User talk:Fuzzyeric

Welcome!

Hello, Fuzzyeric, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. I was delighted to see your responses at the (mathematics) reference desk; We do not have too many regulars there with enough knowledge to tackle the harder questions. I do hope you will create a user page, and more importantly, to see you contributing to articles as well as the RefDesk. When you do, you may find the following pages useful:

• The five pillars of Wikipedia
• How to edit a page
• Help pages
• Tutorial
• How to write a great article
• Manual of Style

And let's not forget the hub of the mathematics community:

• Wikipedia:WikiProject Mathematics
• Wikipedia talk:WikiProject Mathematics

I hope you enjoy editing here and being a Wikipedian! If you need help, check out Wikipedia:Questions or ask me on my talk page.

Again, welcome! -- Meni Rosenfeld (talk) 17:11, 12 September 2006 (UTC)

Msin(i)

Thank you for this response. I offer Msin(i) as a redlink, if you'd like to create an article (your first maybe?). Cheers, Marskell 04:13, 16 September 2006 (UTC)

Sounds like a really, really bad idea. Msin(i) is just the mass times the sine of i. What's next, an article about gcos(θ)? -- Meni Rosenfeld (talk) 07:37, 16 September 2006 (UTC)

Agreed. -- Fuzzyeric 20:46, 17 September 2006 (UTC)

Hey, sorry. Marskell 20:57, 18 September 2006 (UTC)

Nine lemma

Would you happen to know of any interesting applications of the nine lemma? --HappyCamper 04:31, 23 September 2006 (UTC)

I can't think of anything better than the link User:John Baez found. -- Fuzzyeric 05:03, 23 September 2006 (UTC)

(copied to Talk:Nine lemma) -- Fuzzyeric 05:56, 23 September 2006 (UTC)

Invalid division proposition

I don't understand what you mean by "Using the proposed inference form". In general, to show that a universal implication of the form P(x) ⇒ Q(x) is invalid, you have to be able to give a value x in the domain of the implied quantification such that P(x) is true, while Q(x) is false. I defy you to produce a value x such that x³ − x² = 0 but x² − x ≠ 0.

Here is a proof of the implication you are challenging:

Either x = 0 or x ≠ 0.

Case A: x = 0. Then x² − x = 0, so, a fortiori:

x³ − x² = 0 ⇒ x² − x = 0.

Case B: x ≠ 0. Then we may divide both sides of an equation (even according to Fuzzyeric) by x, giving us:

x³ − x² = 0 ⇒ x² − x = 0.

Conclusion: for all possible values of x, it is the case that x³ − x² = 0 ⇒ x² − x = 0.

 --Lambiam^Talk 15:39, 28 September 2006 (UTC)

I really don't get what you are trying to say. You write something on my talk page of the form:

"your claim that P is justified by Q.

1. I did not claim any such thing.
2. In what you wrote, both P and Q are true statements.
3. I gave a proof of Q only because I understood from what you wrote that you challenged the validity of Q, not to justify P.
4. It is easy enough to give an independent proof of P; you do not need the validity of Q for that.
5. Don't you really see that there is a difference between $\implies$ and $\Longleftarrow$?
6. Try to avoid insulting terms like "gibberish", please. Maybe you should first look closer to home for the cause of your lack of ability to understand. -- --Lambiam^Talk 00:41, 29 September 2006 (UTC)

Perhaps you do not understand because you wish not to understand. But in any case, please stop posting incoherent strings of statements on my talk page. I give up.  --Lambiam^Talk 01:27, 29 September 2006 (UTC)

You don't give quickly up, do you? In the discussion of 597 days ago, we were dealing with the field of real numbers, and not some arbitrary algebraic structure. The original issue you raised was that of division by a quantity that was potentially zero. If you allow nilpotent elements, you cannot even in general divide by non-zero numbers, so the "counterexample" fails to make the point.  --Lambiam 10:36, 17 May 2008 (UTC)

For some reason I remember this correspondnce (which emerged here). I will say now what I should have said back then - Lambiam is right, on all counts:

1. Lambiam's original statement was that $x^{3}-x^{2}=0\Leftarrow x^{2}-x=0$. This is true for any ring, as it involves multiplication, not division.
2. Since you (FuzzyEric) failed to understand this, and diverted the discussion to the unrelated implication $x^{3}-x^{2}=0\Rightarrow x^{2}-x=0$, Lambiam has correctly mentioned that this implication also happens to hold (for real numbers). The proof requires more than one step, but is valid nonetheless.

