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December 6

Is there anything that would prevent peforming Weil Descent on binary curves of large characteristics ?

The ghs attack involve creating an hyperlliptic curve cover for a given binary curve. The reason the attack fails most of the time is the resulting genus grows exponentially relative to the curve’s degree.

We don’t hear about the attack on finite fields of large characteristics since such curves are already secure by being prime. However, I notice a few protocol relies on the discrete logarithm security on curves with 400/500 bits modulus resulting from extension fields of characteristics that are 200/245bits long.

Since the degree is most of the time equal to 3 or 2, is there anything that would prevent creating suitable hyperelliptic cover for such curves in practice ? 2A01:E0A:401:A7C0:28FE:E0C4:2F97:8E08 (talk) 12:09, 6 December 2024 (UTC)

December 7

Mathematical operation navigation templates

RDBury is right, this discussion belongs at Wikipedia talk:WikiProject Mathematics
The following discussion has been closed. Please do not modify it.
If anyone with some mathematical expertise is interested, I'd appreciate some additional input at Talk:Exponentiation#funny table at end. The question is whether our articles on various mathematical operations could use a navigational template (aka "{{Navbox}}"). Our Exponentiation article tried to use {{Mathematical expressions}} for this purpose, but it doesn't really work. I've created {{Mathematical operations}} as a potential alternative, but the categorization and presentation I've created is probably naïve. (The whole effort may or not be worth it at all.) —scs (talk) 00:36, 7 December 2024 (UTC) Wikipedia talk:WikiProject Mathematics is a better forum for this kind of thing, since it's focused on Wikipedia's mathematical articles. --RDBury (talk) 04:07, 7 December 2024 (UTC) @RDBury: Excellent point. Thanks. —scs (talk) 13:49, 7 December 2024 (UTC)

December 8

For each positive integer $n$, which primes $p$ are still primes in the ring $Z[e^{\frac {\pi i}{n}}]$?

For each positive integer $n$, which primes $p$ are still primes in the ring $Z[e^{\frac {\pi i}{n}}]$? When $n=1$, $Z[e^{\frac {\pi i}{n}}]$ is the original integer ring, when $n=2$, $Z[e^{\frac {\pi i}{n}}]$ is the ring of Gaussian integers, when $n=3$, $Z[e^{\frac {\pi i}{n}}]$ is the ring of Eisenstein integers, and the primes in the Gaussian integers are the primes $p\equiv 3\mod 4$, and the primes in the Eisenstein integers are the primes $p\equiv 2\mod 3$, but how about larger $n$? 218.187.66.163 (talk) 04:50, 8 December 2024 (UTC)

A minuscule contribution: for $n=4,$ the natural Gaussian primes $3$ and $7$ are composite:

• $3=(1-\omega -\omega ^{2})(1+\omega ^{2}+\omega ^{3}).$
• $7=(2-\omega +2\omega ^{2})(2-2\omega ^{2}-\omega ^{3}).$

So $11$ is the least remaining candidate.  --Lambiam 09:00, 8 December 2024 (UTC)

It is actually easy to see that $7$ is composite, since $2$ is a perfect square:

$2=i(1-i)^{2}=\omega ^{2}(1-\omega ^{2})^{2}.$

Hence, writing ${\sqrt {2}}$ by abuse of notation for $\omega (1-\omega ^{2}),$ we have:

${\begin{alignedat}{2}7&=(3+{\sqrt {2}})(3-{\sqrt {2}})\\&=(2{\sqrt {2}}+1)(2{\sqrt {2}}-1)\\&=(5+3{\sqrt {2}})(5-3{\sqrt {2}})\\&=(4{\sqrt {2}}+5)(4{\sqrt {2}}-5).\end{alignedat}}$

More in general, any natural number that can be written in the form $|2a^{2}-b^{2}|,a,b\in \mathbb {N} ,$ is not prime in $\mathbb {Z} [e^{\pi i/4}].$ This also rules out the Gaussian primes $23,$ $31,$ $47,$ $71$ and $79.$  --Lambiam 11:50, 8 December 2024 (UTC)

