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# October 15

## untilable shape with only right angle corners?

Let A be the set of all polygons with *only* right angle corners. (Everything that isn't a rectangle will be convex and all with have an even number of corners. Let B be the subset of A where the polygons can not tile the plane. What is the smallest number of corners that a polygon in B can have? I know that with 10 corners, I can make an untilable (consider the heptile with the positions 1,2,3,4,6,7 & 8 in a 3x3 and then shrink tile 6 a titch). But I don't know if 6 or 8Naraht (talk) 14:23, 15 October 2014 (UTC)

Are you talking about a convex equiangular polygon that has 10 right angle corners? Is that even possible? "For an equiangular n-gon each angle is 180° - (360°)/n  ; this is the equiangular polygon theorem." I don't understand what you mean about a "heptile" either, but maybe that's my fault :) SemanticMantis (talk) 17:12, 15 October 2014 (UTC)
Heptomino with a hole
U pentomino
It's about a rectilinear polygon. The mentioned "heptile" must mean the heptomino with a hole. A rectilinear hexagon can only have an L-like shape and that can always tile the plane like in [1] regardless of the side lenghts. A rectilinear octagon cannot in general. Consider for example a U pentomino where the gap is made too narrow to fit another part. PrimeHunter (talk) 17:41, 15 October 2014 (UTC)
Ok, that explains my "heptile" confusion, but I'm still at a loss on the convexity claim. I think you're right that the left picture is mostly what the OP describes, but is it convex under any standard definition of convexity? It's certainly not a convex polygon... SemanticMantis (talk) 18:19, 15 October 2014 (UTC)
Right, the poster meant concave. A rectangle is convex and all rectilinear polygons with more than four sides are concave. PrimeHunter (talk) 19:13, 15 October 2014 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── @Naraht:, are you restricting yourself to polyominoes and allowing both rotations and translations? If not:

• If you allow only translations: a simple L-shape polyomino (with 6 corners) won't tile the plane.
• Even allowing rotations, if you don't restrict yourself to polyominoes: I believe, a L-shaped polygon with appropriately chosen (ie, non-commensurate) sides won't tile the plane.

Abecedare (talk) 19:54, 15 October 2014 (UTC)

Did you consider the tiling [2] I linked? PrimeHunter (talk) 21:11, 15 October 2014 (UTC)
Perhaps I'm misunderstanding you but as far as I can see any L shape with right angles will tile the plane with just translations no matter what length the sides are.Dmcq (talk) 21:14, 15 October 2014 (UTC)
You are obviously right. Sorry for the brain-freeze (even I can't replicate what I was thinking). Abecedare (talk) 21:26, 15 October 2014 (UTC)
• Is there some side set of side lengths that will not allow these to tile the plane?
Ok, I think we've mostly gotten the question figured out, and some partial answers. As I now understand the problem, the question is "is there a rectilinear, 8-cornered polygon that cannot tile the plane. We know that there is at least 1 10-cornered that will not, and it seems that any 6-cornered one will. SemanticMantis (talk) 15:30, 16 October 2014 (UTC)
Candidate: take the 'T' tetromino at right. Assume it is made of unit squares. Remove the descender square, and replace it by a rectangle of size 1/3 X 1/4. I don't think that will tile, and it remains an 8-angled rectilinear polygon. SemanticMantis (talk) 17:36, 16 October 2014 (UTC)
Original poster here. Yes, sorry about the convex/concave confusion, you are right. And I think the proposed candidate by is very good - a rectangle with a tiny rectangular "tumor" on the side (not at an end) will have 6 corners and not tile. Thank You!Naraht (talk) 00:56, 17 October 2014 (UTC)
If the "tumor" is not at one end of the side then surely that is 8 corners, not 6 ? Seems to me that any rectilinear 6-cornered polygon must be an L-shape and so can tile the plane as Dmcq points out above (but I could be wrong). Gandalf61 (talk) 11:50, 17 October 2014 (UTC)
Arg. I meant 8. I agree with you on 6 being an L shape.Naraht (talk) 13:48, 17 October 2014 (UTC)
Is something missing in my first post since people continue to discuss this? PrimeHunter (talk) 15:33, 17 October 2014 (UTC)
No, I think your post nailed it. Everything after is just clearing up confusions, typos and terminology. Gandalf61 (talk) 16:21, 17 October 2014 (UTC)

