Division by zero

[wendy]

0 ^ 0 is one, because the product of zero numbers must leave any previous element undisturbed. Therefore 0^0 = 1. The proof does not require division of zero, no more than a count of zero elephants requires the insertion and removal of an elephant.

1/0 is generally taken to be the horotopic infinity U. This is the same as the circumference of a horocycle in both euclidean (straight line) and hyperbolic geometries.

0/0 is generally undefined, not because of division by zero is undefined, but because the equal sign breaks down. However, there are tests that check if the equal sign is stable during the operations. Most of euclidean geometry is the result of a stable 0/0.

[PWrong]

0 ^ 0 is one, because the product of zero numbers must leave any previous element undisturbed. Therefore 0^0 = 1.

There is no previous element. :? We've already shown that 0^0 can be anything.

When we say 1/0 is infinity, which infinity are we generally talking about? Can it be shown that it's the cardinality of the integers, or the reals, or both?

[bo198214]

By assuming that there is only one infinity and it is neither negative nor positive, then there is no problem with saying n/0 = infinity.

With infinity there is always a problem, you can not treat it like normal numbers. For example:
inf * inf = inf then by dividing by inf, you get inf = 1. So if you come up with an additional number like inf, you must define what is the result of the operations with itself and with the other numbers. I guess there is no contradiction free way of doing so with infinity.

0 ^ 0 is one, because the product of zero numbers must leave any previous element undisturbed. Therefore 0^0 = 1. The proof does not require division of zero, no more than a count of zero elephants requires the insertion and removal of an elephant.

Your repetition is boring. Your argument of zero as a counting number leads to contradicition as I already showed. (Not to mention all the other arguments that you are deaf to.) And I really would enjoy a real discussion (which means interaction!) with you instead of speaking with a repetitive machine.

When we say 1/0 is infinity, which infinity are we generally talking about? Can it be shown that it's the cardinality of the integers, or the reals, or both?

interesting question. I think it has nothing do with the cardinalities. Its rather a classification of sequences or functions regarding their behavior for large indexes/arguments. One usual construction of the real numbers is taking Cauchy sequences of rational numbers with the equivalence relation, that their difference tends to 0. If we now admit not only Cauchy sequences but also unbounded sequences, we can despite regard two sequences as equal if their difference tends to 0. For example the unbounded sequences n and n+1/n, would be regarded as equal. So they would represent the same infinity. Another common method (keyword asymptotics) is to regard two sequences as equal if their quotient tends to 1. In both way we would get an extension (i.e. by some non-Cauchy sequences) of the real numbers with various infinities. The real numbers remain included because they are also equivalence classes of (Cauchy) sequences. To not include oscilating non-Cauchy sequences we could demand only to consider strictly increasing sequences (because every real number is also the limit of a strictly increasing sequence).
I can also recommend Hardy's monograph "orders of infinity" of 1924 (which I electronicly possess but dont know about its copyright, so I can only provide it privately).
Further interesting in this topic are the Surreal numbers invented by Conway. And non-standard analysis which works with infinite small and infinite big "real" numbers, see hyperreal numbers.

[wendy]

bo seems to suggest, that because 0^m = 0 for all positieve m, that it must be true for all values of N. Also, the suggestion that 0^0 has no evaluation because 0/0 is indeterminate.

Firstly, 0 is an element of Z, particularly the dividing point between positive and negative. The set Z can be comprised of actions that increase or decrease, eg 0 = +1 -1. The set N is comprised of sets that can only increase.

Zero is the result of a counting exercise, unless we are to beleive that 0 that is in R is different to 0 that is in Z. Bo has not advanced this theory, so we shall ignore it.

The argument that 0^m = 0 for all values, is akin to (1/2)^m < 1, for all positive m, and therefore true for all m. It's a case of appealing to a sample.

The actual reason that 0^0 = 1, is the result of the factorisation theorm. That is, if one writes a number as a product of power-factors (ie a^b), then any factor that is absent, has a count of zero elements, and is therefore f^0 = 0.

Since we can derive 0 in N (which can be done without decrement), we have, 2 = 2 * k^0 = 2 * 0^0. Were 0^0 anything other than 1, then we have 2 = 3 for example. This is why ^0 takes precedence over 0^.

On the other hand, division by zero, requires a decrement. It is this decrement that causes the problem.

[batmanmg]

i've seen the statement made that 0/0 is all real numbers...

well what about (0/0) / (0/0) all real numbers divided by all reall numbers?