So which primes $p$ are still primes in the ring $Z[e^{\frac {\pi i}{4}}]$? How about $Z[e^{\frac {\pi i}{5}}]$ and $Z[e^{\frac {\pi i}{6}}]$? 220.132.216.52 (talk) 06:32, 9 December 2024 (UTC)

As I wrote, this is only a minuscule contribution. We do not do research on command; in fact, we are actually not supposed to do any original research here.  --Lambiam 09:23, 9 December 2024 (UTC)

Moreover, $-2$ is also a perfect square. (As in the Gaussian integers, the additive inverse of a square is again a square.) So natural numbers of the form $2a^{2}+b^{2}$ are also composite. This further rules out $11,$ $19,$ $43,$ $59,$ $67$ and $83.$ A direct proof that, e.g., $11$ is composite: $11=(1+\omega ^{2}+3\omega ^{3})(1-3\omega -\omega ^{2}).$ There are no remaining candidates below $100$ and I can in fact not find any larger ones either. This raises the conjecture:

Every prime number can be written in one of the three forms $a^{2}+b^{2},$ $2a^{2}+b^{2}$ and $|2a^{2}-b^{2}|.$

Is this a known theorem? If true, no number in $\mathbb {Z} [e^{\pi i/4}]$ is a natural prime. (Note that countless composite numbers cannot be written in any of these forms; to mention just a few: $15,1155,2491.$)  --Lambiam 11:46, 9 December 2024 (UTC)

I'll state things a little more generally, in the cyclotomic field $\mathbb {Q} [e^{2\pi i/n}]$. (Your n is twice mine.) A prime q factors as $q=(q_{1}\cdots q_{r})^{e_{r}}$, where each $q_{i}$ is a prime ideal of the same degree $f$, which is the least positive integer such that $q^{f}\equiv 1{\pmod {n}}$. (We have assumed that q does not divide n, because if it did, then it would ramify and not be prime. Also note that we have to use ideals, because the cyclotomic ring is not a UFD.) In particular, $q$ stays prime if and only if $q$ generates the group of units modulo $n$. When n is a power of two times an odd composite, the group of units is not cyclic, and so the answer is never. When n is a prime or twice a prime, the answer is when q is a primitive root mod n. If n is 4 times a power of two times a prime, the answer is never. Tito Omburo (talk) 11:08, 8 December 2024 (UTC)

For your $n$, $n=3$ and $n=6$ are the same, as well as $n=5$ and $n=10$, this is why I use $e^{\frac {\pi i}{n}}$ instead of $e^{\frac {2\pi i}{n}}$. 61.229.100.34 (talk) 20:58, 8 December 2024 (UTC)

Also, what is the class number of the cyclotomic field $Q[e^{\frac {\pi i}{n}}]$? Let $\gamma (n)$ be the class number of the cyclotomic field $Q[e^{\frac {\pi i}{n}}]$, I only know that:

• $\gamma (n)=1$ for $n=1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,24,25,27,30,33,35,42,45$ (is there any other such $n$)?
• If $m$ divides $n$, then $\gamma (m)$ also divides $\gamma (n)$, thus we can let $\varphi (n)=\prod _{d|n}\gamma (n)^{\mu (n/d)}$
• For prime $p$, $p$ divides $\varphi (p)$ if and only if $p$ is Bernoulli irregular prime
• For prime $p$, $p$ divides $\varphi (2p)$ if and only if $p$ is Euler irregular prime
• $\varphi (n)=1$ for $n=1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,24,25,27,30,33,35,42,45$ (is there any other such $n$)?
• $\varphi (n)$ is prime for $n=23,26,28,32,36,37,38,39,40,43,46,49,51,53,54,63,64,66,69,75,81,96,105,118,128,...$ (are there infinitely many such $n$?)

Is there an algorithm to calculate $\gamma (n)$ quickly? 61.229.100.34 (talk) 21:14, 8 December 2024 (UTC)

Can we say anything special about every pair of functions f,g, satisfying f(g(x))=f(x) for every x?

Especially, is there an accepted term for such a pair?

