Mal: Go on. Get in there. Give your brother a thrashing for messing up your plan. River: He takes so much looking after.

'Objects In Space'


Natter 40: The Nice One  

Off-topic discussion. Wanna talk about corsets, duct tape, or physics? This is the place. Detailed discussion of any current-season TV must be whitefonted.


Emily - Nov 20, 2005 7:11:13 am PST #5737 of 10006
"In the equation E = mc⬧, c⬧ is a pretty big honking number." - Scola

Nilly! Would you mind if I ran a math thing by you?

Also, hi!!!

(You know I don't just love you for your math, right?)


Nilly - Nov 20, 2005 7:13:09 am PST #5738 of 10006
Swouncing

Would you mind if I ran a math thing by you?

You know I don't really know math, right? I just know how to use some of it. But I would love to try and help, if I can.

(You know I don't just love you for your math, right?)

(Hee. You know it's true for why I like you, too, right?)


Fred Pete - Nov 20, 2005 7:18:50 am PST #5739 of 10006
Ann, that's a ferret.

Belated Happy Birthday, DX!


Emily - Nov 20, 2005 7:21:25 am PST #5740 of 10006
"In the equation E = mc⬧, c⬧ is a pretty big honking number." - Scola

t MATH

I just know how to use some of it.

Good enough. Check my reasoning, okay? p and q are relatively prime. We have two numbers, p/4 and p^2 + 3q^2 (and we know all of these are integers). If p is not divisible by 3, then these numbers are relatively prime as well, because (this is the part I need to be checked on -- the above is true, I need to prove it) you can divide p^2+3q^2 by p/4 and get 4p + (3q^2 / (p/4)) (I'm kind of using the Euclidean algorithm a bit). Then the problem is reduced to 3q^2 and p/4 -- p/4 can't have any different factors from p, nor can q^2 from q, so the only way 3q^2 and p/4 can have any factors in common is if p is divisible by 3. Yes?

t / MATH Also, I've taken to writing M.Sh.L. at the end of proofs in my notes, just because it makes me think of you every time. Though I suppose, for completeness, I should figure out how to write it in Hebrew?


DavidS - Nov 20, 2005 7:23:38 am PST #5741 of 10006
"Look, son, if it's good enough for Shirley Bassey, it's good enough for you."

Nope, he can't possibly be. Rowlf first appeared on the Jimmy Dean Show (yes, the sausage guy used to be a singer) back in the early sixties, ten years before Waits's first album came out. Waits was only 14 when Rowlf started pounding away on his piano.

I am corrected. However, I am curious if Rowlf was always a piano playing dog, or if that aspect of his character developed later.

Breakfast This Morning: Sauteed mushrooms scrambled with eggs, chopped green onions, cheddar and gouda. Lattes & english muffins.

We're listening to Belle & Sebastian. JZ is reading my Mojo magazine about Kate Bush.

Matt should go to Stockholm. Other people should go to London. Maybe everybody should go to London.


Amy - Nov 20, 2005 7:29:32 am PST #5742 of 10006
Because books.

Maybe everybody should go to London.

Yes, this.

::waits for everyone to start packing::


Jesse - Nov 20, 2005 7:30:28 am PST #5743 of 10006
Sometimes I trip on how happy we could be.

I would like to go to London! Although I should go someplace else first, as I've been to London. How's late May?


Theodosia - Nov 20, 2005 7:59:23 am PST #5744 of 10006
'we all walk this earth feeling we are frauds. The trick is to be grateful and hope the caper doesn't end any time soon"

I vote for a F2F in Bruxelles. At the chocolate shop! ::totally giddy::


Nilly - Nov 20, 2005 8:01:15 am PST #5745 of 10006
Swouncing

Then the problem is reduced to 3q^2 and p/4 -- p/4 can't have any different factors from p, nor can q^2 from q, so the only way 3q^2 and p/4 can have any factors in common is if p is divisible by 3. Yes?

Do you mean that since we get (p²+3q²)/(p/4)=4p+12q²/p and the first term, 4p, is obviously a multiplication of p, we have to show that the second term, 12q²/p is prime with p? Because if that's what you mean, then you show it quite clearly, IMNotMathyHO. The only thing that can connect between p/4 and 3q² is the 3, for exactly the reason you stated.

Though I suppose, for completeness, I should figure out how to write it in Hebrew?

I'm going to find a way to visually show you that, then. I'm thrilled that you prefer to use that!

Oh, if we're all going to meet in London, I can show you there, I guess.


Fred Pete - Nov 20, 2005 8:03:17 am PST #5746 of 10006
Ann, that's a ferret.

I vote for a F2F in Bruxelles. At the chocolate shop! ::totally giddy::

We know a great chocolate shop in Philipsburg, St. Maarten.

IJS.