Wikipedia:Reference desk/Archives/Mathematics/2020 August 29

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August 29

Is 0 a perfect number??

Read the Iteration section of Aliquot sum. There's something in that section that if taken literally it would mean 0 is a perfect number, because a perfect number is its own aliquot sum. Georgia guy (talk) 18:32, 29 August 2020 (UTC)

No. Every number divides zero, so the aliquot sum would be the sum of all natural numbers (other than zero, I guess — whether zero divides zero is another can of worms; the right answer is "yes" but it's a controversial point). --Trovatore (talk) 18:36, 29 August 2020 (UTC)

The can-o-worms. 107.15.157.44 (talk) 20:04, 29 August 2020 (UTC)

Well, no, that's not really the issue. There's no question that, in any ordinary context, you can't divide zero by zero.

The question is how the "divides" relation, denoted |, should be defined on the natural numbers (including zero).

The right answer is not "a divides b if a/b is a natural number".

Rather, the right answer is, "a divides b if there exists a natural number k such that b=ka".

By that definition, clearly 0|0.

Now, why is that the "right" definition? Because it makes divisibility on the natural numbers into a very nice lattice, with 1 being the minimum element and 0 being the maximum element, and in which the greatest common divisor is the meet and the least common multiple is the join.

By this definition, note gcd(0,0)=0 and lcm(0,0)=0. These are the correct definitions. They make Euclid's algorithm run smoothly and convincingly with no special cases. You can find writeups of the algorithm that say things like, we'll just take gcd(0,0) to be 0 for the nonce to make the description of the algorithm cleaner even though it's not really correct. But that's nonsense — it is really correct, when you use the right definitions. --Trovatore (talk) 20:22, 29 August 2020 (UTC)

The funny part is, the aliquot sum is of course the same whether it contains zero or not... 93.136.18.63 (talk) 21:47, 29 August 2020 (UTC)

A very similar situation applies to 0⁰ (see Zero to the power of zero).  --Lambiam 22:26, 29 August 2020 (UTC)

Well, that's a little different. Or rather, it's the same when you're thinking of the natural number 0. The problem is, a lot of people seem to be very invested in the natural number 0 and the real number 0 being the same thing. The arguments for defining 0^0 as 1 when you're considering real-valued exponents are much less convincing. --Trovatore (talk) 22:56, 29 August 2020 (UTC)

Trovatore, what's the difference?? The natural numbers are the positive integers; the real numbers are the rational and irrational numbers combined (they include all numbers that have a point on the number line.) Georgia guy (talk) 23:21, 29 August 2020 (UTC)

The natural numbers (which may or may not include 0, but I think the trend is to include 0) and the real numbers are distinct structures. There's a natural embedding from the naturals to the reals, and for most purposes it works fine if you consider this embedding to be the identity, which would make the naturals a subset of the reals.

But in a few cases, it makes sense to distinguish. The reals and the naturals have very different underlying intuitions. The naturals count things; the reals measure things. It's not obvious that the zero of measuring is the same as the zero of counting.

By the way, it's sort of circular to say that the reals are the rationals and irrationals combined. The irrationals are defined as reals that are not rational. So you have to know what reals are in the first place. --Trovatore (talk) 23:33, 29 August 2020 (UTC)

What's the difference between the 2 zeros you're talking about?? Georgia guy (talk) 00:17, 30 August 2020 (UTC)

One of them is the number of things in the empty set, and the other is the distance from where you are to where you are. --Trovatore (talk) 00:46, 30 August 2020 (UTC)

Which of the two is π − π?  --Lambiam 14:49, 30 August 2020 (UTC)

The real number. --Trovatore (talk) 18:33, 30 August 2020 (UTC)

Then I guess that by your reckoning π/π is held in a different cage at the number zoo than the successor of zero.  --Lambiam 05:35, 31 August 2020 (UTC)

True. --Trovatore (talk) 20:02, 31 August 2020 (UTC)

Just chiming in here - I think Trovatore is correct - in the sense that an overwhelming plurality of mathematicians would recognize the way he's constructing his statements. The fact that some readers find his statements to be non-obvious, or at least unfamiliar, is really a matter of how much time the reader spends formally studying mathematical theory. Many times before, I have recommended this book: Mathematical Proofs: A Transition to Advanced Mathematics. It will help smooth over the weirdness for an intelligent but uninitiated reader. Nimur (talk) 16:41, 3 September 2020 (UTC)

