Comments on: What You Need To Know http://www.chrismclaren.com/blog/2008/05/09/what-you-need-to-know/ As honest as a gambling man can be Wed, 08 Oct 2008 00:21:43 +0000 http://wordpress.org/?v=2.6.2 By: Mr. McLaren http://www.chrismclaren.com/blog/2008/05/09/what-you-need-to-know/#comment-50596 Mr. McLaren Mon, 12 May 2008 14:50:41 +0000 http://www.chrismclaren.com/blog/?p=1226#comment-50596 You never cease to surprise me. Although I am sorely tempted to call fourberie on your supposed familiarity with <a href="http://www2.math.su.se/~jesper/research/wheels/wheels.pdf" rel="nofollow">the works</a> of Jesper Carlström. I've <a href="http://www2.math.su.se/~jesper/jesper2.jpg" rel="nofollow">seen his picture</a>, and he doesn't look like he could make it through the usual MacFarlane "it takes at least half a bottle of absinthe to earn my respect" tests. You never cease to surprise me.

Although I am sorely tempted to call fourberie on your supposed familiarity with the works of Jesper Carlström. I’ve seen his picture, and he doesn’t look like he could make it through the usual MacFarlane “it takes at least half a bottle of absinthe to earn my respect” tests.

]]> By: Neil MacFarlane http://www.chrismclaren.com/blog/2008/05/09/what-you-need-to-know/#comment-50595 Neil MacFarlane Mon, 12 May 2008 14:07:35 +0000 http://www.chrismclaren.com/blog/?p=1226#comment-50595 Oh, sorry, I forgot to mention that I was I presumed we were using a commutative ring of numbers to form a wheel. I should have known you'd all stick to your non-abstract algebraic thought. Commoners. :) Oh, sorry, I forgot to mention that I was I presumed we were using a commutative ring of numbers to form a wheel. I should have known you’d all stick to your non-abstract algebraic thought. Commoners. :)

]]> By: Kira http://www.chrismclaren.com/blog/2008/05/09/what-you-need-to-know/#comment-50564 Kira Sat, 10 May 2008 03:09:09 +0000 http://www.chrismclaren.com/blog/?p=1226#comment-50564 Ah. I see I didn't think it through and you were five steps ahead of me. Ah. I see I didn’t think it through and you were five steps ahead of me.

]]> By: Mr. McLaren http://www.chrismclaren.com/blog/2008/05/09/what-you-need-to-know/#comment-50561 Mr. McLaren Sat, 10 May 2008 01:48:57 +0000 http://www.chrismclaren.com/blog/?p=1226#comment-50561 Either I'm really silly, or he started with that. Although we may both ultimately be saying the same thing, in our own idiom. He goes from A<sup>2</sup>-B<sup>2</sup> to (A+B)(A-B). This is legitimate for all cases except A=B. Unfortunately we know we're in that case as an initial premise. What that amounts to is factoring by dividing by zero. You kind of can't divide by zero. Since we know A and B are the same, they could be any number. Let's say A=B=2. So A<sup>2</sup>-B<sup>2</sup>=AB-B<sup>2</sup> is the same thing as 4-4=4-4, or 0=0. Tautological, but OK. How ever in order to get to (A+B)(A-B) you have to say that 0= 0/0, which you can't say. Or, <a href="http://en.wikipedia.org/wiki/Division_by_zero#Fallacies_based_on_division_by_zero" rel="nofollow">as Wikipedia puts it</a>: <blockquote>"The "proof" above requires the use of the distributive law. However, this requirement introduces an asymmetry between the two operations in that multiplication distributes over addition, but not the other way around."</blockquote> Hence my previous comment. (A+B)(A-B) does equal A<sup>2</sup>-B<sup>2</sup>, <strong>except</strong> when A=B, which is our starting premise. In that case you can't do the reverse distribution. Or rather, if you think you can, you end up proving that 1=2, and consequently that 1+1=4. All this insight, and yet Neil still bluffs me off of pots <em>all the damn time</em>. Either I’m really silly, or he started with that. Although we may both ultimately be saying the same thing, in our own idiom.

He goes from A2-B2 to (A+B)(A-B). This is legitimate for all cases except A=B. Unfortunately we know we’re in that case as an initial premise. What that amounts to is factoring by dividing by zero. You kind of can’t divide by zero.

Since we know A and B are the same, they could be any number. Let’s say A=B=2.

So A2-B2=AB-B2 is the same thing as 4-4=4-4, or 0=0. Tautological, but OK.

How ever in order to get to (A+B)(A-B) you have to say that 0= 0/0, which you can’t say. Or, as Wikipedia puts it:

“The “proof” above requires the use of the distributive law. However, this requirement introduces an asymmetry between the two operations in that multiplication distributes over addition, but not the other way around.”

Hence my previous comment. (A+B)(A-B) does equal A2-B2, except when A=B, which is our starting premise. In that case you can’t do the reverse distribution. Or rather, if you think you can, you end up proving that 1=2, and consequently that 1+1=4.

All this insight, and yet Neil still bluffs me off of pots all the damn time.

]]> By: Kira http://www.chrismclaren.com/blog/2008/05/09/what-you-need-to-know/#comment-50555 Kira Fri, 09 May 2008 22:36:11 +0000 http://www.chrismclaren.com/blog/?p=1226#comment-50555 Silly you. The problem with the proof is that it only works if A=B. Silly you. The problem with the proof is that it only works if A=B.

]]> By: Chris McLaren http://www.chrismclaren.com/blog/2008/05/09/what-you-need-to-know/#comment-50554 Chris McLaren Fri, 09 May 2008 20:21:27 +0000 http://www.chrismclaren.com/blog/?p=1226#comment-50554 If you really think (A+B)(A-B) is the same thing as A<sup>2</sup>-B<sup>2</sup> in this case, then I should be making <em>more</em> money at poker since your calculation of pot and implicit odds should be... um... unique. If you really think (A+B)(A-B) is the same thing as A2-B2 in this case, then I should be making more money at poker since your calculation of pot and implicit odds should be… um… unique.

]]> By: Neil MacFarlane http://www.chrismclaren.com/blog/2008/05/09/what-you-need-to-know/#comment-50552 Neil MacFarlane Fri, 09 May 2008 15:41:37 +0000 http://www.chrismclaren.com/blog/?p=1226#comment-50552 That's stupid. Everyone knows that 1=2. A=B A^2=AB A^2 - B^2 = AB-B^2 (A+B)(A-B)=B(A-B) (A+B)=B A+A=A 2A=A 2=1 Flawless logic. That’s stupid. Everyone knows that 1=2.

A=B
A^2=AB
A^2 - B^2 = AB-B^2
(A+B)(A-B)=B(A-B)
(A+B)=B
A+A=A
2A=A
2=1

Flawless logic.

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