User talk:MvH

Re: 0^0
It is simple high-school algebra to obtain the results of the remaining exponents of 0 and 1 -- except for 0^0, in which case any value works.--Danchristensen (talk) 15:38, 1 May 2015 (UTC) — Preceding unsigned comment added by Danchristensen (talk • contribs)


 * Dan, it is even simpler to start with m=0, then you don't even need high-school algebra.  MvH (talk) 15:44, 1 May 2015 (UTC)MvH


 * Simpler perhaps, but there is no apparent justification for attaching any particular value to 0^0. You could only justify a particular value if you insisted that there was more to exponentiation than just repeated multiplication. Did you read my original posting? Starting from n+2=S(S(n)), n+3=S(S(S(n))) .... you can fully determine n+0 and n+1. Starting from n*2 = n+n, n*3=n+n+n .... you can fully determine n*0 and n*1. Starting from n^2=n*n, n^3=n*n*n ... you can fully determine n^0 and n^1 except for 0^0.--Danchristensen (talk) 16:11, 1 May 2015 (UTC)


 * Dan, the value of 0^0 follows from the notation itself, if we use the simplest possible definition/notation. There is no need to prove that x^2 y^3 = xxyyy. This equation is true in any ring.  Why?  Simply because x^2 y^3 is short notation for xxyyy.  Likewise, if we keep the notation as simple as possible then x^1 y^3 simply denotes xyyy and x^0y^3 denotes yyy.   The equation x^0y^3 = y^3 is true in every ring, simply because both sides denote yyy  (if we use the simplest possible definition/notation)    MvH (talk) 16:39, 1 May 2015 (UTC)MvH


 * Sorry, I don't find that very convincing. Thanks anyway.--Danchristensen (talk) 17:11, 1 May 2015 (UTC)


 * Dan, a yes/no question: if everything else were the same [footnote: that's not the case here, lets get back to that in a later post] wouldn't it be preferable to choose a definition that is shorter/simpler/more natural?
 * What would be the simplest possible definition? x^n = product of n copies of x.  Doesn't get any shorter than that.  So x^0 = product of no copies of x.  Before one has even written down an interpretation for that, it is already clear from the beginning that if "product of no copies of x" has a meaning, it'll have to be something that is independent of x because x doesn't even appear in a list of zero copies of x. So if x^0 has a value for any x, then it should have the same value for every x.
 * Now of course, simplicity and brevity aren't the only requirements for a good definition, there are other, more important, requirements. We'll look at that in a later post, but let me start that discussion with asking: what would you say are the most important requirement(s) for a good definition?   MvH (talk) 01:40, 2 May 2015 (UTC)MvH


 * I would require that any definition would accurately model what you might call "numerical reality" (to coin a phrase). If you are not absolutely sure what value 0^0 should be, you shouldn't just assume some arbitrary value for convenience or "simplicity." --Danchristensen (talk) 03:17, 2 May 2015 (UTC)


 * Let me propose some: A definition in math (a) should be consistent, and (b) must describe what mathematicians actually do. Can we agree that those are key requirements? (PS. you haven't answered the other yes/no question yet).  MvH (talk) 03:31, 2 May 2015 (UTC)MvH


 * This won't answer your question directly, but, for an exponentiation function ^ as strictly repeated multiplication on N, ^ must satisfy two conditions: 1) For all n in N, n^2 = n*n, and 2) for all n, m in N, n^(m+1) = n^m * n. (Can you think of any others?)  Why start with exponents of 2? Why not start at 0 or 1? Simply because to multiply, you need at least two numbers. You can't multiply one or zero numbers. We can ask, however, of all the binary functions ^ on all of N, which of them satisfy these two conditions? There are infinitely many such functions, but they differ ONLY for the value assigned to 0^0. So, these two conditions determine all but the value of 0^0. What do you make of that? --Danchristensen (talk) 13:20, 3 May 2015 (UTC)


