@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).
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
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
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?
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)-
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.
so you wouldn't even know a definition of the determinant for, say, 5x5 matrices? or nxn matrices?
maybe they gave you something
but yeah, if they didn't, i'd grab a book that does give you a definition. det(ab) = det(a) det(b) is something that you will see proved pretty soon after any definition.
which isn't to say that the proof is always nice to look at
@Koro this is an interesting question. offhand i would guess that the dimensions do have to be the same, but also that it matters that your field is R and not some arbitrary field
i'm thinking, given some more general field k, if there is a field endomorphism f: k to k that is not surjective, then {1} is a basis for (k,+,*), but if you give k the funny scalar multiplication k.x = f(k)*x, then {1} is not a basis for (k,+,.). which is not all by itself a proof that (k,+,.) isn't one-dimensional, is suggestive that maybe it doesn't have to be
e.g. if k is something like Q(t) and f is the automorphism sending p(t) to p(t^2), it feels like maybe {1,t} could be a basis for (k,+,.)
but R doesn't have any endomorphisms that aren't automorphisms
i guess maybe actually the fact that {v} isn't a basis for a nonzero vector v in V is a proof that V isn't one dimensional. it's just, this fact all by itself wouldn't tell us what the "not one dimensional" dimension is
Hey! I've tried to solve this problem for quite a few times now and I'm just not getting it. I looked up the solutions online and most of them don't even get the answer 41. Any help would be appreciated!
Let $A$ be set such that $n(A)=5$. How many functions can we define on $A$ with the property $(f\circ f)(x)=x$ ?
I think the identity function works but what about others? Should $f$ have an inverse? I think permutations may be involved, but I am not sure how to progress.
@leslietownes Man I had gotten 25 itself in my first attempt (I had forgotten to map the numbers to themselves). It's so weird they have not made any clarification about if this question was given bonus or something. Such a wastage of time and effort
@SoumikMukherjee There are more errors (or I guess ambiguity) in the biology section. (You have to study PCMB to get the top ranks). And they don't even have an official answer key on their site.
Let $A$ be set such that $n(A)=5$. How many functions can we define on $A$ with the property $(f\circ f)(x)=x$ ?
I think the identity function works but what about others? Should $f$ have an inverse? I think permutations may be involved, but I am not sure how to progress.
