0^0 • Hacker News Board

Re: 0^0

#151

Post by j2kun » Thu, Apr 03, 2014, 3:42 AM UTC

mathgrad wrote: post #128

j2kun wrote: post #119

Earlier quoted context omitted.

You're just pushing the arbitrariness of defining things one step further to the definition of a number. It makes it no less arbitrary that you've defined it and force us to accept the definition to get to your conclusion.

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.

Re: 0^0

#152

Post by einhverfr » Thu, Apr 03, 2014, 3:43 AM UTC

j2kun wrote: post #147

einhverfr wrote: post #139

Earlier quoted context omitted.

I see it as somewhat complex but in the end I sort of agree that it is a definition. Let's start with a problem: 0/0 The problem with this is not that the equation itself is meaningless. the limit of n/x as x->0 is infinity for any positive real number, and negative infinity for any negative real number. In essence 0/0 ends up reducing to 0 * infinity, which isn't very helpful. I would argue that discontinuity in a f…

But what if you're in a context where you're not reasoning about continuous functions at all? Why would you have to be subject to reasoning that doesn't apply to your situation?

That's actually an interesting point. And actually it occurs to me that even in continuous equations you have a problem:

Does it matter if we are talking 0^x or x^0?

I would think that in the context of 0^x, you'd have a constant function of 0, but x^0 you'd have a constant function of 1. These have different limits as x -> 0.

I think you have just convinced me that 0^0 is undefined.

Edit: Ouch. 0^x can't be defined for a negative x, so that doesn't work. I am back to siding with 1.

Re: 0^0

#153

Post by dllthomas » Thu, Apr 03, 2014, 3:50 AM UTC

baddox wrote: post #94

dllthomas wrote: post #85

Earlier quoted context omitted.

Yes, but getting there steps out of the realm of "intuition" for me.

Yes, perhaps "intuition" isn't the best word. Formal limits certainly aren't "intuitive" to me, at least by one definition of the word. I suppose I used "intuitive" to mean "according to my mathematical understanding, ignoring the mathematics explicitly dealing with 0^0."

Sure. I was limiting it to things I could do in my head in tens of seconds.

Re: 0^0

#154

Post by rickhanlonii » Thu, Apr 03, 2014, 3:51 AM UTC

chacham15 wrote: post #121

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…

> 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)?

This case isn't as exciting as the parallel postulate case; you're not going to get a whole new branch of mathematics, just a change in a lot of formulas/definitions/notations.

The example in the article is the binomial theorem. Another example is the definition of Taylor Series which defines 0! and 0^0 as 1 so that the series come out cleaner.

For reference, we use Taylor Series to define e, sin, cos and there's a really nice proof for e^(ipi)+1=0 using the Taylor Series for these three.

Re: 0^0

#155

Post by ndeine » Thu, Apr 03, 2014, 3:59 AM UTC

einhverfr wrote: post #139

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…

I see it as somewhat complex but in the end I sort of agree that it is a definition. Let's start with a problem: 0/0 The problem with this is not that the equation itself is meaningless. the limit of n/x as x->0 is infinity for any positive real number, and negative infinity for any negative real number. In essence 0/0 ends up reducing to 0 * infinity, which isn't very helpful. I would argue that discontinuity in a f…

The relationship f(x)=x/x is only defined for x /= 0 and thus is not equivalent to 1. We can see this in [a] when we ask for the domain of the function, and thus we can redefine some function f' as a piecewise function which is defined to have f'(0)=1, but in proofs we must thus make sure to first prove that using f' as a substitute for f does not affect our result.

In one of the below posts we have the suggestion

> But what if you're in a context where you're not reasoning about continuous functions at all? Why would you have to be subject to reasoning that doesn't apply to your situation?

In this case you could either do the above, if you have to concern yourself with e.g. a domain of the set of real numbers arbitrarily close to the undefined location. Alternatively you could just define our original function f only for real numbers greater than 0, in which case you escape the necessity of redefining functions to be easier to work with.

[a]: http://www.wolframalpha.com/input/?i=domain+of+f%28x%29+%3D+...