 * Yes, if you ignore requirement (b) then there are infinitely many choices. If you define a new function, then you can define it anyway you like. But this is not true if you define a function that is already widely used. Then you can either (1) match current usage, or (2) use different notation (unless of course your goal is: (3) cause confusion! There are situations where (3) is the final result, for instance, there are two definitions of natural numbers, and since neither definition is dominant, both definitions have become useless.).  MvH (talk) 17:21, 3 May 2015 (UTC)MvH


 * Or I could just say, at the beginning of any of my proofs that I am assuming 0^0 is also undefined on N. Thanks.--Danchristensen (talk) 19:12, 3 May 2015 (UTC)


 * But why would others want to follow this when the alternative is much more convenient, and better in the sense of requirement (b) mentioned before? MvH (talk) 01:28, 4 May 2015 (UTC)MvH


 * Someone might want to use this approach if they were uncomfortable making assumptions based on little more than convenience.--Danchristensen (talk) 02:12, 4 May 2015 (UTC)


 * There is no reason to be uncomfortable, math is very consistent, so there is no need to worry. But to those that do have irrational fears of two zeros and a ^ sign, it would also be irrational for those fears to be alleviated by undefining 0^0 (this won't solve anything because there are plenty of proofs out there (of widely used theorems) that use x^0 = 1 unconditionally (i.e. for all x, including 0)).   MvH (talk) 02:38, 4 May 2015 (UTC)MvH


 * I don't worry much about inconsistency, but I do worry about being wrong. --Danchristensen (talk) 03:19, 4 May 2015 (UTC)

Here is one way to look at it: should we write down the simplest possible definition, or a more complicated one that is specifically designed to keep 0^0 out? (and then add things like: in polynomials, x^0 is 1, in power-series, x^0 is 1, in the binomial theorem, x^0 is 1, etc etc etc) MvH (talk) 15:52, 1 May 2015 (UTC)MvH


 * I don't mean to intrude on a private discussion, but it seemed like it might not be entirely so since it started on an article talk page. I can see Danchristensen's point, as well as MvH's. I thought I'd make the observation that even coming from Dan's perspective, the linked page omits a restriction that one could reasonably make: that we expect the further rules such as ab+c = abac and (ab)c = acbc to hold universally for all values on which the expressions are defined. Putting a = b = c = 0, there are not an infinitude of possible values for z = 00, but rather only values that are idempotent under multiplication. This means that in any ring without nonzero zero divisors, the only consistent possibilities are z = 0 and z = 1. Not unique, as Dan's perspective would require, but closer to it. —Quondum 22:31, 3 May 2015 (UTC)