[moonlord]

(0/0) / (0/0) all real numbers divided by all reall numbers?

That makes no sense. 0/0 is undefined, and therefore cannot be used in algebraic operations. If you want to extrapolate, though, I'd say that's the R^2 set. Anyway, extrapolation is ... well ... idiot in this case.

[PWrong]

To be completely general, you need a new variable for every zero, then send all the variables to zero.

(0/0)/(0/0) is the limit of (x/y)/(z/w) as (x,y,z,w) tends to (0,0,0,0)
That is, (xw)/(yz). This is simply equal to 0/0.

[batmanmg]

don't you elliminate some of the information about the limit when you simplify the equation?

[pat]

i've seen the statement made that 0/0 is all real numbers...

There is a difference between "could be any real number" and "is all real numbers". And, there's no reason to stick to real numbers.

[bo198214]

bo seems to suggest, that because 0^m = 0 for all positieve m, that it must be true for all values of N.

Absolutely not. So argumenting against 0^0=0 is like shooting into the air.

I suggest that it depends on the context what can be defined and what leads to contradictions. So in the context of real numbers 0^0 can not be defined (as 1) because there are the following sequences

(1/n)^(1/n) -> 1
(1/n^n)^(1/n) -> 0

I even chose them completely positive, because you mentioned that the flaw would be them crossing positive/negative. But till now no response from you. In the contrary you even insist that 0^0=1 on the real numbers instead of restricting it to natural, interger or rational numbers. Though by your argument that the 0 of Z is also the 0 of R (which I however dont support) it would follow that 0^0 is undefined throughout.

The other counter argument I still got no response from you is about your most stated (in various disguises though) zero as a counting number.

0 * (1/0) = 0 (identity of +) but of course 0/0=x/x=1 too.

Maybe though thats a bit cryptic, so I will explain. If 0 is a counting number then must 0*a = 0 no matter what a. If I now let a=1/0 then would 0=0*(1/0)=0*1/0=0/0. But you at least admit that 0/0 is undefined. This shows that we must be cautious about a. And in the same way we must be cautious about the a in a^0. Because multiplicatively 0 and 1/0 (or 1/infty and infty) are opposite infinities.

Though I think its already a heavy burden of proof, I am still interested what you mean by your sometimes mentioned stability (I intuitively would think that the above two sequences show an instability at 0). And what you meant by this:

Since we can derive 0 in N (which can be done without decrement),
...
On the other hand, division by zero, requires a decrement.
It is this decrement that causes the problem.

[houserichichi]

1/0 is generally taken to be the horotopic infinity U. This is the same as the circumference of a horocycle in both euclidean (straight line) and hyperbolic geometries.

I'm not entirely sure what you're trying to say here. Care to elaborate?

0/0 is generally undefined, not because of division by zero is undefined, but because the equal sign breaks down.

Division is a special case of multiplication. Assuming this discussion is focused on algebraic operations and the properties of fields (which I have all along, perhaps incorrectly) then 0/0 is generally undefined precisely because division by zero is undefined. Zero is strictly removed from the definition of multipicative inverse in a field because fields are special cases of integral domains which, by definition, have no zero divisors.

Whatever point you have been trying to make is being lost on me and I suspect either I'm missing a key point in your argument (I suspect the fact that I'm talking about fields and only fields and maybe you're not) or it's simply a matter of fighting a losing battle. I am putting my money on the former.

Many of the arguments put against an undefined x/0 or 0/0 through examples are nothing more than mathematical conventions. In its purest form, division in fields is defined (and I stress that word) as the multiplication of one element with the inverse of another (or perhaps the same) except when the inverse is that of the additive identity. In R and C that additive identity happens to be the integer zero.

[wendy]

The argument that 0^0 = 1, is not based on division by zero, and therefore is free from the indeterminateness of 0/0.

The argument runs like this.

suppose P(a,b,c,...) = a*b*c*... , ie P is a product of a list.

We see that P(a,a,a...) = n@*a = a*@n, is the product element applied to n members. In convention, this is a^n, ie P(a)=a, p(a,a)=a^2 and so forth.

Next we question what P() might mean, that is, a product of zero numbers. We know that S() [a sum of zero numbers] = 0, and therefore it is not absurd to consider the result of a zero list.

We further note that the list can be freely broken, eg

P(a,b,...,p,q,...) = P(a,b,...) * P(p,q,...)