Re: 0^0

#156

Post by zodiac » Thu, Apr 03, 2014, 4:04 AM UTC

einhverfr wrote: post #139

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…

I see it as somewhat complex but in the end I sort of agree that it is a definition. Let's start with a problem: 0/0 The problem with this is not that the equation itself is meaningless. the limit of n/x as x->0 is infinity for any positive real number, and negative infinity for any negative real number. In essence 0/0 ends up reducing to 0 * infinity, which isn't very helpful. I would argue that discontinuity in a f…

I see that you are defining 0/0 to be a value such that f(x) = x/x has no discontinuity at x=0. But your chosen value means that the function g(x) = x^2 / x is discontinuous at x=0, and the function h(x) = -x/x also has a discontinuity, etc.

Re: 0^0

#157

Post by Retric » Thu, Apr 03, 2014, 4:04 AM UTC

rickhanlonii wrote: post #141

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…

I agree with the point you're making, but I want to be a little pedantic: It's true that addition is a definition, but 1+1=2 is not--it logically follows from the definition of addition.

Not quite, just as in calculus you end up rediscovering some old geometry formulas there are branches of mathematics where you end up rediscovering 2 such that 1 + 1 equals it. Other options include 1 + 1 = 0 or 1+1 = 1.

Re: 0^0

#158

Post by einhverfr » Thu, Apr 03, 2014, 4:08 AM UTC

zodiac wrote: post #156

einhverfr wrote: post #139

Earlier quoted context omitted.

I see it as somewhat complex but in the end I sort of agree that it is a definition. Let's start with a problem: 0/0 The problem with this is not that the equation itself is meaningless. the limit of n/x as x->0 is infinity for any positive real number, and negative infinity for any negative real number. In essence 0/0 ends up reducing to 0 * infinity, which isn't very helpful. I would argue that discontinuity in a f…

I see that you are defining 0/0 to be a value such that f(x) = x/x has no discontinuity at x=0. But your chosen value means that the function g(x) = x^2 / x is discontinuous at x=0, and the function h(x) = -x/x also has a discontinuity, etc.

I am defining continuity such that single point that is undefined does not break continuity if and only if it has the same limit when approached from both sides. This is because, as I say, a point is arbitrarily small and therefore if the limit is the same on both sides, it's simpler to treat it as still a continuous function.

Otherwise I think you get some problems which break algebra, because multiplying of dividing by variables can break the domains of functions.

If we have f(x) = 1, then this should be the same even if we multiply both sides by x. x * f(x) = x, which results in f(x) = x/x. You break algebra by allowing such straight-forward manipulations (which really are the core of algebra) to tamper with the original domain of the function.

Just as you can't just treat a square root as a core algebraic process (because it is not a function), you would no longer be able to divide by a variable, since that would be undefined wherever the variable is 0.

Edit Also it occurs to me that treating x/x as undefined at 0 would also man that the first derivative of f(x) = x would also be undefined at 0, which can't work either.

So therefore treating x/x as discontinuous at 0 breaks both algebra and calculus.

Re: 0^0

#159

Post by cbright » Thu, Apr 03, 2014, 4:09 AM UTC

EGreg wrote: post #83

Mathematics is about generalizing concepts and principles to ever larger domains. In this case, x^y is defined for all pairs (x, y) of real numbers except (0, 0). The question is what limit is "closer" to the set of outputs in the neighborhood of (0, 0) than any other. 0^x is defined for all x except 0, and same for x^0. We can define 0^0 as the limit of one or the other as x goes to 0. and one is constant and more "…

Correction: x^y is defined for all pairs of real numbers with x!=0. 0^x is not defined for negative x.

Re: 0^0

#160

Post by rickhanlonii » Thu, Apr 03, 2014, 4:09 AM UTC

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.

That's a great point! Under that definition of a^b, 0^0=1 follows from the definition and is not itself defined. It's exactly the same as 1+1=2.

However, what I said is absolutely true for the definition of a^b that the author of the article is using. In that system, 0^0 does not follow from the definition, and must be defined itself (or not at all a la Cauchy). I should have made that more clear.