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8:24 PM
I can guarantee a const remainder from division algorithm of polynomials right
like $f(x) = q(x)p(x) + r(x)$ where $\text{deg}(r(x)) = 0$
I know it's b/t $\text{deg}(q(x))$ and $0$, but I can make it 0?
specifically, can I guarantee existence of some $q(x)$ such that $r(x) = 1_F$
 
Huh?
 
I don't actually need it to be $1_F$ I need it to be constant so I can choose a $g(x)$ such that $f(x)g(x) = v(x)p(x) + 1_F$ for some $v(x) \in F[x]$ and $p(x)$ is coprime with $f(x)$
 
You only get a constant remainder if the degree of $q$ is $1$.
You're confusing various things.
 
I've made a mess of my latex document but I basically wanted to use division algorithm to make this $g(x)$
 
If the gcd of $f$ and $p$ is $1$, then you get polynomials $g$ and $v$ so that $g(x)f(x)+v(x)p(x)=1$. This has nothing to do with the division algorithm per se.
You can do the Euclidean algorithm to find such $g$ and $v$.
 
8:29 PM
Ohh right I forgot about that linear combination thing
 
The Euclidean algorithm may be the most important thing you learn in this course. You will use it over and over and over and over.
 
so can I use the euclidean algorithm to get constant polynomials $r(x),h(x)$ s.t. $f(x)=p(x)q(x) + r(x)$ and $g(x) = p(x)m(x) + h(x)$ and $h(x)r(x) = 1_F$
so that $f(x)g(x) = p(x)[...]+h(x)r(x) = p(x)s(x) + 1_F$
hence $f(x)g(x) \equiv 1_F$
where $s(x)$ was the $[...]$ we get from multiplying the terms with $p(x)$
I was going in that direction but I didn't know if I could guarantee $h(x)r(x) = 1_F$
I shoulda done the euclidean algo to get it down to degree 0 remainders tho, gonna have to tweak it.
 
8:47 PM
You're writing stuff that's wrong again, Obliv. Understand the difference between the division algorithm and the Euclidean algorithm.
I assume you've actually had homework to do some of these computations. If not, you should have.
 
 
1 hour later…
10:09 PM
Or try and make up your own simple examples :^)
 
go sit in the corner and compute gcd(28657,46368)
 
Trivial. We’re on polynomials. Pay attention!
 
it takes like 20 steps of the algorithm, ted :)
i chose the worst possible numbers of that size
in hell they make you compute the fibonacci sequence backwards from infinity
 
10:37 PM
that would be infinitely torturous
 
10:58 PM
@leslietownes Surely that's hyperbole.
 
@TedShifrin no, Fibonnaci numbers are a known worst case for Euclidean algorithm
 
if you can find a, b with max(|a|,|b|) <= 46368 and gcd(a,b) taking more steps in the euclidean algorithm than the above, be my guest
 
Fibonacci, poor italian names
 
okay, Alesanndro
 
Fibonnaci is the other mathematician in town who changed his name so he can get the business of confused townsfolk who use the yellow pages
he also owns AAAAA Mathematical Services
 
11:09 PM
@Jakobian You'd be surprised by how many germans wrote my name as Allesandro
 
two?
 
Alles of them?
 
remind my of a video "explanation in five levels why 0.99999... = 1"
one of them was an infinite series... the other one was a limit :P
 
alles andro oder was?
 
idk what the author was smoking
limit was after infinite series by the way
 
11:14 PM
this isn't the level of rigor i expect in viral videos
 
and then he started at level 5: "oh yeah but in hyperreals..."
pretentious
 
there should have been a sixth level of a guy explaining why (well, basically just stating that) all numbers are the same, and telling people to repent of their sins before it's too late
 
another video I've watched was someone supposedly claiming that Darboux integrals are superior to Riemann integrals
I didn't see any real argument in that video
 
all integrals are the same too. repent
 
definition of the Riemann integral is way better
 
11:21 PM
a=a if a is real.
 
Almost never needed instead of Darboux, and one additional quantifier is a pain.
@user85795 And if it’s not?
 
Then i repent.
 
Will the world be ablaze?
 
it already is
 
Good point. But you’re not bothering to repent?
 
11:26 PM
not quite yet
 
@TedShifrin not an argument
 
poetry rarely is
 
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