What the fuck?

Oh yeah, I forgot you live in Brazil. Thats still some freaky shit man.

I'm not. I don't know what you're saying anymore. I was pointing out that 1 is a valid GCD, since you said it had "no GCD" if the algorithm gave you 1. I have no idea where you're going with the "lower common divisor". If you want to prove that no other number can be the GCD, you'd have to prove that there's no BIGGER divisor.

Also, a|b means "a divides b". I'm not sure if you're saying otherwise.

By the way, it's valid up to a unit (i.e. a divisor of 1) or something like that. Probably. I'd have to think about it.

2k and 2k+1 are relatively prime. 1 is an acceptable GCD. To clear myself up, I was meaning to say that there is no way to simplify 2k and 2k+1 to find a greater common divisor than one, so they are relatively prime and the LCM of 2k and 2k+1 is 2k(2k+1).

This was just a step into working with polynomials and the Algorithm, it was just an observation, perhaps worded poorly. Something to look at I guess.

This post has been edited by Elnendil: Oct 5 2010, 06:28 AM

Well, let's try it and see what happens.

I generated two somewhat-random polynomials:
f = x^6 + 5x^5 + 7x^3 + 7x^2 + 2
g = x^8 + 6x^6 + 7x^4 + 2x^2

g = f * (x^2-5x+31) -162 x^5+35 x^4-182 x^3-217 x^2+10 x-62

Can't really go any further than that. But I did find out that q and r are unique.

On another topic of math, my Calculus II professor showed how to figure square roots using arithmetic. Looking it up at another source makes me want to ask him for some more information, but all i'm really doing is trying to find a pattern. I think I found one, but i'll just discuss it with the professor later.

:U Square roots by hand.

PHYSICS MIDTERM AGH


Saw a doctor today and got some antibiotics for my bronchitis or whatever. She never actually told me what I had. But I have my voice back and I'm already feeling better.

This post has been edited by Hayate: Oct 6 2010, 02:17 AM

Hmm:
http://www.google.com/search?q=euclidean+a...thm+polynomials

So the thing is, you're not restricted to integers in the Euclidean Algorithm for polynomials. It's so obvious in retrospect, since you're not looking for a constant GCD.

Also:
http://en.wikipedia.org/wiki/Fast_inverse_square_root

That makes sense. Still, it was neat to fiddle with, although I haven't hit all of the good parts of the concept.

Also, the whole thing with me trying to figure out what would happen with two different variables using the Euclidean Algorithm just led me back to a very basic point thats mentioned about the Algorithm: If a=bq+0, then b|a. And thats kind of obvious, since there's some q that b multiplies into that equals a, so b|a. Although I actually just had to retract the statement I was going to make, because I was going to imply that there was nothing to show, but either way it does show something (I made it so a=b(a/b)+0, and using an example, 2000=1000(2)+0 "doesn't imply that 2 is the greatest common divisor". Well, 1000 does, and I actually proved that in my head earlier today, so go me missing the point on that).

Anyways, thats enough for tonight.

wat

Whats the "wat" for? If its about this, "a=bq+0, then b|a", then mull it over I guess. Its one of the basic principles of the theorem. It makes perfect sense. I don't know what else to tell you now. Do you need more examples? Here:

Of the form a=bq+r:

15=3(5)+0. Then we can safely assume that b|a.
100=10(10)+0. We can safely assume that b|a.
60=30(2)+0. We can safely assume that b|a.
12,504,698,990 = 5(2,500,939,798)+0. We can safely assume that b|a.

Or should Just mention that its about integers to help?

This post has been edited by Elnendil: Oct 6 2010, 04:24 PM

I DON'T KNOW WHAT YOU'RE SAYING

Oh. Yeah. Actually, the definition of divisible is that:
a|b iff \exists q s.t. b = aq.

I took math in high school.

I'll be picking it up later, but after finding out a few things, I decided to finish up this half semester of this german intensive course and drop the second half. My semester is pretty heavy, and i'm in courses in which a lot of concepts are very new to me, and German isn't my major. It'll be something to consider, obviously, if I want to get a Ph.D, but right now I need to focus on whats important. Doesn't mean i'll slack off though, I want to pass the class with a grade that is as high as it possibly can be. It'll last until next week, after that things will be more relaxed. I've barely gotten any sleep, the only thing i'm running on is being entertained, be it listening to upbeat music in the morning like its my coffee or making mathematical observations.

Aside from that, things are starting to get better all round. Discussed some observation I made with my professor after class, got a slight refresher on Trigonometry too, because its one of my weak points. Next week i'll be catching up on my math homework so i'll understand all of these concepts better.



Yep. If a|b, there has to be some integer q that, when multiplied by a, equals b. I've been throwing some other concept around recently, has to do with using arithmetic to find square roots. This isn't a theorem, just a plausible observation:

q=lim(x->1+) sqrt(b)/x

When solving, it makes sense in a way that well, if x=1, then q=sqrt(b). But the observation is that you can get an approximation if q is an integer, when sqrt(b)/x and x approaches 1 from the right. Like how sqrt(33)=5.744... So there is some sqrt(x) that you can multiply into 5 that will equal sqrt(33) [5*sqrt(x)=sqrt(33)]. Then I divided both sides by sqrt(x), and I noticed that every time I pulled out a new decimal place, sqrt(x) approached 1 from the right, hence the observation. I showed this to the professor and he mentioned some other theorem...Newtons...gah, I forget the name. Its in my calculus book, i'll look into it later; has to deal with constantly finding approximations to find roots. I'll also say from observation that 1<=x<2 and q is the greatest integer of the root. Anyways, what do you think Rai?

This post has been edited by Elnendil: Oct 7 2010, 08:02 PM