Likewise, we note that a break of list can split out P(), as

P(a,b,...) = P(a,b,...) * P() [where P() is a power of absent q]

Since we have P() as an identity of multiplication, the conclusion that P() = 1,

Also, since P() = p*@0 = 1 , for all p, we have

p^0 = 1, for all P

0^0 = 1, QED.

We now consider the implication of P() <> 1, for 0^0.

The product 2*3*5 contains 0 values, each equal to zero, and therefore we can write, without generality of loss, 2*3*5*0^0. Let 0^0 = 2, then 30 = 60. This is obsurd and would destablise multiplication. Therefore 0^0 = 1.

Note that this argument does not rely on division at all, but rearangement of present multiplication factors, and therefore is to regarded as proven.

[wendy]

Zero is not a counting number per se, but the result of a count. Specifically, it is the value of a count that ends before it begins.

For example, you can have one, two or three elephants, or you can have no (ie zero) elephants. You can't have minus-one or one-third of an elephant.

Zero, then is a legitimate whole number: a number that represents a count of whole things. This is the notion behind the set N (ie 0,1,2...).

I am still mistified on why you are unable to see that this is relevant even in mathematics.

[wendy]

I even chose them completely positive, because you mentioned that the flaw would be them crossing positive/negative. But till now no response from you. In the contrary you even insist that 0^0=1 on the real numbers instead of restricting it to natural, interger or rational numbers. Though by your argument that the 0 of Z is also the 0 of R (which I however dont support) it would follow that 0^0 is undefined throughout.

The other counter argument I still got no response from you is about your most stated (in various disguises though) zero as a counting number.

0 * (1/0) = 0 (identity of +) but of course 0/0=x/x=1 too.

Maybe though thats a bit cryptic, so I will explain. If 0 is a counting number then must 0*a = 0 no matter what a. If I now let a=1/0 then would 0=0*(1/0)=0*1/0=0/0. But you at least admit that 0/0 is undefined. This shows that we must be cautious about a. And in the same way we must be cautious about the a in a^0. Because multiplicatively 0 and 1/0 (or 1/infty and infty) are opposite infinities.

Though I think its already a heavy burden of proof, I am still interested what you mean by your sometimes mentioned stability (I intuitively would think that the above two sequences show an instability at 0). And what you meant by this:

Since we can derive 0 in N (which can be done without decrement),
...
On the other hand, division by zero, requires a decrement.
It is this decrement that causes the problem.

U is the horotopic infinity. The limit of circles or pseudocycles of ever increasing radius is a horosphere, or Euclidean line. Such a space has zero curvature, and events of these, in terms of the whole-space, is of the order 1/U.

Typically, U is taken to be not less than we can resolve discrete points in the horizon, each representing the size of the visitable real world.

What allows U to work here is that the smallest resolvable angle (a) in units of the circle, and the largest visitable length (r), is that U > r/a.

Since in practice we can not discriminate values below r/a, lengths of this order can not be measured, even by parallax, and therefore we should regard U as a real measure of this order.

That even at this distance, space is still "real" in the local neighbourhood, we note that space far away is not breaking down, but it is a kind of breakdown in the equality.

Counting Number

One must realise that Zero is the result of a legitimate count, or it is not. Further, the equality of x/x = x^0 is only true where division by x is defined. So in the statement 0/0 = x^0, the LHS contains an undefined statement. The RHS is a statement that can be evaluated without recourse to division, and its value is 1.

ie: 0/0 is undefined, because division by zero is undefined
but: 0^0 is defined, because 0 is a counting number, and a product of 0 numbers of any kind, leaves the pre-existing value undisturbed, and therefore must be the identity of multiplication (1). It is true in matrix theory as well.

0 * 1/0

In practice, even if we set 1/0 to U, this statement tells us nothing: it is still undefined. Even for values of U, there are sometimes that you can be sure that it's the same number on both places.

For example, one can work with polynomials in U, and extract the 1/U term out as Euclidean geometry. The model for circle-drawing geometry resolves around a kind of expression that just does that.

For example, the general form of the pythagerous theorm is

2.(2-c) = (2-a)(2-b), where a,b,c are the squares of the euclidean chords on a sphere. [change the sign to +, and you get the hyperbolic case, measured in arcs on the horocycles]

Allow a, b, c to go to zero, and you get:

4-c/u = 4-a/u-b/u+ab/u²

One can then cancel out the 4 (ie 4/0), and ab.0, to get

c = a+b (on squares of lengths)

c²=a²+b² (on natural lengths)

(this relies on the terms of 0¹x, and drops terms in 0²x.)

