@EE18 Well, Bėzout's identity implies Euclid's lemma, which leads to the Fundamental Theorem of Arithmetic (that every natural number has a unique prime factorisation).

@PM2Ring I think this is what I was thinking about!

I think my book went with Euclidean division algorithm implies FTA, and then got Euclid's Lemma as a corollary

So I See Thorgott's point about order of all this being arbitrary

is any one good at differential equations here?

if $b_t(s)$ satisfies a pde where t is time and s is space

and you add up discrete space slices to obtain a function strictly of time and look at what happens at $t \to \infty$

I wonder if that has a differential equation intepretation

How important is it to know elementary number theory for analysis?

I feel like there are a lot of facts out there in the ether which I don't know

Bezout, Chinese Remainder, Extended Euclid, and I feel like so many more

I figured out how to post 2 questions simultaneously without waiting the 40minutes

I conjecture that it works for $N$ simultaneous questions

implying that I could fill the entire front page

with all my questions

from my calculations there are 100 questions shown at any given time on the front page

Well I guess I will start writing

will I get banned?

I conjecture that 'no' I won't get banned

as it's just a loophole in the system

@EE18 for analysis, none of these matter

OK, will not get sidetracked then

Will I guess leave them for some other study at another time

BTW Thorgott, what do you think the authors have in mind with the below. Exercise 6.3 was just that there are $n \choose k$ $k-element$ subsets of an $n$-element set, and Theorem 8.4 is the binomial theorem. Obviously putting the two together gives me $2^n = \sum_{k=0}^n n \choose k$ but I don't see how that gets me where they're suggesting. I was going to just start by induction but that doesn't seem like what they're suggesting?

so if I'm constructing a map from a quotient group to another quotient group, I'm mapping congruence classes to other congruence classes right

e.g. $f:\mathbb{Z}_{10}\to\mathbb{Z}_5$

defined by $f([x]_{10}) = [x]_{5}$

the kernel is $\{[0]_{10},[5]_{10}\}$ I think

how would one describe $SL_2(\mathbb{Q})$? Like $$\left\{A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\bigg| a,b,c,d\in \mathbb{Q}, ad-bc=1 \text{ and }A^{-1}\in GL_2(\mathbb{Q})\right\}$$?

I guess I don't have to really write that since it's implied in the definition

w/e

idunno

how bad does $A\in B \ni C$ look :D

when I'm too lazy to erase and rewrite $A,C \in B$

@XanderHenderson might smite me for that

@Obliv looks correct.

@Obliv A^{-1} has rational entries if A does. Moreover, since A A^{-1}= I, you also have det ( A^{-1})= 1 so actually A^{-1} is in SL_2.

oh ok

how do inverses work in $SL_2(\mathbb{Q}$? Like, if I wanna show $SL_2(\mathbb{Q})\subset GL_2(\mathbb{Q}$ I should show for $A,B \in SL_2(\mathbb{Q})$ that $AB^{-1} \in SL_2(\mathbb{Q})$ but it's matrix multiplication and addition in $\mathbb{Q}$

oh wait maybe this is a ring not a group?

anyway, this seems so trivial but I have to show it :\

i don't mean to be annoying, but the details of this depend on how you have defined SL_2(Q). i would not assume that because there is a standard notation for that object, there is a single standard definition of it

isn't it just the set of 2x2 matrices with entries in $\mathbb{Q}$ that have det = 1 and invertible

i.e. your question is likely entirely about how to get it out of your definitions, and not just something you can ask someone who knows what SL_2(Q) is, because they don't know what you know, or are "allowed to use," about that object

obliv: that's one definition. is it your definition?

what do you know about determinants?

It's the definition on wiki and I just assume that's the only one I guess

(i also have bad news re: "it depends on how you have defined the thing" and determinants, but maybe we don't need to go there)

well probably, but isn't it typical to show $H \subset G$ via $ab^{-1} \in H$ for some $a,b \in H$?

to show H is a subgroup of G

So if (V,+,.) is an n dimensional vector space over R, where . is scalar multiplication, then the vector space (V,+, p), where p is another scalar multiplication need not have dimension n right?

OK. it might help to preface a question contextualized like that with "i'm just reading stuff randomly on wikipedia and don't have any particular set of definitions in mind, or any particular limit on what i am allowed to use. if SL_2(Q) is defined as in [web page], [question]"

to answer your question, if you have something in SL_2(Q), it is (among other things) an invertible 2x2 matrix, and its inverse in the sense of the usual algebra of 2x2 matrices is also going to be in SL_2(Q)

you also seem to be maybe asking if SL_2(Q) is a subgroup of GL_2(Q), but i can't be sure

GL_2(Q) is pretty much SL_2(Q) but with the invertible matrices that don't just have det=1

so it's pretty obvious (at least to me) that it'll be a subgroup. I just have to show that for any A,B in SL_2(Q) the det=1 property is preserved under matrix mult.

I was going to type out all the matrix operations algebraically

but the message was too long lol

let A,B be in SL_2(Q) with AB=(a_1,a_2,a_3,a_4)(b_1,b_2,b_3,b_4) arranged s.t. a_1,a_2 are the top row entries and a_3,a_4 are the bottom (same for B), then AB=(a_1b_1+a_2b_3,a_1b_2+a_2b_4,a_3b_1+a_4b_3,a_3b_2+a_4b_4) and since det=1 for A,B separately, we know a_1a_4-a_3a_2 = 1 and b_1b_4-b_3b_2 = 1 so for AB we have (a_1b_1+a_2b_3)(a_3b_2+a_4b_4)-(a_3b_1+a_4b_3)(a_1b_2+a_2b_4) multiply it out to get (a_1b_1+a_2b_3)(a_3b_2)+(a_1b_1+a_2b_3)(a_4b_4)-(a_3b_1)(a_1b_2+a_2b_4)-

(a_4b_3)(a_1b_2+a_2b_4)

then manipulate that stuff and hopefully det=1

obliv: this is where not starting with some set of definitions might hurt you. it isn't particularly clear from an "expanded out" formula for det as an express polynomial in the matrix entries that det(AB) = det(A) det(B), which is basically what you're wanting to use here.

it helps to add at least one layer of abstraction to prove that property of the determinant. although there's no reason why the idea of the argument you sketch couldn't go through.