Division by zero

[wendy]

I still struggle with exactly what bo sees with the different kinds of zero.

Infinity is not a number, the issue here is that it is an equality issue. That is, infinity is more a statement of how we can tell different large numbers apart. The recriprocals of these all converge on zero. The whole point of infinity is not that it is a "place", but a "location".

One might suppose that "here" is always here, but its size changes relative to "there". For example, when talking of long distances, here might be the town you are in. When talking about your own house and yard, here might be a room. But there is always a here that includes your personal space. There is always somewhere remote with some effort to get to. It depends on how you get there, too.

Zero is zero is zero. That is, while there may be many kinds of infinity, there is only one zero. It is the only number of that sign (neither positive nor negative), and arguments based on the properties of positive numbers can not really be used to draw conclusions on other signs unless there is some argument along these lines.

We note further that my argument of 0^0=1 is roundly dismissed, yet the very function of zero is the empty column. We hear silence on the implications of 0^0 = 0, in the expression that 6 = 6.m^0 = 6.0^0 = 0.

One should understand that the reals don't exist in nature, but are the underlying properties of any sequence of numbers. You simply can not construct a real number, because to do so, you have to give it a name, and that name brings extra properties to the number.

One can, indeed define P() for a real list of arguments, via the ability to construct fractions &c. That is, while x^n is a product of n terms x, it is still possible to superimpose fractions, and other things, to turn n from natural to real. In no case, is the original case of the natural number disturbed.

What i find more tiring is that the argument that zero's properties as a counting number ought be dismissed, with little to no proof, except a faulty argument about approaches from the positive side (which show that as one approaches an exponent of zero, the value is more closer to 1), the fault is that there is a change of sign when moving along 0^x.

The argument is that 0^x has different values when x is positive, zero or negative, and only for these signs. One can not draw conclusions on one from the other.

We are then left with some imposible task of trying to prove something that is unprovable (ie R^R), since any number in R has a construction, and thus is endowered by the properties of the construction. You simply can not generate a continium. You generate a series of discretes.

There is a clear demonstration of the nature of P() and p^0, which is based on the property of zero, that represent the absence of effect. That is what zero is. There are no other examples of numbers of this sign. It is surely a bitter pill.

One must also understand that if the absence of effect (0^0) can have an effect (ie != 1), then this would distabilise all numbers, because one can create m^0 as a factor in any equation, and then set m=0, and then distabilise that equation.

W

[Keiji]

Right now, I really wish that phpbb had a merge topics feature.



This is a really old topic about division by zero. Take a look at it.

Also, you might want to read this old topic, as part of it contains the "circular number line" ideas.

[batmanmg]

im not sure if this was said before or not... but the limit as x aproaches zero of 1/x from the negative side equals negative infinity, but from the positive side it equals poditive infinity. soooo is this why there is no definition for 1/0 becuase it would involve the leap between all real numbers. but if you write it in a different way. lim as x aproaches infinity of 1/(1/x - 1/n) with n=infiinty. do you still get this dual aproach? on the one side of infinity you have it aproaching negative infinity, and on the other side you have it aproaching positive infinity?


i say that zero isn't a number and should no longer be alowed in mathmatics.

[PWrong]

I don't see the point of this argument at all. It should be obvious by that 0^0 is undefined. This should be surprising since x^y isn't very well defined for most of the complex numbers (100% of them in fact, but not all of them :wink:)

I can accept that in some situations, like the binomial theorem, it's useful to substitute 1 for 0^0. But it suffices just to change the binomial theorem slightly.

Is it correct to say that 1^sqrt(2) is technically the entire unit circle in the complex plane?

i say that zero isn't a number and should no longer be alowed in mathmatics.

A field needs an additive identity.

[wendy]

x^y is well defined for x in C, y in N. Since it is also well defined for complex y (eg one can evaluate i^i uniformly, you could say that x^y is well defined when y is a gaussian or eisenstein integer.

Since ultimately, one can define y in any cyclotomic number system, it is well defined in the set ZZ (ie the span of chords of all unit-edge polygons). That is, i^sqrt(5) is well defined.

But note here that this does not correspond to the set R, because the set ZZ does not break down into multiple instances of 1, 2, 3 times around the circle, as do the fractions.

This is because the set ZZ does not generate an ambiguious segment by division, and therefore one can not have more than one solution to an equation.

Since we see that 0^0 = 1 is already evaluated in R×N, then this is its true value even in C×ZZ.

[PWrong]

I still struggle with exactly what bo sees with the different kinds of zero.

A zero is just an additive identity for a group, ring or field. You can't mix elements of different fields. You can't claim that the zero in the integers is the same as the zero in the reals, anymore than it's the same as the zero vector in R^2.

[wendy]

I still struggle with exactly what bo sees with the different kinds of zero.

Having demonstrated to bo that 0^0 is 1 for indices in N, he then refuses to accept the same point when 0 is in R. That is, he suggests that the 0^0 is somehow faulty in R, because 0 in reals is different to 0 in integers.

[PWrong]

Having demonstrated to bo that 0^0 is 1 for indices in N, he then refuses to accept the same point when 0 is in R.

That's because 0 in N isn't the same number as 0 in R. Just like it's not the same number as 0 in R^2.

[wendy]

So i take it, that N is not a subset of R? You must explain this one day!

[PWrong]

So i take it, that N is not a subset of R? You must explain this one day!

I'm not actually sure. But isn't R a subset of R^2? Anyway, even if the zero is the same, the operations are different. N isn't a field, or even a group (although Z is), so it can't be said to have the same operations as R.