a1a » Thu, Apr 03, 2014, 11:48 AM UTC

pavelrub wrote: post #193

rntz wrote: post #105

Earlier quoted context omitted.

In a certain sense, "1+1 is 2" is also merely a definition, in the same sense that "0^0 is 1" is a definition. Addition can be formally defined in mathematics; we habitually omit this definition because it is tedious, and because addition is such an intuitive operation that we do not require a definition in order to reason about it. Much as the question "what if the parallel axiom didn't hold?" leads to alternative g…

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…

>There is no popular set of axioms where 1+1=2 is taken as an axiom - it is always proven.

I would love to see a proof of 1+1=2. For example in the case of elliptic curves, as far as I know addition is simply an axiom. E.g. A+A'=0=inf , where inf is the point at infinity and A' is the reflection of A.

Math is not related to nature. Try to reason your way to matrix multiplication with apples: A * B != B * A