 * Putting two zeros in a row makes people do strange things. He has an entire web-page devoted to a triviality presented as a discovery, and, as you point out, he didn't even fully cover that. Unrelated, but even more absurd, I found linear algebra texts that said that A^0 is undefined if A is the zero-matrix, and the identity matrix if A is a non-zero matrix. In other words, if A is a non-zero nilpotent 2x2 matrix, then (A^0)^2 is the identity matrix, but (A^2)^0 is undefined! Oh, the ambiguity indeed! MvH (talk) 03:04, 4 May 2015 (UTC)MvH
 * Heheh. Yes, I sort of enjoy noticing inconsistencies like this. I'd put this matrix example down to a slip, and assume that had the person thought it through, they would have said to make A0 undefined for any singular matrix A. They might not have put much thought into it. Anyhow, I thought you might like the idea that there are, in a wide range of contexts, only two algebraically consistent choices: 1 and undefined (see my last edit to Exponentiation). —Quondum 03:24, 4 May 2015 (UTC)
 * Here's another one that is remarkably common (see e.g. wolfram math world).
 * Argument[*]: f(0,a) = 0 for all a>0 implies f(0,0) = 0
 * I think the issue is more psychological than mathematical. If something has been learned a long time ago then most people will strongly resist any changes, becoming unable to judge rational arguments. In most situations, the people at wolfram are smart enough to see a gap in argument[*], however, apply the same argument[*] to a deeply held belief, then the brain suddenly becomes unable to see such gaps. The same issue is behind the Pluto/planet conflict. People learned in kindergarten that Pluto is a planet, but for many people it is impossible to accept that something learned a long time ago is suddenly no longer true.
 * PS. About your edit, x = 1/x only shows that x = 1 or -1. You need to add x^2 = x to rule out x=-1.  MvH (talk) 15:07, 4 May 2015 (UTC)MvH
 * You describe the meta-picture nicely. It is very prevalent, including amongst most mathematicians. On the PS: We do not suddenly "forget" rules such as x^(n+m) = x^n*x^m, so x=-1 is ruled out anyway.
 * True, x=-1 is ruled out by the other rules (it was just a reminder that one rule isn't enough). Your argument applies to the Riemann sphere (not a ring!) if one drops the requirement that b is not zero in section negative exponents. In the Riemann sphere, 0^x = infinity for x 0, so neither value (0 or infinity) is a natural (i.e. symmetric) choice for x=0. Symmetric choices here are 1 and -1, with -1 eliminated by the other rules.
 * About logic among mathematicians, sometimes you see proofs like this: "Since x^0 = 1 (for x nonzero), it follows that ...." where the ... is a statement that uses x^0=1 for every x.  Even when the proof requires "for all x" people still write "for nonzero x". This was the case (I fixed that) in our page on the binomial theorem. MvH (talk) 16:54, 4 May 2015 (UTC)MvH
 * Yeah, I see that. I guess that's something people get wrong because it is so "automatic" – we do not prove that 1 + 1 = 2, we just use it. And some things people get confused, like when we are assuming that x_0 = 1. That binomial theorem error is a good example. —Quondum 17:27, 4 May 2015 (UTC)


 * As an afterthought, I translated this to wheel theory, which can be enlightening with exactly this type of algebraic question, since it provides a consistent framework in which to investigate ∞ and "undefined" (0/0) algebraically. In general, one would seek to define any function in terms of operations on the values, rather than as the solution to some constraints – e.g., one would not define y = 1/x as the solution to the equation yx = 1 (which is not always unique), but rather as a primitive operation. There are now four possible values for 00, corresponding to the four idempotents of the wheel: 0, 1, ∞ and 0/0. It seems that we should require that aba−b = 1 (or, as in typical in wheel theory, an operation is allowed to produce 0/0 without satisfying the equation as stated). The only consistent values now are 1 and 0/0: we can eliminate 0 and ∞ as values, leaving only two options: 1 and 0/0 ("undefined" when we translate back to a field). We have not beaten the uniqueness requirement, but we have a fair argument for eliminating the pesky contender z = 0 (as well as the hidden z = ∞). That is to say, any integral domain in which we define a consistent operation of division (read field), the only algebraically "sensible" choices are 1 and "undefined". —Quondum 23:35, 3 May 2015 (UTC)


 * I can't comment on wheel theory since I know nothing about it, but whether you have infinite possibilities, or only two, you still have ambiguity. Note that the addition function + on N is uniquely determined by: 1) For all n in N, n+2 = S(S(n)), and (2) for all n,m in N, n+S(m)=S(n+m). Likewise, the multiplication function * on N is uniquely determined by: 1) For all n in N, n*2=n+n, and (2) for all n, m in N, n*(m+1) = n*m + n. Similarly, the exponentiation function ^ on N is uniquely determined (except for 0^0) by 1): For all n in N, n^2=n*n, and (2) for all n, m in N, n^(m+1) = n^m * n. --Danchristensen (talk) 02:56, 4 May 2015 (UTC)