Here are three simple examples, for two functions f,g, satisfying the above, and defined for every natural number:

Example #1:

f is constant.

Example #2:

f(x)=g(x), and is the smallest even number, not greater than x.

Example #3:

f(x)=1 if x is even, otherwise f(x)=2.

g(x)=x+2.

2A06:C701:746D:AE00:ACFC:490:74C3:660 (talk) 09:31, 8 December 2024 (UTC)

One way to consider such a pair is dynamically. If you consider the dynamical system $x\mapsto g(x)$, then the condition can be stated as "$f$ is constant on $g$-orbits". More precisely, let $D$ be the domain of $f,g$, which is also the codomain of $g$. Define an equivalence relation on $D$ by $x\sim y$ if $g^{a}(x)=g^{b}(y)$ for some positive integers $a,b$. Then $f$ is simply a function on the set of equivalence classes $D/\sim$ (=space of orbits). In ergodic theory, such a function $f$ is thought of as an "observable" or "function of state", being the mathematical analog of a thermodynamic observable such as temperature. Tito Omburo (talk) 11:52, 8 December 2024 (UTC)

After you've mentioned temprature, could you explain what are f,g, as far as temprature is concerned? Additionally, could you give another useful example from physics for such a pair of functions? 2A06:C701:746D:AE00:ACFC:490:74C3:660 (talk) 19:49, 8 December 2024 (UTC)

This equation is just the definition of function g. For instance if function f has the inverse function f⁻¹ then we have g(x)=x. Ruslik_Zero 20:23, 8 December 2024 (UTC)

If f is the temperature, and g is the evolution of an ensemble of particles in thermal equilibrium (taken at a single time, say one second later), then because temperature is a function of state, one has $f(x)=f(g(x))$ for all ensembles x. Another example from physics is when $g$ is a Hamiltonian evolution. Then the functions $f$ with this property (subject to smoothness) are those that (Poisson) commute with the Hamiltonian, i.e. "constants of the motion". Tito Omburo (talk) 20:33, 8 December 2024 (UTC)

Thx. 2A06:C701:746D:AE00:ACFC:490:74C3:660 (talk) 10:43, 9 December 2024 (UTC)

Let $f$ be a function from $X$ to $Y$ and $g$ a function from $X$ to $X.$ Using the notation for function composition, the property under discussion can concisely be expressed as $f\circ g=f.$ An equivalent but verbose way of saying the same is that the preimage of any set $B\subseteq Y$ under $f$ is closed under the application of $g.$  --Lambiam 08:54, 9 December 2024 (UTC)

Thx. 2A06:C701:746D:AE00:ACFC:490:74C3:660 (talk) 10:43, 9 December 2024 (UTC)

IEEE Xplore paper claim to acheive exponentiation inversion suitable for pairing in polynomial time. Is it untrustworthy ?

I just found https://ieeexplore.ieee.org/abstract/document/6530387. Given the multiplicative group factorization in the underlying finite field of a target bn curve, they claim to acheive exponentiation inversion suitable for pairing inversion in seconds on a 32 bits cpu.

On 1 side, the paper is supposed to be peer reviewed by the iee Xplore journal and they give examples on 100 bits. On the other side, in addition to the claim, their algorithm 2 and 3 are very implicit, and as an untrained student, I fail to understand how to implement them, though I fail to understand things like performing a Weil descent.

Is the paper untrustworthy, or would it be possible to get code that can be run ? 2A01:E0A:401:A7C0:152B:F56C:F8A8:D203 (talk) 18:53, 8 December 2024 (UTC)

About the paper, I agree to share the paper privately 2A01:E0A:401:A7C0:152B:F56C:F8A8:D203 (talk) 18:54, 8 December 2024 (UTC)

December 9

If the Mersenne number 2^p-1 is prime, then must it be the smallest Mersenne prime == 1 mod p?