# October 16

## Algorithm

I am searching for an algorithm to break down the acquisition cost of a 100.000$house to a number of people. The goal of the algorithm is to get people into it at the early stage. I want to convince other people to pay a part of the cost. So I thought of an incentive. The prize when a person joins should go down fast at the start for every addition person to speed up the seed phase in which I am willing to take a greater share. At the start there are mostly enthusiast who are willing to pay a higher price at the start to get the collection running. When I am buying the house I would pay the total 100.000$.

I convince a person to pay for it together, I pay 2/3 (-> 66.666$) and the other person 1/3 (-> 33.333$). Perfect, my percentage dropped to 66% and the other person had the incentive to pay lesser than I and also lesser than the "normal distribution" (100.000$/2= 50.000$).

A next (third) person joins to lower the cost for us and the person should pay lower than the previous person (me and the second person with whom I shared cost initially).

Every addition person should lower the price for all the others already in and pay less than the previous ones.

The last step repeats until 40 people are in. Now enough people are in and the threshold is low enough to share the costs equally (-> Prize/People = 100.000/40 = 2.500) without the initially prohibitive high cost which were chilling finding other people than enthusiast.

Is there a name for that algorithm? 87.78.103.137 (talk) 21:59, 16 October 2014 (UTC)

Don't know a name, but if after the first person joins you reduce every additional payment to 93.6% of the previous one, you end up down around the $2500 you want: $33,333.33 x 0.93639 = \$2527.18

But to me it looks like the incentive is to wait, as the early birds risk more money and get no more benefit. Also, this sounds close to a pyramid scheme. Just what is the benefit to owning 1/40th of a house, anyway ? Will they be set up as time shares ?  :StuRat (talk) 14:25, 17 October 2014 (UTC)

# October 17

## Does a solution exist?

Firstly, not a homework question, I'm long past school age, I just dabble in mathematics recreationally. I'm playing around with something and I've narrowed it down to one problem, basically I need to find two integers $a, b$ which are not equal such that $a, b, \frac{a}{a+b}, \frac{b}{a+b}$ are all positive integers. I'm beginning to suspect that no solutions exist, but can anyone help me either find a solution, or show that there are none? Organics talk 11:14, 17 October 2014 (UTC)

If $a$ and $b$ are positive then $a+b > a$, so $\frac a{a+b}<1$ can not be a positive integer. Similar for $b$. --CiaPan (talk) 11:21, 17 October 2014 (UTC)
(ec) Well, $\frac{a}{a+b} + \frac{b}{a+b} = 1$, so it's not possible for both terms to be positive integers. If you allow non-negative integers then the terms have to be 0 and 1 in some order, so a = 0, b = any positive integer (or vice versa) will work, and these are the only solutions. AndrewWTaylor (talk) 11:26, 17 October 2014 (UTC)
That seemed too easy. If you meant $a, b, \frac{a+b}{a}, \frac{a+b}{b}$ then note that (a+b)/a = 1 + b/a, and (a+b)/b = 1 + a/b. If a ≠ b then either b/a or a/b will be below 1 and not an integer. PrimeHunter (talk) 11:34, 17 October 2014 (UTC)

## Is there a easy way to do this math?

PS:Nope, this is not homework, its for some musical thing I am doing.

Is there a easier way to do this math? Imagine this. Two drummers are drumming (at the same time) during one second. One at 1 bps and other at 1.5 bps. A stick will hit a drum at 2 points points of time, at 1 second, and at 2/3 of a second.

With that info, now imagine 3 drummers, one playing at 1.5 bps, the other at 1 bps, and the other at 2 bpm (the number we previously got). A stick will hit a drum 3 points of time, at half second, at 2/3 of a second and at 1 second.

With that info now imagine 4 drummers, one playing at 1 bps, the other playing at 1.5 bps, the third at 2 bps and the fourth at 3 bps (the number we previously got). A stick will hit a drum 4 points of time, at half second, at 1/3 of a second and at 2/3 of a second and at 1 second.

