Real Number Operations Caveat

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When you declare it undefined, you define it as undefined; i.e., we are to understand it as an undefined term having the appearance of a useful term, but deemed meaningless.
It is defined as undefined so as not to give the appearance of having been overlooked. I anticipate your response that no, it is not defined, but that is a minor point to me.




Yes, you did, in your 'proof' that 0^0 = 0^0+1-1 = 0/0, which is fallacious in that it takes the original term 0^0 and multiplies it by zero (+1 exponent), and divides by zero (-1 exponent) to 'prove' that it is 0/0. You commit 2 errors purposefully to 'prove' that the original term commits precisely those 2 errors. Does anyone else see the problem here?

To answer the question of 0^0 it makes sense to go back to first principles on defining the function Z^R where Z and R are any complex number.

First you can define ln(x) = integral_(1 to x) 1/s ds where x is a positive real number. The domain of ln(x) is the set of positive real numbers and the range is the set of all real numbers. It's obvious that ln(x) as defined here is monotone increasing and thus a one-to-one function with a unique inverse. The proof that ln(x) as defined here is unbounded as x approaches either zero or infinity (and thus covers a range of all real numbers) is given in real analysis texts.

Now exp(x) can be defined as the inverse of ln(x). The domain of exp(x) is the set of all real numbers and the range is the set of positive real numbers.

Now, for a complex number Z = a + i*b, define exp(Z) = exp(a) * (cos(b) + i*sin(b)). The domain of exp(Z) is the entire complex plain and the range is the complex plane excluding zero. The inverse of exp(Z) isn't unique anymore unless you restrict the domain so that the imaginary part covers exactly 2*pi radians. Usually the restriction is 0 <= b < 2*pi. Then ln(Z) can be defined as the inverse of exp(Z), namely...

ln(Z) = ln(a + i*b) = ln(sqrt(a^2+b^2)) + i * atan2(b,a). This definition holds for the entire complex plane except for the case where a and b (i.e. the real and imaginary parts) are both zero.

Now that ln(Z) and exp(Z) are defined for complex numbers we can define Z^R in general as exp(ln(Z) * R).

This definition of Z^R works for any Z and R with the exception of Z = 0. Also it's pretty straightforward that the usual rules of Z^(X+Y) = Z^X * Z^Y and (X*Y)^R = X^R * Y^R hold. The only issue is 0^R must be defined separately.

My general opinion is that 0^R should be defined as limit_(Z -> 0) Z^R. This would make the function f(Z) = Z^R continuous. The problem is this limit doesn't always exist depending on R. If we let Z = x + i*y and R = a + i*b, then Z^R = exp( ( ln(sqrt(x^2+y^2)) + i*atan2(y,x) ) * ( a + i*b ) ). If a and b are both zero the limit is one. If a is positive the limit is zero. Otherwise the limit is undefined.

Therefore I would say 0^R is only defined when either R is zero (in which case 0^0 = 1) or when the real part of R is positive (in which case 0^R = 0). However, the rule Z^(X+Y) = Z^X * Z^Y doesn't necessarily hold when Z is zero, so writing 0^0 = 0^(1 + -1) = 0^1 * 0^(-1) = 0 / 0 is not valid. 0^(-1) isn't defined as when the base is zero you must have that either the real part of the exponent is positive or the exponent is zero.

Last edited by marshall on 15 Jan 2012, 11:39 am, edited 2 times in total.

In summary, 0^R is only defined for R where the complex function f(Z) = Z^R has a removable singularity at Z=0. The Z=0 singularity is removable only when either 1.) the real part of R is positive, or 2.) R is zero. When the singularity is removed we get 0^R = 0 whenever the real part of R is positive and 0^0 = 1.

Last edited by marshall on 14 Jan 2012, 6:41 pm, edited 1 time in total.

Ahem. 0^(0 + 0) = 0^0 * 0^0 and that is correct whether 0 ^0 = 1 0r 0^0 = 0.

ruveyn

Ruveyn: you must have the patience of a monk.

Yes. I see someone who has no idea what proof by contradiction means.

If you're so sure that 0^0 is not well-defined, do write and submit your proof to the Nobel committee or the Clay Mathematics Institute. Let us know how that turns out.