If the Mersenne number 2^p-1 is prime, then must it be the smallest Mersenne prime == 1 mod p? (i.e. there is no prime q < p such that 2^q-1 is also a Mersenne prime == 1 mod p) If p is prime (no matter 2^p-1 is prime or not), 2^p-1 is always == 1 mod p. However, there are primes p such that there is a prime q < p such that 2^q-1 is also a Mersenne prime == 1 mod p:

• 2^19-1 is Mersenne prime == 1 mod 73 (p=73, q=19)
• 2^31-1 is Mersenne prime == 1 mod 151 (p=151, q=31)
• 2^61-1 is Mersenne prime == 1 mod 151 (p=151, q=61)
• 2^17-1 is Mersenne prime == 1 mod 257 (p=257, q=17)
• 2^31-1 is Mersenne prime == 1 mod 331 (p=331, q=31)
• 2^61-1 is Mersenne prime == 1 mod 331 (p=331, q=61)
• 2^127-1 is Mersenne prime == 1 mod 337 (p=337, q=127)
• 2^89-1 is Mersenne prime == 1 mod 353 (p=353, q=89)
• 2^89-1 is Mersenne prime == 1 mod 397 (p=397, q=89)

but for these primes p, 2^p-1 is not prime, and my question is: Is there a prime p such that 2^p-1 is a prime and there is a prime q < p such that 2^q-1 is also a Mersenne prime == 1 mod p?

• If 2^11-1 is prime, then this is true, since 2^11-1 is == 1 mod 31 and 2^31-1 is prime, but 2^11-1 is not prime
• If 2^23-1 or 2^67-1 is prime, then this is true, since 2^23-1 and 2^67-1 are == 1 mod 89 and 2^89-1 is prime, but 2^23-1 and 2^67-1 are not primes
• If 2^29-1 or 2^43-1 or 2^71-1 or 2^113-1 is prime, then this is true, since 2^29-1 and 2^43-1 and 2^71-1 and 2^113-1 are == 1 mod 127 and 2^127-1 is prime, but 2^29-1 and 2^43-1 and 2^71-1 and 2^113-1 are not primes
• If 2^191-1 or 2^571-1 or 2^761-1 or 2^1901-1 is prime, then this is true, since 2^191-1 and 2^571-1 and 2^761-1 and 2^1901-1 are == 1 mod 2281 and 2^2281-1 is prime, but 2^191-1 and 2^571-1 and 2^761-1 and 2^1901-1 are not primes
• If 2^1609-1 is prime, then this is true, since 2^1609-1 is == 1 mod 3217 and 2^3217-1 is prime, but 2^1609-1 is not prime

Another question: For any prime p, is there always a Mersenne prime == 1 mod p? 220.132.216.52 (talk) 19:03, 9 December 2024 (UTC)

Neither question is easy. For the first, relations $2^{q-1}\equiv 1{\pmod {p}}$ would imply that the integer 2 is not a primitive root mod p, and that its order divides $q-1$ for the prime q. This is a sufficiently infrequent occurrence that it seems likely that all Mersenne numbers could be ruled out statistically, but not enough is known about their distribution. For the second, it is not even known if there are infinitely many Mersenne primes. Tito Omburo (talk) 19:23, 9 December 2024 (UTC)

I found that: 2^9689-1 is the smallest Mersenne prime == 1 mod 29, 2^44497-1 is the smallest Mersenne prime == 1 mod 37, 2^756839-1 is the smallest Mersenne prime == 1 mod 47, 2^57885161-1 is the smallest Mersenne prime == 1 mod 59, 2^4423-1 is the smallest Mersenne prime == 1 mod 67, 2^9941-1 is the smallest Mersenne prime == 1 mod 71, 2^3217-1 is the smallest Mersenne prime == 1 mod 97, 2^21701-1 is the smallest Mersenne prime == 1 mod 101, and none of the 52 known Mersenne primes are == 1 mod these primes p < 1024: 79, 83, 103, 173, 193, 197, 199, 227, 239, 277, 307, 313, 317, 349, 359, 367, 373, 383, 389, 409, 419, 431, 443, 461, 463, 467, 479, 487, 503, 509, 523, 547, 563, 587, 599, 613, 647, 653, 659, 661, 677, 709, 727, 733, 739, 743, 751, 757, 769, 773, 797, 809, 821, 823, 827, 829, 839, 853, 857, 859, 863, 887, 907, 911, 919, 929, 937, 941, 947, 971, 977, 983, 991, 1013, 1019, 1021 220.132.216.52 (talk) 20:51, 9 December 2024 (UTC)