With that info now imagine 5 drummers, one playing at 1 bps, the other playing at 1.5 bps, the third at 2 bps and the fourth at 3 bps and another playing at 4 bps (the number we previously got). A stick will hit a drum 6 points of time, at half second, at 1/3 of a second and at 2/3 of a second and at 1 second 1/4 of a second and 3/4 of a second.

And this goes on.

Is there a easier way to do that, or some formula that continue this...?
— Preceding unsigned comment added by 201.78.201.199 (talk) 19:12, 17 October 2014 (UTC)

I'm not following this. If a drummer is playing at 1 bpm (beats per minute), why would his stick hit the drum at 1 second in? Did you mean beats per second? OldTimeNESter (talk) 20:27, 17 October 2014 (UTC)
Do what? You're not doing any math, just establishing facts. I'm not sure why you keep saying "the number we previously got". That number doesn't somehow follow from the previous scenario.--80.109.80.31 (talk) 20:30, 17 October 2014 (UTC)
The 1.5 bps drummer doesn't really fit into the pattern of 1 bps, 2 bps, 3bps etc. Apart from that, the sequence of numbers that you are describing starting 1,2,4,6 is sequence A002088 at OEIS. For more information see our article on Farey sequences. Gandalf61 (talk) 11:49, 18 October 2014 (UTC)
I already tried wolfram alpha to find the results and already saw your oeis link. The number 1 and 1.5 are the first 2 numbers used to make the thing (to ge the number 2). — Preceding unsigned comment added by 201.78.160.103 (talk) 13:06, 18 October 2014 (UTC)
So, how do 1 and 1.5 give 2? And again, what are you trying to do? Make a list of the rational numbers with denominator up to a given limit? —Tamfang (talk) 07:38, 19 October 2014 (UTC)
I think the n+1th drummer has a number of bps equal to the total number of drumbeats each second from the previous n drummers, where only a single beat is counted if two drummers hit their drum simultaneously (so the next in the sequence would be 1bps, 1.5bps, 2bps, 3bps, 4bps, 6bps). Not sure whether there is a good general form for the series, but this seems to be what the OP is getting at. MChesterMC (talk) 12:00, 20 October 2014 (UTC)

## Is the following proof of Goldbach Conjecture correct?

Keywords: π(*):= Odd Prime Counting Function and Fundamental Theorem of Arithmetic (FTA) Goldbach conjecture states every positive even integer is the sum of two prime numbers. (We count one as prime in the sense of additive number theory outside of the FTA.)

Proof of Goldbach Conjecture:

Suppose there exists a positive even integer, e > 4, that is not the sum of two odd prime numbers or 1. e ≠ p + q

over S = {all odd prime numbers less than e} and where k = card(S) = π(e). Therefore, e ≠ p + q over S, (p,q є S) , implies the following system of equations over S, 1 = e - n1 * q1, 3 = e - n2 * q2, ..., pk = e - nk * qk, according to the Fundamental Theorem of Arithmetic where 1 < qj ≤ (nj * qj)^.5 ≤ nj for 1 ≤ j ≤ k where pj, qj є S and nj is a positive integer. Note: If qj = 1, then nj є S, or nj is an odd prime less than e.

Therefore, 2. Probability(e ≠ p + q over S) = ∏(j=1 to k→∞)[Probability (qj ≠ 1 | e - pj = nj * qj over S) * Probability (e - pj = nj * qj over S)]

= ∏(j=1 to k→∞)[(π((nj * qj)^.5) - 1)/π((nj * qj)^.5)] → 0 (This is an increasingly fast convergence for this almost everywhere monotonic non-increasing expression. This implies that the expected value of e ≠ p + q over S is practically zero, or E[e ≠ p + q over S] = e * Probability(e ≠ p + q over S) ≈ 0 for all e ≥ 100.)

Note: Probability (e - pj = nj * qj over S) = 1 for 1 ≤ j ≤ k.