0^0 if defined can only be 0 or 1. Either way produces problems. So let it be undefined and there are no problems.

ruveyn

Ugh, thanks for pointing that out, I was tired after writing all that and put a plus instead of a times. The rule Z^(X+Y) = Z^X * Z^Y holds when Z is zero as long as no negative exponents are involved.

I'm still going to insist that it makes no sense to write 0^(-1+1) = 0^(-1) * 0^1 as 0^(-1) is undefined in all cases.

The problem is one must prove that the rule Z^(X+Y) = Z^X * Z^Y holds. Making an a priori assumption that it must hold in all cases isn't proper mathematics. It isn't a valid proof by contradiction. You have to start with definitions and then derive the rules. Writing 0^(-1 + 1) = 0^(-1) * 0^1 makes no sense unless you define 0^(-1). There's no consistent way to define 0^(-1) in the real or complex number systems. It can be defined as infinity (i.e extending the real numbers to include symbols for infinity and negative infinity) but there are major caveats to that.

Try this. Try defining 0^0 as lim (x -> 0) (0^x). If one insists on continuity the 0^0 = 0 On the other hand

x^0 = 1 for x != 0 If one insists on continuity lim(x->0) (x^0) = 1 so one should define 0^0 = 1.

Since these are the only values possible for 0^0 something is loose. The solution: leave 0^0 undefined.

ruveyn

That's fair.

Nope. I do not define it. I state that by saying it is "undefined". I have no idea why you wish to play a game of infinite recursion.


Yes. I do see the problem. You are not quoting me. I will never give a proof such as you attribute to me. I do not introduce extra complexity into a "proof". E.g. see "obfuscation" in Mathematical fallacy. You, on the other hand, have done that repeatedly. In fact, I would never give a "proof" that 0^0 is undefined. The concept of such a proof is meaningless. I will happily give examples of proofs stemming from any attempt to attach a value to "0^0" that result in self-contradiction. I hope you can accept "reductio ad absurdum" as a valid proof?

I'd like to see such a proof. 0^0 = 0^(1 + -1) = 0^1 * 0^(-1) = 0 / 0 isn't a valid proof because 0^(-1) and 0 / 0 are always undefined. A proof by contradiction has to at least compare defined terms.

It's no more valid than saying 0 = 0^1 = 0^(2 + -1) = 0^(2) * 0^(-1) = 0 / 0. Only in this case the value on the left is always defined while the value on the right is undefined. The third and fourth inequalities are problematic because they are stating equalities of values that result from undefined operations. The undefined status of 0^(-1) is completely independent from the status of 0^0.

Ruveyn: possible misunderstanding; I'm with you on this one. The quoted quoted quote was originally attributed to jackmt.

No. The proof was given by you to show that 0^0 = 0/0 and therefore entails a contradiction and therefore must be undefined. If 0^0 = 0/0 it must be so for another reason because your proof is fallacious: it takes the base (0^0), which is the term under investigation, and multiplies it by 0/0 ((0^+1) * (0^-1)) to prove that the original term is 0/0. By that method we can prove that everything is 0, 1, or undefined.

To maintain truth in a proof involving multiplication, one must ensure to multiply by 1 in some form. 0/0 -= 1.

Last edited by jackmt on 20 Jan 2012, 7:10 pm, edited 3 times in total.

If 0/0 must be undefined else a contradiction follows. If 0^0 is left undefined then they cannot be equal. The equality relation cannot exist between undefined things.

0/0 and 0^0 are just marks of a page or screen and have no mathematical meaning.

However if one insists on defining 0^0 it must be either 1 or 0.

ruveyn

I've no idea why you should want to construct such proofs.

I'd go for something more like:
Assume 0^0 = 1.
We know 1^0 = 1
Therefore 0^0 = 1^0
Take the 0'th root of each side.
0 = 1

All such proofs are essentially nonsense.

Actually, it really is only if you wish to have a well behaved, continuous behaviour to your exponentiation function that it is necessary to avoid attaching a value to 0^0.

I do find it interesting that the "dogma" actually used to be that "0^0" was unity. It is more recent that it was recognised as being badly defined.
http://en.wikipedia.org/wiki/Exponentia ... zero_power

In some contexts, especially those involving purely integer interpretation, defining "0^0" as one makes some sense, although even then, I find the justification is still pretty much "hand waving in the air" - as in the "obvious" set-theoretic interpretation, which I find quite unconvincing.