We could easily split 0 as 0*n, and then write 0*n * 1/0 = n. However, one needs to know if these 0 point to the same inverse before making some sort of statement of this order.

[bo198214]

I mean its fine to set 0^0=1 and with your P() and S() you restrict yourself of course to N, i.e. ^ defined on RxN, thatswhy I already suggested several posts before

We can define ^ on RxN by induction (as above taking the multiplication on RxR). And have again that 0^0 can only be defined as 0 or 1 if the operation on the other values of RxN should be left unchanged.
We can define ^ on RxN by Cauchy sequences (taking ^ defined on QxN). Then 0^0 can be defined as 1 and 1 is the only value that it can take.

This follows simply from the definition, we dont need to apply zero-as-counting-number-argument, which is somewhat unreliable as we see from 0*U undefined. But your reply was

0^0 is the same thing throughout.
The limit of 0^x as x-> 0 differs as x starts negative or positive, and is inconclusive.

And this useless grimmness is the only reason for this thread to continue. I mean is it that difficult to admit the restriction to RxN facing contradictions on RxR?!

[PWrong]

This reminds me of something interesting I learned in Algebra. Apparently the intersection of zero sets is the set of everything. It works like this:

Let J be an "index set", so that for every j in J, we have a set U_j. Then the intersection of all the U_j is also a set, and x is in the intersection iff for every j in J, x is in U_j.

Now if J is the empty set (that is, there are no sets to intersect), then it's always true that for every j in J, x is in U_j, no matter what x is. So for every x, x is in the intersection. So the intersection of nothing is everything.

[bo198214]

yes thats the 1 in the boolean algebra of sets, i.e. if you consider all subsets of say X. Then intersection and union form with the constants emptyset and X form a boolean algebra.
And yes the general intersection of the emptyset is X and the general union of the emptyset is emptyset, as well as the general intersection of all subsets of X is the emptyset and the general union of all subsets of X is X
And this works quite well because there is no inverse operation neither of intersection nor union (no group structure).

[wendy]

The argument that 0^0 = 1, is based on the set N, and therefore must carry through to R. One can of course, suggest alternate limits from using different directions, but in every case, 1 is a valid substitute.

The further point that if 0^0 is not 1, then it makes all mathematics unstable, since, eg 7 is a product of one 7 and zero 0, and therefore can be written as 7^1 * 0^0. Since 0^0 has an indefinite value, then so does 7.

The argument about the various infinities, makes the fatal mistake that "infinity" is a place. The reality is that "infinity" is a sense of /far away/. That is, one can never really be "far away", but that "here" and "there" shift.

Infinity is more a statement of our abilities to resolve things, and has an onset. That is, the properties of infinity settle on even finite numbers.

The proposition of 0 = 1/U is not a mathematically precise statement, but a statement of our abilities to resolve 1/U from 1/(U+1) or something like this.

The proof that 0^0 = 1, does not invoke division by zero, but rather the effect of letting 0^0 equal anything other than one, and therefore is not limited by the implications of division by zero, or the nature of infinity.

That is, 0^0 = 1, but 0/0 is undefined.

[bo198214]

The argument that 0^0 = 1, is based on the set N, and therefore must carry through to R.

The direction is different, if you see the 0 is the same in N and R, then you cant define 0^0 on RxN because you can not define it on RxR.

One can of course, suggest alternate limits from using different directions, but in every case, 1 is a valid substitute.

What a sense does it make to define an operation only by approximation from one certain direction?!
Its no doubt that you can/should define the function f(x)=x^x as f(0)=1, its because for *every* sequence (x_n)->0 is f(x_n) -> 1.

The further point that if 0^0 is not 1, then it makes all mathematics unstable, since, eg 7 is a product of one 7 and zero 0, and therefore can be written as 7^1 * 0^0. Since 0^0 has an indefinite value, then so does 7.

Wendy dont you see that this is tautologic? You can also omit the 7 and say that 1 is the product of 0 times the 0, its then a shorter "proof". *sarkastic*

The argument about the various infinities, makes the fatal mistake that "infinity" is a place. The reality is that "infinity" is a sense of /far away/. That is, one can never really be "far away", but that "here" and "there" shift.