Also,

• 2^19937-1 is Mersenne prime == 1 mod 2^16+1
• 2^521-1 is Mersenne prime == 1 mod 2^13-1
• 2^3021377-1 is Mersenne prime == 1 mod 2^17-1
• 2^2281-1 is Mersenne prime == 1 mod 2^19-1
• 2^21701-1 is Mersenne prime == 1 mod 2^31-1
• 2^19937-1 is Mersenne prime == 1 mod 2^89-1
• 2^86243-1 is Mersenne prime == 1 mod 2^107-1

but none of these primes p has 2^p-1 is known to be prime, the status of 2^(2^89-1)-1 and 2^(2^107-1)-1 are still unknown (see double Mersenne number), but if at least one of them is prime, then will disprove this conjecture (none of the 52 known Mersenne primes are == 1 mod 2^61-1 or 2^127-1), I think that this conjecture may be as hard as the New Mersenne conjecture. 220.132.216.52 (talk) 20:55, 9 December 2024 (UTC)

Also, for the primes p < 10000, there is a prime q < p such that 2^q-1 is also a Mersenne prime == 1 mod p only for p = 73, 151, 257, 331, 337, 353, 397, 683, 1321, 1613, 2113, 2731, 4289, 4561, 5113, 5419, 6361, 8191, 9649 (this sequence is not in OEIS), however, none of these primes p have 2^p-1 prime. 220.132.216.52 (talk) 02:23, 10 December 2024 (UTC)

December 10

More on the above conjecture

Above I posed:

Conjecture. Every prime number can be written in one of the three forms $a^{2}+b^{2},$ $2a^{2}+b^{2}$ and $|2a^{2}-b^{2}|.$

If true, it implies no natural prime is a prime in the ring $\mathbb {Z} [e^{\pi i/4}]$.

The absolute-value bars are not necessary. A number that can be written in the form $-(2a^{2}-b^{2})$ is also expressible in the form $+(2a^{2}-b^{2}).$

It turns out (experimentally; no proof) that a number that can be written in two of these forms can also be written in the third form. The conjecture is not strongly related to the concept of primality, as can be seen in this reformulation:

Conjecture. A natural number that cannot be written in any one of the three forms $a^{2}+b^{2},$ $2a^{2}+b^{2}$ and $2a^{2}-b^{2}$ is composite.

The first few numbers that cannot be written in any one of these three forms are

$15,$ $21,$ $30,$ $35,$ $39,$ $42,$ $55,$ $60,$ $69,$ $70,$ $77,$ $78,$ $84,$ $87,$ $91,$ $93,$ $95.$

They are indeed all composite, but why this should be so is a mystery to me. What do $2310=2\times 3\times 5\times 7\times 11,$ $5893=71\times 83$ and $7429=17\times 19\times 23,$ which appear later in the list, have in common? I see no pattern.

It seems furthermore that the primorials, starting with $5\#=30,$ make the list. (Checked up to $37\!\#=7420738134810.$)  --Lambiam 19:23, 10 December 2024 (UTC)

Quick note, for those like me who are curious how numbers of the form $-(2a^{2}-b^{2})$ can be written into a form of $2a^{2}-b^{2}$, note that $2a^{2}-b^{2}=(2a+b)^{2}-2(a+b)^{2}$, and so $2a^{2}-b^{2}=-p\Rightarrow p=2(a+b)^{2}-(2a+b)^{2}$. GalacticShoe (talk) 02:20, 11 December 2024 (UTC)

A prime is expressible as the sum of two squares if and only if it is congruent to $1\!\!\!{\pmod {4}}$, as per Fermat's theorem on sums of two squares. A prime is expressible of the form $2a^{2}+b^{2}$ if and only if it is congruent to $1,3\!\!{\pmod {8}}$, as per OEIS:A002479. And a prime is expressible of the form $2a^{2}-b^{2}$ if and only if it is congruent to $1,7\!\!{\pmod {8}}$, as per OEIS:A035251. Between these congruences, all primes are covered. GalacticShoe (talk) 05:59, 11 December 2024 (UTC)

