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0^0

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Re: 0^0

#191

Post by Dewie »

Dylan16807 wrote: post #65
mason55 wrote: post #44

Earlier quoted context omitted.

The symbols we use to represent math are arbitrary but that doesn't mean the rules behind them are. Many concepts in math are fundamental truths.

There are an infinite number of fundamental logical truths out there, but we arbitrarily picked useful ones to make a system of math.

But if you're unlucky, you pick a set of axioms that makes the whole system inconsistent, which means that you can prove anything, which means that the whole system is useless. To say that it is purely arbitrary is in a sense right, but it seems to undermine the care that you have to go through in order to be reasonably sure that the system is not set up to fail.

Re: 0^0

#192

Post by Perseids »

trentmb wrote: post #185
SamReidHughes2 wrote: post #173

Earlier quoted context omitted.

It's a definition of what "2" means.

I was always under the impression that 2 is shorthand for s(s(0)).

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 with 1 and define 2 as 1+1, 3 as 1+1+1…

Re: 0^0

#193

Post by pavelrub »

rntz wrote: post #105
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…

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 2 is not defined as 1+1.

Conceptually this reflects the fact that knowing what 2 means is different from knowing what 1 means and what addition means. In principle a computer can correctly recognize 2 apples as being 2 apples without knowing anything about addition and without being able to recognize 1 apple.

Re: 0^0

#194

Post by lmm »

baddox wrote: post #8
mcweaksauce wrote: post #2

This is ridiculous. Doesn't this seem counter-intuitive for it to be anything else besides 0 or undefined?

If we don't have a specific mathematical context, then saying it's undefined is intuitive to me. Without context, 0 is no more intuitive to me than 1. These two statements are equally intuitive to me, but they give different results for 0^0: "Zero raised to any power is still just zero." "Any number raised to the zeroth power is one."

0^2 and 0^1 may be zero, but 0^-1 and 0^-2 are infinite. So I don't think that intuition leads to 0^0=0. If anything 0^0=1 preserves the symmetry.

Re: 0^0

#195

Post by t__r »

To me the matter looks similar to taking the conjunction of the empty set, which is defined to be TRUE. This, however, starts to make sense when you see that the taking the conjunction of a set X can be interpreted as "forall x in X, x is TRUE". If X is empty then this statement is trivially true.

Re: 0^0

#197

Post by pavelrub »

Perseids wrote: post #192
trentmb wrote: post #185

Earlier quoted context omitted.

I was always under the impression that 2 is shorthand for s(s(0)).

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.

Re: 0^0

#198

Post by tomp »

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…

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.

Re: 0^0

#199

Post by tomp »

einhverfr wrote: post #189
lotharbot wrote: post #181

Earlier quoted context omitted.

> "x * f(x) = x cannot be equivalent to f(x) = x/x" Sure it's equivalent, over a domain not including x=0. This does not break algebra any more than, say, restricting the domain of the square root (when working in the reals) to non-negative numbers. We work in restricted domains in mathematics all the time. > " f'(x) = 2x/x" f'(x) = lim (h->0) [2(x+h)-2x]/h. Since h is approaching (and therefore not equal to) zero, t…

> This does not break algebra any more than, say, restricting the domain of the square root (when working in the reals) to non-negative numbers. Sure it does, because if that is the case, you restrict your domain when you divide by a variable expression. If you divide both sides by x-1, then you effectively rule out 1 from the domain. That's the problem. Now this is not the same as 0/0. The point is that 0/0 is only…

> Sure it does, because if that is the case, you restrict your domain when you divide by a variable expression. If you divide both sides by x-1, then you effectively rule out 1 from the domain. > That's the problem.

That's the problem that a mathematician must handle. The solution to the equation

  x * f(x) = x
is very simple: x is either 0 or such that f(x) = 1.

Re: 0^0

#200

Post by thomasahle »

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…

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.
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