Sorry Wendy, but this is the only reality *in your mind*. Non-standard Analysis is a well-established part of mathematics and does not use doubtful philosophical statements like the 0 as a counting number.

The proposition of 0 = 1/U is not a mathematically precise statement, but a statement of our abilities to resolve 1/U from 1/(U+1) or something like this.

Your arguments arent mathematicly precise either. I mean even you see 0*U as undefined, so you must admit that your *philosophical* principle is not reliable.

I anyway think its enough said. I gave you opportunity to relax your fixed attitude, without nearly any costs. Its your choice to live in this only world. And the discussion reached in the meantime a nearly unbearable repeating factor, so I probably stop attending it, if no new ideas emerge.

[PWrong]

The further point that if 0^0 is not 1, then it makes all mathematics unstable, since, eg 7 is a product of one 7 and zero 0, and therefore can be written as 7^1 * 0^0. Since 0^0 has an indefinite value, then so does 7.

I see what you're trying to do here, but it doesn't work. There is a theorem that says a number can be expressed as a product of its prime factors, but any good mathematician would be careful enough to say "except zero". The factorisation stuff depends on more fundamental work, like division and exponentiation, not the other way round.

[wendy]

The point here is that unless 0^0 = 1. Whatever propertities you want to apply to zero, it is also the result of a count.

One can, for example, write 120 as 2^3 * 3^1 * 5^1, or that same product by, eg 7^0. Since this number is clearly not equal to zero (for which we can write 0^1 as a factor), and since it is finite, then the only option for the power of zero is 0^0.

More over, 0^0 can not change the value of any product to which it is applied, and therefore 0^0 = 1.

Note that this is not dependent on the nature of infinity, or division by zero, but rather the process of counting, and the meaning of the empty count. Oh well...

[moonlord]

Okay, you DO have a problem with comunication.

How the hell do you count the real numbers!? We're not talking about counting here. Your definition works for RxN, as bo198214 said, but try to apply it to RxR.

By the way, you don't convince us if you repeat yourself for a hundred times.

[wendy]

In the first instance, zero is a member of the set Z, which is a subset of R. The assumption that zero looses its properties it acquires from Z when it becomes a member of R presupposes that zero element Z is different to zero element of R.

Secondly, the statement i make is that 0^0 does not involve a division by zero: that is, one can take the product of zero elements, even in matricies, without invoking a division.

Thirdly, one can not draw any valid conclusion simply by looking at the approaches to x^r, for r < 0, because the sign of x^r depends on parity of r, which is undefined for real numbers. One can only use magnitude, which approaches 1, for fuzzy values of x and r.

The argument that x^0 = 1 for all x, comes from the natural role of zero as a count of absences.

Fiftly, one can do effective division by zero, and compare this to real outcomes. This is true for both numbers and matrices. The nature of euclidean geometry is effectively a repetition of divisions by zero, and assorted second and third remainders ie (1/0 + x + 0.y) - (1/0) => x.

The question that if it does not apply to R×R, then your assumption is that 0 in N is not 0 in R. When you show that 0 in R is different to 0 in N, then i should consider a proof that 0^0 = 1 in R×R as well as in R×N.

[bo198214]

Thirdly, one can not draw any valid conclusion simply by looking at the approaches to x^r, for r < 0, because the sign of x^r depends on parity of r, which is undefined for real numbers. One can only use magnitude, which approaches 1, for fuzzy values of x and r.

So what do you say then about the approaches

(1/n)^(1/n) -> 1
(1/n^n)^(1/n) -> 0

?

[wendy]

If you write the equation as eg exp( k.ln(u)/(u)), then this suggests that the value goes to one as u goes large and k is constant.

What this says is that in a given base (exp(k)), the a large number has typically more digits than it is in size: 2^64 is greater than all of the number of grains on the earth, but is only 20 digits.

This produces an approach from an angle given by the line of exp(k). For k=2.3&c, it means that the number of digits in a decimal number is vanishingly small when compared to the number itself.

When k is set to extremely low numbers, eg base 1+(1e-100), then the number of digits keeps pretty close with the actual number for a good while. Even so, at 1e100 digits, the number is 2.718 times this.

If you write something in the form of 1/(n^n)/(1/n), this is pretty much something of the form of 1/(n)^(n/n) = 1/n. This particular line, suggests that the asymtope is someting in the order of the number of digits in the base (1+1/n), for which the logrithm is already 1/n. The asymtope is scaresly divisible from that of approaching along the line of 0^n = n.