More generally, a number is not expressible as:

1. $a^{2}+b^{2}$ if it has a prime factor congruent to $3\!\!\!{\pmod {4}}$ that is raised to an odd power (equivalently, $3,7\!\!{\pmod {8}}$.)
2. $2a^{2}+b^{2}$ if it has a prime factor congruent to $5,7\!\!{\pmod {8}}$ that is raised to an odd power
3. $2a^{2}-b^{2}$ if it has a prime factor congruent to $3,5\!\!{\pmod {8}}$ that is raised to an odd power

It is easy to see why expressibility as any two of these forms leads to the third form holding, and also we can see why it's difficult to see a pattern in numbers that are expressible in none of these forms, in particular we get somewhat-convoluted requirements on exponents of primes in the factorization satisfying congruences modulo 8. GalacticShoe (talk) 06:17, 11 December 2024 (UTC)

Thanks. Is any of this covered in some Wikipedia article?  --Lambiam 10:06, 11 December 2024 (UTC)

All primes? 2 is not covered! 176.0.133.82 (talk) 08:00, 17 December 2024 (UTC)

$2$ can be written in all three forms: $2=1^{2}+1^{2}=2\cdot 1^{2}+0^{2}=2\cdot 1^{2}-0^{2}.$  --Lambiam 09:38, 17 December 2024 (UTC)

I don't say it's not covered by the conjecture. I say it's not covered by the discussed classes of remainders. 176.0.133.82 (talk) 14:54, 17 December 2024 (UTC)

Odd prime, my bad. GalacticShoe (talk) 16:38, 17 December 2024 (UTC)

Assume p is 3 mod 4. Suppose that (2|p)=1. Then $x^{4}+1\equiv (x^{2}+\lambda x+1)(x^{2}-\lambda x+1){\pmod {p}}$ where $\lambda ^{2}-2\equiv 0$. Because the cyclotomic ideal $(p,\zeta ^{2}+\lambda \zeta +1)$ has norm $p^{2}$ and is stable under the Galois action $\zeta \mapsto 1/\zeta$ it is generated by a single element $a\zeta ^{2}+b\zeta +a$, of norm $(2a^{2}-b^{2})^{2}$.

If (2|p)=-1, then the relevant ideal is stable under $\zeta \mapsto -1/\zeta$ and so is generated by $a\zeta ^{2}+b\zeta -a$, of norm $(2a^{2}+b^{2})^{2}$. Tito Omburo (talk) 14:43, 11 December 2024 (UTC)

December 11

Unique normal ultrafilter

So I'm supposed to know the answer to this, I suppose, but I don't seem to :-)

"Everyone knows" that, in $L[U]$, Gödel's constructible universe relative to an ultrafilter $U$ on some measurable cardinal $\kappa$, there is only a single normal ultrafilter, namely $U$ itself. See for example John R. Steel's monograph here, at Theorem 1.7.

So I guess that must mean that the product measure $U\times U$, meaning you fix some identification between $\kappa \times \kappa$ and $\kappa$ and then say a set has measure 1 if measure 1 many of its vertical sections have measure 1, must not be normal. (Unless it's somehow just equal to $U$ but I don't think it is.)

But is there some direct way to see that? Say, a continuous function $f:\kappa \to \kappa$ with $\forall \alpha f(\alpha )<\alpha$ such that the set of fixed points of $f$ is not in the ultrafilter no singleton has a preimage under $f$ that's in the ultrafilter? I haven't been able to come up with it. --Trovatore (talk) 06:01, 11 December 2024 (UTC)

December 15

What is the cause of this paradox?

I recently completed a calculus term, in which one of the last units involving how much one aspect of an object was changing in relation to time at a certain point, given the rate of change of another aspect. Many specific questions could be analyzed as a right triangle with one leg (the x) remaining constant and the other leg (the y) growing at a specified rate. When it came time to solve for the value of the dz/dt (the rate of the hypotenuse’s growth with respect to time) at a certain point, it ended up as less than the provided dy/dt. Here’s an illustration:

The x is the distance from me to a tower. This remains constant.