In addition, empirical evidence has confirmed the validity of the conjecture for all positive even integers up to at least an order of 10^18. Therefore, we conclude the conjecture is true. Euler was right! Thank God! Praise God! --David Cole, primesdor@gmail.com — Preceding unsigned comment added by Primesdegold (talkcontribs) 20:04, 17 October 2014 (UTC)

You have a couple of misconceptions about the Goldbach Conjecture: first, we do not consider 1 to be prime for this purpose; second, we do not require the primes to be distinct. Those aren't the real problems with your argument, though.
As far as I can tell, you're putting a uniform distribution on the integers less than $e$ and then assuming that $m$ and $e-m$ being prime are independent events. There's no reason to make that assumption.
You can find the same thinking in our Goldbach's Conjecture article under heuristic justification.--80.109.80.31 (talk) 21:01, 17 October 2014 (UTC)
(edit conflict) This is not a proof. You make arguments similar to Goldbach's conjecture#Heuristic justification. Such arguments can only show that if we make certain unproven assumptions that numbers behave sufficiently "randomly" and independently, then the chance of a large counter example is extremely small. We don't know whether the assumptions are correct and even if they are in som sense, each large untested number would still have a tiny chance of being a counter example. That chance would quickly approach 0 for e > 4×1018 (the current search limit), but it would never actually be 0. PrimeHunter (talk) 21:15, 17 October 2014 (UTC)
First, as PrimeHunter notes, a "proof" using heuristic arguments is not considered a proof. Goldbach's Conjecture is thought by most mathematicians to be almost certainly true, but there is the annoying possibility that it may be true but unprovable, an example of Gödel's incompleteness theorems. (If so, its unprovability is itself unprovable. It isn't like the Halting Problem, which was proved to be undecidable.) The same possibility was advanced with regards to Fermat's Last Conjecture until its proof by Wiles. The challenge is to find a formally valid proof of Goldbach's Conjecture, which may require branches of mathematics not in existence in Euler's time, as is the case with Wiles's proof of Fermat's Last Conjecture. Robert McClenon (talk) 15:29, 20 October 2014 (UTC)
Any proof is probably beyond the scope of "classical" number theory. Otherwise Gauss would have found it if Euler didn't find it. Robert McClenon (talk) 15:45, 20 October 2014 (UTC)
Please don't add wording to your proposed proof after their have been responses. It causes confusion because the replies may not appear to address the updated version of the proof. However, it doesn't change the original objections. A "proof" using heuristic arguments is not a rigorous proof, and only a rigorous proof is acceptable. Robert McClenon (talk) 18:19, 20 October 2014 (UTC)

## Is the following proof of Riemann Hypothesis correct?

Riemann Hypothesis states that the real part of all non-trivial zeros of the Riemann zeta function, or ζ(s) = Σ(k=1 to ∞) 1/k^s = 0, equals one-half. For the non-trivial zero, s, a complex number, we have s = a + bi where Re(s)= a = 1/2.

Proof of Riemann Hypothesis

Fact I: The real part of all non-trivial zeros of the Riemann zeta function are located in the critical strip, [0, 1], according to a Riemann Theorem.

Fact II: There are infinitely many non-trivial zeros of the Riemann zeta function whose real part equals one-half according to a Hardy Theorem.

Fact III: The sum of the complex conjugate pairs of non-trivial zeros, s = a + bi and s' = c + di where ζ(s) = Σ(k=1 to ∞) 1/k^s = 0 and ζ(s') = Σ(k=1 to ∞) 1/k^s' = 0, of the Riemann zeta function equals one according to the Fundamental Theorem of Arithmetic and the Harmonic Series (H):(Note: Euler and others have proven that there exists an infinite set of primes in H. And that the divergence of H is a key reason for that result.)

H = Σ(k=1 to ∞) 1/k = Σ(k=1 to ∞) (1/k^s)(1/k^s') = Σ(k=1 to ∞) 1/k^(s+s') = ∝. Therefore, according to Facts I, II, and III, we have the following properties for all non-trivial zeros of the Riemann zeta function:

s + s' = 1 which implies a + c = 1 and b + d = 0 such that 0 ≤ a ≤ 1 and 0 ≤ c ≤ 1.

And of course, k^s = k^(a + bi) implies k^(s-bi) = k^a, and k^s' = k^(c + di) implies k^(s-di) = k^c.

Fact IV: For all k > 1, k is a positive integer, there exists a prime number, p, so that p|k such that p = k or p ≤ k^(1/2).