 * Ignore the bit about wheels – I just use it as a guide for my intuition. A simpler argument is this: it is reasonable to require the other identities such as ab+c = abac must not be violated when the expressions involved are all defined. The only values that satisfy this law are 00 = 0 and 00 = 1. Hardly an infinitude! In the context where we want a−b = 1/ab (which will be the case when exponentiation with negative exponents is defined), we are down to a unique value: 00 = 1. On the real numbers with any type of exponent (discrete or continuous), this approach completely kills the non-uniqueness argument. —Quondum 03:13, 4 May 2015 (UTC)


 * You can formulate all the usual Laws of Exponents even if 0^0 is undefined e.g. if x=/= 0 then x^(y+z) = x^y * x*z, just like they taught us in high school. Yours is just a variation on the convenience argument. --Danchristensen (talk) 03:35, 4 May 2015 (UTC)


 * Yes. I'm not arguing about that: my position is that the choice is between leaving it undefined or defining it as 1. I am only pointing out that the uniqueness argument shoots itself in the foot, in the sense that there is a unique value that satisfies the laws in most cases of interest, and on its premises it would conclude that 00 should be 1, because it is the unique value that satisfies the exponentiation laws. The argument fails to consider that even when the solution is unique, leaving it undefined is equally valid. —Quondum 03:41, 4 May 2015 (UTC)


 * 0^0=0 also satisfies your Laws of Exponents. --Danchristensen (talk) 04:10, 4 May 2015 (UTC)


 * No, it doesn't, in the context of negative exponents being defined, where we would have the law that b−n = 1/bn, or equivalently, that b−nbn = 1 if bn is defined. Putting b = 0, n = 0, we get 0−000 = (00)2 = 1, or 00 = ±1. Together with 00+0 = 0000, the only solution is 00 = 1. To claim that 1/0 is undefined and hence that this (i.e. the identity 0 = 1/0) is not a problem would be horribly kludgey, and would be algebraically untenable in the projectively extended real numbers where this is defined. —Quondum 05:02, 4 May 2015 (UTC)


 * I think you will find, for example, that if you set 0^0=0, then you will also get x^y * x^z = x^(y+z) for all x,y,z in N. IIRC, this is true for the other Laws of Exponents on N as well. --Danchristensen (talk) 05:49, 4 May 2015 (UTC)


 * I'm aware of that several of the "laws" are satisfied (b1=b, bm+n=bm⋅bn, bmn=(bm)n, (bc)n=bn⋅cn) all hold. In my book, there are exactly four distinct choices on the value of 00 for which these hold, as I have already indicated: 0, 1, ∞, undefined. This is true in any ring. (I think you'll concede that this is not an infinity of possibilities, so we can say that the blog page can safely be disregarded as giving a respectable argument.) This is why I specifically qualified this for the context of negative exponents, where it is reasonable to expect that the identity bn⋅b−n=1 holds if the individual terms are defined. The uniqueness argument is a bad argument (there is no logical basis for it), we are left with that we can leave 00, 01, 02, 03 up to any power we choose undefined, and all the laws still hold. So the argument that we should only constrain the function to what satisfies the initial laws also does not fly: we cannot determine the "correct" function domain. The "best" choice is to leave 0n undefined for all n. Without the requirement bn⋅b−n=1, this argument goes the other way too: the choice of 0n=0 for all n (including n<0) satisfies all the laws. It is decidedly nicer than the existing accepted definition, because it defines bn consistently for all pairs (b,n). When b=0, we are left with an inherent arbitrariness – we are forced to make a choice. —Quondum 14:07, 4 May 2015 (UTC)


 * We seem to be going in circles here. I will summarize my position and leave it at this. If the ONLY requirements for the exponentiation function ^ on N are that (1) for all n in N, n^2=n*n, and (2) for all n, m in N, n^(m+1) = n^m * n, then there are infinitely many possibilities for ^ on ALL of N. If you add other conditions, you may well be able to narrow down the possibilities. Interestingly though, ^ is uniquely determined on ALL of N by only these two conditions, except for the value assigned 0^0 for which any value in N will work. It may then make sense to represent ^ as a partial function on N, leaving only 0^0 undefined. Using such a partial function, we can derive the usual Laws of Exponents with certain restrictions to avoid a zero base and exponent -- just like we learned in high school. Nothing new or radical here. A novel argument for an old idea perhaps, but nothing more. --Danchristensen (talk) 15:04, 4 May 2015 (UTC)