The y is the distance from the tower to a flying bird.

The dy/dt is the speed at which the bird is flying from the tower.

The z is my distance from the bird.

In this illustration, the distance between me and the bird is increasing at a slower rate than the speed at which the bird itself is flying. What is the cause of this paradox? Primal Groudon (talk) 19:43, 15 December 2024 (UTC)

I do not see any paradox here. Ruslik_Zero 20:30, 15 December 2024 (UTC)

If the bird is between you and the tower (0 ≤ y < x), the distance between you and the bird is even decreasing: dz/dt < 0. By the time it flies right overhead (y = x), the distance is momentarily stationary: dz/dt = 0. After that, it increases: dz/dt > 0. This rate of increase will asymptotically approach dy/dt from below as the bird flies off into an infinite distance.  --Lambiam 00:34, 16 December 2024 (UTC)

I think the issue here is that even though the rate of change of z is less than the rate of change of y, z never actually becomes less than y. You can see this graphically, for example, by comparing the graphs of y=x and y=√(x²+1). The second graph is always above the first graph, but the slope of the first graph is x/√(x²+1), which is always less than 1, the slope of the first graph. But this is typical behavior when a graph has an asymptote. As a simpler example, the slope of 1/x is negative, but the value never goes below 0 (at least for x>0). Similarly, the slope of x+1/x is always less than 1, but the value of x+1/x is always greater than x (again, for x>0). The graph of y=√(x²+1) is one branch of a hyperbola having y=x as an asymptote, and this looks very much like the x>0 part of y=x+1/x. In general the difference in rates of change can imply that that two quantities get closer and closer to each other, but this does not mean they ever become equal. This phenomenon is, perhaps, counterintuitive for many people, but the math says it can happen anyway. I don't know if this rises to the level of a paradox, but I can see that it might be concerning for some. --RDBury (talk) 09:39, 16 December 2024 (UTC)

For x > 0, the graph of y=√(x²+1) looks even more like that of y=x+1/(2x). For example, when x = 5, √(x²+1) = √26 ≈ 5.09902 is approximated much more closely by 5.1 than by 5.2.  --Lambiam 18:35, 16 December 2024 (UTC)

December 19

Who is the following unknown?

When asked "WHO IS YOUR X?" (X still being unknown to me but is known to the respondents), here are the answers I get:

A answers: "A"

B answers: "C"

C answers: "C"

D answers: "F"

E answers: "F"

F answers: "F"

To sum up, the special phenomenon here is that, everybody has their own X (usually), and if any respondent points at another respondent as the first respondent's X, then the other respondent must point at themself as their X.

I wonder who the unknown X may be, when I only know that X is a natural example from everyday life. I thought about a couple of examples, but none of them are satisfactory, as follows:

X is the leader of one's political party, or X is one's mayor, and the like, but all of these examples attribute some kind of leadership or superiority to X, whereas I'm not interested in this kind of solution - involving any superiority of X.

Here is a second solution I thought about: X is the first (or last) person born in the year/month the respondent was born, and the like. But this solution involves some kind of order (in which there is a "first person" and a "last person"), whereas I'm not interested in this kind of solution - involving any order.

Btw, I've published this question also at the Miscellaneous desk, because this question is about everyday life, but now I decide to publish this question also here, because it's indirectly related to a well known topic in Math. 79.177.151.182 (talk) 13:27, 19 December 2024 (UTC)

Head of household comes to mind as a fairly natural one. The colours then correspond to different households which can be just one person. One objection is that "head of household" is a fairy traditional concept. With marriage equality now being the norm it's perhaps outdated. --2A04:4A43:909F:F9FF:397E:BBF9:E80B:CB36 (talk) 15:11, 19 December 2024 (UTC)

I have already referred to this kind of solution, in the example of "my mayor", see above why this solution is not satisfactory. 79.177.151.182 (talk) 15:31, 19 December 2024 (UTC)

The question has been resolved at the Miscellaneous reference desk.