Therefore, according to Facts I, II, III, and IV, we have:

k^(1/2) ≤ k^a ≤ k, k^(1/2) ≤ k^c ≤ k, and a + c = 1.

Hence, k^a = k^c = k^(1/2) which implies a = c = 1/2. Riemann Hypothesis is true! Riemann was right! Thank God! Praise God!

Note the Importance of the Harmonic Series (H) with regards to prime numbers:

H = Σ(k=1 to ∞) 1/k = 2Σ(k=1 to ∞) [1/(2k) = 1/(q - p)] = ∝ where q > p, and p,q are odd prime numbers. H1 = 2( 1/(5-3) + 1/(11-7) + ...) = ∝; There is a infinite set, P1, of primes, p, generated from H1, and there is a infinite set, Q1, of primes, q, generated from H1. H2 = 2( 1/(7-5) + 1/(17-13) + ...) = ∝; There is a infinite set, Q2, of primes, q, generated from H2. H3 =2( 1/(13-11) + 1/(23-19) + ...) = ∝; There is a infinite set, Q3, of primes, q, generated from H3. ... H∞ = 2(...) = ∝; There is a infinite set (Q∞) of primes, q, generated from H∞. Therefore, the infinite set of all odd primes or ℙ\{2} = P1 ∪ Q1 ∪ Q2 ∪ Q3 ∪ ... ∪ Q∞. (Note: If we accept one as prime in the sense of additive number theory outside of FTA, this inclusion of 1 as a prime will change H1 through H∞, slightly. For example, H1 = 2( 1/(3-1) + 1/(11-7) + ...) = ∝, and H2 = 2( 1/(5-3) + 1/(17-13) + ...) = ∝, ...,H∞ = 2(...) = ∝ ; Thank God! Praise God!

Fact/Proposition: There are infinitely many more positive integers than there are prime numbers, or prime numbers have a zero density relative to the positive integers, and prime numbers generate the positive even integers efficiently so that gaps between two consecutive prime numbers increase without bound. Thank God! Praise God!

Fact/Proposition: π(e = mg = 1 + p2n) = 2π(g = 1 + pn)= 2n ∈2ℕ where π():=Prime Counting Function, p2n, pn ≥ 3 ∈ ℙ\{1,2}, and 2 < m ∈ ℚ ≤ 3. As g → ∞, m → 2. Thank God! Praise God! --David Cole, primesdor@gmail.com

 — Preceding unsigned comment added by Primesdegold (talk • contribs) 22:29, 17 October 2014 (UTC)

My guess is that it is about as right as your proof of Goldbach's conjecture above. Bubba73 You talkin' to me? 00:54, 18 October 2014 (UTC)
I don't see what implies your 2nd to last line, I don't believe that anything does. If you wouldn't mind elaborating on your reasoning, we could through things a bit more thoroughly and see what is going on; at the moment, it just appears as jump. I'd love to discuss more if you're willing. Though, I do want to point out, while it can be rewarding to ponder over, and workout, our own notions about such problems, the odds that conjectures such as Goldbach and Riemann being solved in a few lines of elementary mathematics is, essentially, nil. If you would like to learn more about such questions, I can definitely recommend a few books and articles that would be of use. Best of luck - there is something captivating about these problems, to be sure:-) Phoenixia1177 (talk) 03:59, 18 October 2014 (UTC)
Fact III is wrong. By itself this would trivially imply the Riemann hypothesis if true. I don't see what the harmonic series has to do with it, but to me it seems that you have assumed what you wanted to prove here. Sławomir Biały (talk) 12:21, 18 October 2014 (UTC)
Despite recent incomprehensible attempts to clarify the role of the harmonic series in proving "fact" III, the claim that s+s'=1 follows from the divergence of the harmonic series remains a non-sequitur. Sławomir Biały (talk) 14:00, 19 October 2014 (UTC)

# October 18

## Norm identity for a normed *-algebra with B*-identity

Hi,

I'm trying to show a claim for a normed *-algebra A that is not necessarily unital or commutative but is non-zero and satisfies the B*-identity: $||x^*x||=||x||^2\;\forall x \in A$:

$||x||=\sup_{||y||\leq 1}\left\{||xy||\right\}$ for all $x\in A$.