 * Even starting from these premises, your conclusion is incorrect: strictly speaking, none of the choices for n^m for m<2 are uniquely determined by these conditions: leaving them out of the domain is a valid choice (this is something you assume by allowing a partial function). Using the fuzzy guideline of preferring a uniquely determined defined value that extends the formulas over leaving a point out of the domain, we still cannot determine 0^1 if 0^0 is undefined. Thus, we must leave both 0^0 and 0^1 out of the table unless you actually choose to have a defined value for 0^0. It is fallacious to say that "if 0^0 was a finite value, then 0^1 is always 0", then conclude that 0^1=0 and then proceed to leave 0^0 undefined. This is a logical fallacy. If we can use it for 0^1, we can conclude that 0^0=0 too. Your choice of premises is also arbitrary, so the whole argument says nothing about what b^n "should" be. —Quondum 16:27, 4 May 2015 (UTC)


 * Whatever value you may assign to 0^0, when you apply (2) for n=m=0, you obtain 0^1 = 0^(0+1) = 0^0 * 0 = 0. So, in every one of the functions ^ on all of N that satsify these conditions, you must have 0^1=0. But you are correct in the sense that when defining the partial function ^, we must specify 0^1=0. --Danchristensen (talk) 16:51, 4 May 2015 (UTC)


 * Only if you do assign some finite value to it. You don't assign a value to it.  Your premise is not satisfied, so your conclusion fails. As I pointed out, the same (fallacious) logic proves that 0^0=0: whatever value you assign to x−1=0−1, we obtain x0=x−1⋅0=0.


 * You are confused, perhaps rightly so. I am overloading the ^-symbol in some sense. First, I said that there are infinitely many functions ^ on all of N that satisify the above conditions. For each of these, a value is indeed assign to 0^0. Then I talk about another function ^, a partial function this time for which 0^0 is undefined. We could define it as follows: (1) For all n in N, if n=/=0 then 0^n=1, (2) for all n, m in N, if ~[n=m=0] then n^(m+1) = n^m * n, and (3) 0^1=0.--Danchristensen (talk) 17:33, 4 May 2015 (UTC)


 * But Dan, there is absolutely no good reason to do any of this. You mentioned earlier that you worry that something is wrong, but in math that's not a valid argument. Math is based on proofs. Either there is a proof, or there isn't (in both cases, worries won't count as an argument). And if there really were a proof that defining 0^0=1 is wrong, simply undefining 0^0 wouldn't even begin to solve the problem because 0^0=1 is already used in other proofs anyway. You'd have to find and "correct" all of those too. You can continue to believe that there's something wrong with 0^0 but keep in mind, it's just a belief, there's no rational justification for it.  Plenty of people worry about the number zero but that doesn't make it a valid argument!  MvH (talk) 12:47, 5 May 2015 (UTC)MvH


 * Is 0^0 undefined or not? Oh, that depends on whether you are talking about 0 the real number or 0 the natural number. How "rational" is that? I just think there is a solid, number-theoretic rationale (see above) for leaving 0^0 undefined for the natural numbers as well. Yes, there may be an element of subjectivity here -- the "usefulness" of one definition vs. another -- so readers will have to judge for themselves. Thank you for your time and patience. --Danchristensen (talk) 14:41, 5 May 2015 (UTC)

Proof by contradiction
The reason why I deleted the argument for the infinitude of primes from the article on here. Michael Hardy (talk) 03:22, 12 April 2015 (UTC)

Exponentiation
Thank you for your fine work regarding 00. Bo Jacoby (talk) 21:22, 24 April 2015 (UTC).

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