Resolved

79.177.151.182 (talk) 15:48, 19 December 2024 (UTC)

X may well be 'the oldest living person of your ancestry'. --CiaPan (talk) 20:46, 19 December 2024 (UTC)

Resolved or not, let's try to analyze this mathematically. Given is some set $S$ and some function $f:S\to S.$ For the example, $S=\{{\mathsf {A}},{\mathsf {B}},{\mathsf {C}},{\mathsf {D}},{\mathsf {E}},{\mathsf {F}}\},$ with $f({\mathsf {A}})={\mathsf {A}},$ $f({\mathsf {B}})=f({\mathsf {C}})={\mathsf {C}},$ $f({\mathsf {D}})=f({\mathsf {E}})=f({\mathsf {F}})={\mathsf {F}}.$

Knowing that "everybody has their own X (usually)", we can normalize the unusual situation that function $f$ might not be total in two ways. The first is to restrict the set $S$ to the domain of $f,$ that is, the set of elements on which $f$ is defined. This is possible because of the condition that $f(u)=v$ implies $f(v)=v,$ so this does not introduce an undue limitation of the range of $f.$ The second approach is to postulate that $f(u)=u$ whenever $f(u)$ might otherwise be undefined. Which of these two approaches is chosen makes no essential difference.

Let $T=f(S)$ be the range of $f$, given by:

$T=\{v\in S:(\exists u\in S:f(u)=v)\}.$

Clearly, if $f(v)=v,$ we have $v\in T.$ We know, conversely, that $v\in T$ implies $f(v)=v.$

Let us also consider the inverse image of $f$, given by:

$f^{-1}(v)=\{u\in S:f(u)=v\}.$

Suppose that $f^{-1}(v)\neq \emptyset .$ This means that there exists some $u\in f^{-1}(v),$ which in turns means that $f(u)=v.$ But then we know that $f(v)=v.$ Combining this, we have,

$f(v)=v\Leftrightarrow v\in T\Leftrightarrow f^{-1}(v)\neq \emptyset .$

The inverse-image function restricted to $T,$ to which we assign the typing

$f^{-1}:T\to \mathbb {P} (S),$

now induces a partitioning of $S$ into non-empty, mutually disjoint subsets, which means they are the classes of an equivalence relation. Each class has its own unique representative, which is the single element of the class that is also a member of $T$. The equivalence relation can be expressed formally by

$u\sim v\Leftrightarrow f(u)=f(v),$

and the representatives are the fixed points of $f.$

Applying this to the original example, $T=\{{\mathsf {A}},{\mathsf {C}},{\mathsf {F}}\},$ and the equivalence classes are:

• $\{{\mathsf {A}}\},$ with representative ${\mathsf {A}},$
• $\{{\mathsf {B}},{\mathsf {C}}\},$ with representative ${\mathsf {C}},$ and
• $\{{\mathsf {D}},{\mathsf {E}},{\mathsf {F}}\},$ with representative ${\mathsf {F}}.$

Conversely, any partitioning of a set defines an equivalence relation; together with the selection of a representative for each equivalence class, this gives an instance of the situation defined in the question.  --Lambiam 20:47, 19 December 2024 (UTC)

FWIW, the number of such objects on a set of size n is given by OEIS: A000248, and that page has a number of other combinatorial interpretations. If you ignore the selection of a representative for each class, you get the Bell numbers. --RDBury (talk) 00:35, 21 December 2024 (UTC)

December 20

Give a base b and two base b digits x and z, must there be a base b digits y such that the 3-digit number xyz in base b is prime?

Give a base b and two base b digits x and z (x is not 0, z is coprime to b), must there be a base b digits y such that the 3-digit number xyz in base b is prime? 1.165.207.39 (talk) 02:10, 20 December 2024 (UTC)

In base 5, $3Y1_{5}=76+5Y$ is composite for all base-5 Y. GalacticShoe (talk) 03:39, 20 December 2024 (UTC)

$3Y1_{7}$ also offers a counterexample. While there are many counterexamples for most odd bases, I did not find any for even bases.  --Lambiam 09:58, 20 December 2024 (UTC)