One direction is trivial; the required property of a normed *-algebra gives:

$\sup_{||y||\leq 1}\left\{||xy||\right\}\leq \sup_{||y||\leq 1}\left\{||x||\cdot ||y||\right\}\leq ||x||$

But I can't seem to do the other.

Cheers,

Neuroxic (talk) 14:32, 18 October 2014 (UTC)

Are you allowed to use $\|x\|=\|x^*\|$? If so, try $y=x^*/\|x\|$. Then $\|y\|=1$ and $\|xy\| = \|x\|$, which gives the opposite inequality. Sławomir Biały (talk) 21:13, 18 October 2014 (UTC)
Yes, one can derive that that B* condition implies the involution is isometric. Many thanks!
Neuroxic (talk) 22:45, 18 October 2014 (UTC)

# October 19

## Smallest Polymino whose Convex Hull is not tilable...

What is the Smallest Polymino whose convex Hull is not tilable? is it the quadmino "T", that one has a convex hull with sides 3-1-root2-1-root2? Naraht (talk) 03:00, 19 October 2014 (UTC)

It looks like all the tretrominos have tilable convex hulls. But the convex hull of the Y-pentomino does not tile. It has one side of length √5 and this can only be placed next to another tile which is rotated 180 degrees. When you then try to match the side with length √2, there are two ways but both leave a bay that can't be filled. There are probably other pentominoes as well, I haven't checked them all, but that's enough to answer the question as to the smallest. --RDBury (talk) 07:28, 19 October 2014 (UTC)
PS & correction. Actually the convex hull of the T-tetromino does not tile. When you place a tile next to the side of length √2, there are two ways to do it but both leave a bay that can't be filled. After a bit of doodling I found that the convex hulls of 5 of the 12 pentominoes, Y, F, T, X and W do not tile, while the convex hulls of the remaining 7, I, L, N, V, P, U, Z, do. Perhaps someone could check this as it would probably take more effort than it's worth to turn my doodles into something rigorous. --RDBury (talk) 07:57, 19 October 2014 (UTC)

## Real Analysis: Bounded Sequences

The question I am having trouble with is Prove that sn is bounded where sn=sin(n)/n.

I know I have to use the definition of bounded sequence. I know sin(n) converges and therefore is bounded and 1/n is bounded, but I am not really sure where to go with this. A suggestion about where to begin would be very helpful. — Preceding unsigned comment added by Pinterc (talkcontribs) 13:44, 19 October 2014 (UTC)

sin(n), although bounded, does not converge. The sequence 1/n converges to zero and so is bounded. sin(n)/n is the product of two bounded sequences, so it's bounded. (In fact, you can prove that it converges to zero.) Sławomir Biały (talk) 14:31, 19 October 2014 (UTC)

## Is the following elementary proof of Fermat's Last Theorem correct?

The theorem states there exists no integral (all positive integers) solutions for the equation,


1. x^n+y^n=z^n for n > 2. I begin the proof by assuming there exists an integral (positive integer) solution to equation one for some n > 2. Equation one becomes with some algebraic manipulation, 2. x^n=z^n-y^n = (z^(n/2)+y^(n/2))*(z^(n/2)-y^(n/2)).

Now that I have factored the right side of equation two, Fermat, the great French mathematician and respectable jurist, made I believe the next logical and crucial step. He factored the left side as well, x^n, with the help of an extra real variable, Ɛ, such that 0 < Ɛ < n . I have the following equation, x^n = x^(n/2+Ɛ/2)* x^(n/2-Ɛ/2) = (z^(n/2)+y^(n/2))*(z^(n/2)-y^(n/2) ). This equation implies x^(n/2+Ɛ/2)= z^(n/2)+y^(n/2) and x^(n/2-Ɛ/2) = z^(n/2)-y^(n/2).

And next I have, 3. x^(n/2+Ɛ/2)/ x^(n/2-Ɛ/2) = x^Ɛ = (z^(n/2)+y^(n/2) )/(z^(n/2)-y^(n/2) ). Using equation 3 and applying some algebraic manipulation and simplification to it, I generate the equation, 4. z^n=y^n *((x^Ɛ+1)/(x^Ɛ – 1))^2.