Please re-read the discussion before commenting further. -- Meni Rosenfeld (talk) 17:55, 17 May 2008 (UTC)

Incorrect. Lambiam explicitly writes in the initial context (which you reference):

$\pi r(r+{\sqrt {r^{2}+h^{2}}})=\pi r^{2}h/3.$

Divide both sides by πr/3:

$3(r+{\sqrt {r^{2}+h^{2}}})=rh.$

which is quite clearly a division and is so described by him. The action that was used was left cancellation by an element without an inverse. (The algebra of polynomials does not have inverses for the "variable".) I attempted to point this out in-discussion and then on his Talk page using the simpler sentence $x^{3}-x^{2}=0\Rightarrow x^{2}-x=0$. Use of left cancellation by x can be used to derive a contradiction even in the field of the reals, as I demonstrated using the simpler statement.

On his talk, he makes two incorrect claims. The first is that the only way to prove his inference false is by stating values (and a model, although he doesn't mention this) which makes the statement false. Of course, there are other methods, including derivation of a contradiction from the proposed statement (the one I started with) and the somewhat more esoteric proof of dependence of the statement on the set of axioms with a proof of consistency of the negated statement. Regardless of the discursively incorrect assertion that only one form of argument of falsehood was valid, the previous question was still adequately dismissed by me.

The correct inference in the original context has always been:

$\pi r(r+{\sqrt {r^{2}+h^{2}}})=\pi r^{2}h/3.$

$\Rightarrow (3(r+{\sqrt {r^{2}+h^{2}}})=rh)\lor (r=0).$"

... and the additional solution ignored by Lambiam in his original demonstration should be explicitly checked and/or rejected by the student in the context of the modeled problem. -- Fuzzyeric (talk) 22:17, 17 May 2008 (UTC)

... And on the off-chance you care at all why I took ~600 days to reply, I have lately been constructing a representation of a monoid ring with nilpotent elements and I was reminded of the challenge to find a value that proves the statement false. -- Fuzzyeric (talk) 22:38, 17 May 2008 (UTC)

I admit that Lambiam's original choice of words was poor. The "Divide both sides", in retrospect, should not have been taken literally, and Lambiam clarified his intention in later comments. He sought to produce a sufficient condition (a solution, not necessarily all solutions), and for this the relevant implication is $\pi r(r+{\sqrt {r^{2}+h^{2}}})=\pi r^{2}h/3\Leftarrow 3(r+{\sqrt {r^{2}+h^{2}}})=rh,$ which is trivially true.

To find all solutions, you need a two-sided implication - the one you gave doesn't really tell you that what you have found is a solution.

It was assumed that a statement as simple as $\forall x\in \mathbb {R} (x^{3}-x^{2}=0\Rightarrow x^{2}-x=0)$ ought to be decidable, thus if it is not true, it must be false and a counterexample must exist. Considering the finer points of logic is an unneeded digression. -- Meni Rosenfeld (talk) 23:02, 17 May 2008 (UTC)

This I understood, but where's the proof that the set of solutions in the original problem is not reduced by one when the degree is reduced by one. Because while $x^{3}-x^{2}=0\Rightarrow x^{2}-x=0$ is valid in a field without zero divisors for a perhaps inobvious reason, the statement $x^{2}-x=0\Rightarrow x-1=0$ will appear equally valid to the student and is of course wrong. Using Lambiam's inference, but one degree lower gives $x^{2}-x=0\Leftarrow x-1=0$ is valid, but has introduced additional solutions, which must be detected and deleted in the context of the original problem. There's no way around it: when cancelling or multiplying by an expression that may or may not have an inverse, one must verify that the solution set is unchanged, not merely contained. -- Fuzzyeric (talk) 23:09, 17 May 2008 (UTC) (edited very slightly by Fuzzyeric (talk) 23:26, 17 May 2008 (UTC))

But Lambiam has said very clearly that he had no interest in keeping the solution set unchanged. By removing an r, he obtained an equation which lacks some solutions of the original (those involving r=0), but which still has solutions and all of those solutions are also solutions to the original problem. This is what Lambiam aspires to have found, and that much is correct. -- Meni Rosenfeld (talk) 10:03, 18 May 2008 (UTC)

Probability

If you don't mind, I'd like to continue the discussion from the Reference Desk. You sound like you know what you're talking about. I would have continued it then, but I've been offline for three days.

I like the idea of using cellular automata to figure this out. I'll see what I can do with that.

I'm not familiar with the word "ontological", but I looked it up and it apparently means "An explicit formal specification of how to represent the objects, concepts and other entities that are assumed to exist in some area of interest and the relationships that hold among them." That fits with what you said. For me, in this case, it's not very complicated. I'm not concerned exactly with some general definition of "random sequence", I'm interested in seeing how our world (or something essentially similar, like perhaps the cellular automota) produces something that looks for all the world as though it were random. By random, I mean something along the lines of "all instances are independent and all possibilities are equally likely". You know, random. I know the answer should be simple and straightforward, it's just damnably hard to get across to other people what I'm asking. So, please be patient. Black Carrot 19:50, 16 December 2006 (UTC)