I'm well aware of how real numbers are formally introduced as equivalence classes of Cauchy sequences or as Dedekind cuts, but it is common practice among working mathematicians to identify the member of $\mathbf {Z}$ that is the equivalence class of pairs of the form $(n,n)$, $n\in \mathbf {N}$, with the element $0$ of $\mathbf {N}$, and likewise the member of $\mathbf {Q}$ that is the equivalence class of pairs of the form $(a,a)$, $a\in \mathbf {Z\setminus \{0\}}$, with the element $1$ of $\mathbf {Z}$ and thus of $\mathbf {N}$. A category theorist will not like this and keep the injections (or in category speak monomorphisms) $\mathbf {N} \hookrightarrow \mathbf {Z} \hookrightarrow \mathbf {Q}$ explicit. But I bet Trovatore is not a category theorist. It is likewise common practice to leave the usual injection $\mathbf {Q} \hookrightarrow \mathbf {R}$ implicit, which has many practical advantages. Unless you think that $1_{\mathbf {N} }\times \pi$ is type-incorrect, it is perfectly reasonable to maintain that $\pi /\pi =1\in \mathbf {N}$. I see no reason to suddenly renege on this common practice in the case of exponentiation, pace Cauchy, (and thereby throw away the considerable advantage that $x^{0}$ is well defined for all $x$).  --Lambiam 17:31, 3 September 2020 (UTC)

Yeah, it's usually fine to identify the naturals with their images under the natural inclusion into the reals. It's not fine when you're talking about exponentiation. --Trovatore (talk) 17:42, 3 September 2020 (UTC)

Or another way to put it, if you want the naturals to be a literal subset of the reals, is to say that there are (at least) two different exponential functions, having the same name and notated the same way (see operator overloading). The one with domain NxN satisfies 0⁰=1. The one with domain R⁺xR does not satisfy that equation, because the point (0, 0) is not in its domain. Both of these functions are defined naturally and with no special cases; when you try to jam them together into a single function, it doesn't work smoothly anymore, and there's actually no good reason even to try. --Trovatore (talk) 17:47, 3 September 2020 (UTC

The point (-π,2) is not in the domain of either; how many different exponentiation functions do we need, and should we need to specify which one is intended when we write down a formula using exponentiation?  --Lambiam 08:13, 4 September 2020 (UTC)

Well, lots, actually, and sometimes you need to decide by context. For example ordinal exponentiation and cardinal exponentiation are completely different functions, though they happen to agree on the finite ordinals.

But the ones where the exponent is a natural number all have a lot in common. Conceptually, they're all repeated multiplication. (−π)² is the product of 2 copies of −π (provided this is in a context where 2 is taken to be a natural number). And if the base is taken from a structure that has a multiplicative identity, then it's pretty canonical to assign that identity to the product of no copies of the base, even if the base is 0.

Once the exponent is a real or complex number, the idea of "repeated multiplication" makes no sense whatsoever. Really really none at all. I think this is the reason that some people are overly impressed with the equation e^iπ=−1. It would be very mysterious if it meant that the product of iπ copies of e is −1. But it doesn't mean that at all. In fact that statement is plain nonsense and completely meaningless. You can't multiply iπ copies of anything.

We do have a function that's reasonable to call "exponentiation", that makes sense with real (or complex) exponents. It's the one defined by x^y=e^y log x, where the exponential function is defined (say) by its power series.

But the fact that this function happens to agree with the one defined by repeated multiplication, on their common domain, is ... sort of a coincidence. I mean, not exactly a coincidence; you can prove it after all. But it's conceptually completely different. Really really completely. Almost nothing in common at all. Kind of like the relationship between factorials and the Gamma function.

So I don't see any justification whatsoever to force-define a value for 0⁰ for this second function. --Trovatore (talk) 02:27, 5 September 2020 (UTC)

A perfect number is defined to be a positive number with a certain property. 0 is not positive so it can't be a perfect number, it doesn't matter if it satisfies the other property or not. Similarly Aliquot sum is only defined for positive number. One reason for restriction the definitions to positive numbers is a avoid confusion about what happens with 0. --RDBury (talk) 22:21, 30 August 2020 (UTC)

That's an excellent point, formally, but the OP's question is still a reasonable direction to inquire: "What happens if I remove this seemingly arbitrary restriction?" Why the OP thought 0 had no proper divisors I'm not quite sure, but if that were true, then it would be reasonable to ask why 0 isn't considered a perfect number. --Trovatore (talk) 22:55, 30 August 2020 (UTC)