And finally, by combining equations one and four, I generate the following equation after some more algebraic manipulation and simplification, 5. y = (1/4)^(1/n)*(x^(n-Ɛ))^(1/n)*(x^Ɛ – 1)^(2/n).

However, (1/4)^(1/n) is not a rational number, a ratio of two whole numbers, for n > 2. This implies the right side of equation five is not a positive integer. This contradicts my assumption that y is a positive integer. Thus, Fermat’s Last Theorem is true, and Fermat was right! Thank God! Praise God!--David Cole, primesdor@gmail.com Primesdegold (talk) 19:59, 19 October 2014 (UTC)

There are several very elementary errors in this proof. Are you a troll, or just an idiot? Sławomir Biały (talk) 20:06, 19 October 2014 (UTC)
WP:AGF. There are many people who think they've proved some open problem in math, and presenting their work here is not necessarily a sign of trolling or idiocy. These questions could just as easily be based in naivety and zeal ;) If anyone wants to help OP find their mistakes, I don't think that's a problem. SemanticMantis (talk) 14:45, 20 October 2014 (UTC)
We should not tolerate trolls like this. I think we should just tell the idiot to go away and bother some other corner of the internet. Sławomir Biały (talk) 15:06, 20 October 2014 (UTC)
If any elementary proof were correct, then it would have been found in the eighteenth century (or Fermat could have written it in the margin in the seventeenth century). Robert McClenon (talk) 15:21, 20 October 2014 (UTC)
I absolutely agree! I once worked in Graduate admissions in the math dept, and we had people claiming to trisect angles in their applications, which is even worse, because that is not merely open, but provably impossible... The unsatisfying thing about this reasoning though, is it ends up being basically an appeal to authority, which is of course not how math really works, and not something we really should rely on when answering questions about math. Though I don't have the time or interest to find errors on these so-called 'proofs', I still think others should be allowed to do so if they wish. Sławomir seems certain this is a troll, but I think a crank need not be a troll. It's a distinction of intent. I try not to ascribe malice to online users where simple ignorance and naivety may provide sufficient explanation. Maybe I'm in the minority, but I see it as our goal here to help educate the naive. SemanticMantis (talk) 15:29, 20 October 2014 (UTC)
First, SemanticMantis is obviously referring to trisection with Euclidean tools, where he or she correctly notes that the construction has been proved impossible. Trisection of an angle is easy if one is not limited to Euclidean tools. Archimedes trisected the angle using an expanded toolset. Doubling a cube is similarly impossible with Euclidean tools and easy with better tools. Squaring the circle is a different, more difficult problem. Robert McClenon (talk) 15:42, 20 October 2014 (UTC)
Second, I agree with SemanticMantis that the OP is not a troll, but an editor whose enthusiasm exceeds his knowledge and his knowledge of the limits of his knowledge. Robert McClenon (talk) 15:42, 20 October 2014 (UTC)
I find it funny that the OP has the skill to carry out some elaborate algebraic manipulations correctly, and yet seems to think that $ab=cd$ implies $a=b,\ c=d$.
This could be a result of a phenomenon that is all too common among students, of treating math too formally - they learn by rote manipulations and problem solving techniques, but they have no idea what the concepts actually mean. -- Meni Rosenfeld (talk) 05:10, 21 October 2014 (UTC)

## Request permission to add a category to the Fractal subheading 'Common techniques for generating fractals'

If you don't mind, I'd like to add 'Sirsty-Firsty' to the Fractal subheading 'Common techniques for generating fractals'.

Sirsty-Firsty, as in www.sirsty-firsty.org This is a method for generating designs based on moving points around in spirals.

Any thoughts?

Thanks, --InverseSubstance (talk) 22:40, 19 October 2014 (UTC)

That subsection of the article is not about specific software implementations, but general methods. Is the method used in the software that you linked to discussed in the peer reviewed literature? If not, then it should not be added to the article. Sławomir Biały (talk) 12:53, 20 October 2014 (UTC)
You could perhaps add a link in the section Fractal#Fractal-generating_programs. You could also post this question at Talk:Fractal. SemanticMantis (talk) 14:49, 20 October 2014 (UTC)