It also shows that there is this culture that if there is a common symbol for any plentifella big number [to use the Pidgin], we can freely substitute one for another without scarsely butting an eyelid. It is as if to say, because one can not count things on a glance of 100, or 1000000, that these two numbers are identical, and one can freely replace a hundred by a million, without loss of error.

W

[bo198214]

If you write the equation as eg exp( k.ln(u)/(u)), then this suggests that the value goes to one as u goes large and k is constant.

If I write which equation?

If you write something in the form of 1/(n^n)/(1/n),

Do you mean 1/(n^n) ^ (1/n) ?

this is pretty much something of the form of 1/(n)^(n/n) = 1/n.

Not only pretty much but exactly (1/(n^n))^1/n = 1/n.

This particular line, suggests that the asymtope is someting in the order of the number of digits in the base (1+1/n), for which the logrithm is already 1/n. The asymtope is scaresly divisible from that of approaching along the line of 0^n = n.

Sorry Wendy but thats a cheap excuse. 1/n^n is nowhere 0, its simply one of many possible ways to approach 0. What ever you mean by fuzzy values, it surely should contain the approach (1/n^n, 1/n).

I mean there are various other approaches. A friend suggested if we can not achieve a common position, we simply should agree on 0^0=1/2, as a kind of diplomatic solution *ggg*. But indeed there is also an approach of x^y to 1/2:

(1/2^n)^(1/n) = 1/2

A look at the graph of x^y shows anyway that each of but only the numbers between (and including) 0 and 1 can be approached (from the positive quadrant).

[wendy]

There is a proposition in quantum mechanics that a vacuum might be inhabited by a large number of virtual particles, creating and destroying on the instance. This would have an observable effect, and one can thus search for it.

The proposition that 0^0 is not equal to one, would disstabalise every value, since one can always add the factor m^0 to it, and set m = 0, and were m^0 .NE. 1 for any value, then any number could equal any other number.

Since we suppose that this does not happen, in R or in Z or C, then the case is m^0 = 1 for all values of m, including 0.

The bulk of arguments presented to support m^0 = 1 do not depend on the division by zero.

The bulk of arguments presented to support m^0 = 0 (or something else), depend on the division by zero.

In the main, one is not restricted to R×N or R×R. One can use the set B2, (which is the closure of Z, with the closure of division by 2). This allows us to take square roots etc. The arguments that apply to 0 as a counting number in Z, apply equally to that of B2.

We are then lead to believe that zero in any discrete set (eg base-2, geometric numbers, base-10), is singularly different to zero in the reals?

I don't think so?

W

[bo198214]

There is no bulk of arguments. Your only argument was zero as counting number which we saw already being unrealiable. Instability is the same, ie. if you already assume that 0^0=1 then of course any other value would collapse the numbers. Dont know why you multiple times repeat the non-division thing, it wasnt talked about for many posts already.

Further we can not take arbitrary square roots in the 2-division closure of Z. Because it is the finite binary fractions, while sqrt(2) is an infinite binary fraction. Further is B2 not discrete, it is dense and we can define numbers being limits already there. And, there are non-1 approaches to 0^0 already in that B2:

(1/2^(2^n))^(1/2^n) = 1/2

1/2^(2^n) is in B2 and 1/2^n is also in B2 and their power is constant 1/2 and hence the limit is also 1/2 which is again in B2.

[wendy]

You don't have to define anything other than the list operator to get 0^0=1. Moreover, it is valid for any constructable set.

Suppose P(x,y,...) gives the product of the arguments. The principle that multiplication is communitive means that P(a,b,...) × P(c,d,...) = P(a,b,...,c,d,...).

This allows us to derive meanings for both P(a) and P(), the former can be found from P(1,1)×P(a) = P(1,1,a) = a.

The latter can be found by P()×P(x,y,z) = P(x,y,z), for all x,y,z, and therefore P() is the identity element of the ×.

This same argument applies without failure with S()=0 and S(a)=a, the first being the identity of the summation operator +.

There is also a compound operator, for several repeated items in the list, such that S(a,a,a...) = n @+a = a +@ n (n times, summation of a), where the + is the repeated operator, and the @ is the number of times repeated. For +@, we can write ×, that n×a = a×n.

For the product operator, we can write n @× a = a ×@ n = a ^n. Note that this is not n ^ a. The value of P() = a +@ 0 = 0 @+a = a^0 for all a.

This holds true, not only for a in the integers, because we can apply meaning of fraction, in that a^(1/n) = x, implies P(x,x....,x) = x ×@ n = a. Since the process remains valid as we extend the naturals to the reals, the argument that 0^0 = 1 is therefore true.

The Cauchy arguments simply say that either 0 or 1 is a possible value, but there is one that changes sign in the process (going to positive to unsigned). All the approach from zero tells us is that as one approaches x^y as y approaches 0, is that the expression becomes closer to 1.

It is possible to extend any integer sequence to cover the real number line, by extracting roots (square roots, cube roots), that ever approximate the real values. In any of these cases, the argument of P() = 1 holds true, and therefore this must be the case in the reals also (unless we are to believe that the set Z is not a subset of R).

The argument that 0^0 = 1, does not make any statement regarding division by zero. What it says, is the absence of a given number in a product does not unsettle the product. The argument is based on the ability to write 0 to represent the empty column, even amongst products.

Were this not the case, one should argue that 10 and 10.0 are different numbers, because we are adding an indefinite value to 10.

One of the arguments presented (the limit of (1/n)^n), is so incorrect that even Cantor distinguishes between infinity (aleph_0), and 2^infinity (aleph_1).

However, the argument that 0^0 = 1 does not resolve on division, but on the property that 0 is also the result of a count, and it has never been demonstrated that 0 in R and 0 in N are different.

One needs only to consider the effect of 0^0 <> 1 to realise that setting this reduces all mathematics to instability.

[bo198214]

Wendy you really dont need always to repeat your argument of counting 0. I understood it thoroughly and anyway its no reliable argument, as I already showed. Its merely a reason why it could be *useful* to regard 0^0=1 (in contexts of natural exponents).

This holds true, not only for a in the integers, because we can apply meaning of fraction, in that a^(1/n) = x, implies P(x,x....,x) = x ×@ n = a. Since the process remains valid as we extend the naturals to the reals, the argument that 0^0 = 1 is therefore true.

The real numbers (and also the rational numbers) allow an additional aspect: convergence and continuity. An aspect that is only in a trivial form possible on discrete sets like N. I dont see where your above statement relates to this.

but there is one that changes sign in the process (going to positive to unsigned).

What exactly are you referring to? The sequences that I gave in the previous posts are strictly greater than 0.

All the approach from zero tells us is that as one approaches x^y as y approaches 0, is that the expression becomes closer to 1.

This becomes quite boring: in the same way x^y comes closer to 0 if x approaches 0. So whats your point here????? That y is the better part of the operation and therefore takes precedence????

It is possible to extend any integer sequence to cover the real number line, by extracting roots (square roots, cube roots), that ever approximate the real values. In any of these cases, the argument of P() = 1 holds true, and therefore this must be the case in the reals also (unless we are to believe that the set Z is not a subset of R).

P() is a product of a natural number of arguments, isnt it? So what has this to do with real/fractional exponents?
I would regard it as proper discussion style to not put pseudo arguments.

The argument that 0^0 = 1, does not make any statement regarding division by zero.
...
However, the argument that 0^0 = 1 does not resolve on division

Just in the previous post I wondered why you always repeating this argument though nobody else talks about it. And now you even repeat it another time!!! Did you read my posts at all?

One of the arguments presented (the limit of (1/n)^n),

And I would also ask you to be a bit careful with your formulas, do you mean here (1/n^n)^(1/n) or (1/2^n)^(1/n) or (1/2^(2^n))^(1/2^n) or what?

otherwise is so incorrect that even Cantor distinguishes between infinity (aleph_0), and 2^infinity (aleph_1).

Wendy! The infinities of Cantor are cardinalities of sets. Two sets have the same cardinality if there is a bijection between them. Indeed there is also an arithmetics of cardinalities. But we regard here sequences and not cardinalities of sets (as I already stated in this thread before). The infinities of functions and sequences are regard in Hardy's monograph, orders of infinity.

If I think your objection to the end, you say that there are different infinities and so different 0's. One zero that is the limit of 1/n^n say 0₀ and a zero that is the limit of 1/n, say 0₁, and another zero that is the limit of 1/2^n, say 0₂. And then
0₀ ^ 0₁ = 0 and 0₁ ^ 0₁ = 1 and 0₂ ^ 0₁ = 1/2 ?