I suppose we'll all try to get through these first six exercises and then we'll move on to chapter 2 when we're all good.
Collaboration should be encouraged here, but if you already know the answer, hints and Socratic lines of questioning would be far more useful to us than direct answers.
But then again we all know that--just wanted to make it common knowledge.
Though while I'm in general supportive of using machinery when available, I think that the puzzle-ish nature of the subject might mean that it's better to stick more to techniques that have been introduced up to that point
I'm more concerned about whether there's an insight that is involved in solving a problem using less machinery, but then once you invoke more it becomes immediate
so just to thrash out Araske's argument on I5: i guess, given an N, you expand it in base 3 to get $N = \sum a_i 3^i$ where $a_i = 0, 1, 2$. rearrange so that $N = \sum b_i 3^i$ and $b_i = -1, 0, 1$
and we want $N$ to be of size $n$, right? because the expansion goes till $3^{n-1}$?
I am not sure I get that. You can expand all $N$'s such that $1 \leq N \leq 10^n - 1$ as $N = \sum_{i = 1}^{n-1} a_i 10^i$, right? ($a_i = 0, \cdots, 9$)
I was thinking, like, if you have some $a_i = 2$, you can rearrange to get all of them to be $-1$ (because $3 - 2 = -1$) but that might add a $3^n$ somewhere in the expansion
Well the base case is obvious: $1$ can be represented as $1 \cdot 1$
For the inductive step, I'm not so sure, but I think it would work to show that any integer between $(3^n - 1)/2$ and $3^n$ is of the form $3^n - k$ where $k$ is an integer less than $(3^n - 1)/2$, and then beyond $3^n$ is of the form $3^n + k$.
Since $k$ has a balanced ternary representation by inductive assumption, it holds.
Err, so if $N$ is so that $3^n \leq N \leq (3^{n+1} - 1)/2$, then $N - 3^n$ is less than $(3^n - 1)/2$? Hmm, it's bounded by $(3^{n+1} - 1)/2 - 3^n = (3^{n+1} - 1 - 2 \cdot 3^n)/2 = (3^n - 1)/2$. Ok, yea
Nice.
i should probably write rough calculation stuff on pen and paper than on this chat :P
exactly. if you do it for homogeneous poly's, you're counting terms of the form $X_0^{i_0} X_1^{i_1} \cdots X_n^{i_n}$ such that $i_0 + i_1 + \cdots + i_n =d$
so this is a partition question; how many unordered partitions are there of $d$ into $n$ parts?
Ok, let's see. So define $\Bbb{RP}^n = \Bbb R^{n+1} - 0/x \sim y$ like Daminark said, where $x \sim y$ if $x = \lambda y$ for some $\lambda \neq 0$. That means a "point" in $\Bbb{RP}^n$ is a whole line through origin in $\Bbb R^{n+1}$
You can write a point in $\Bbb {RP}^n$ as $[x_0 : x_1 : \cdots : x_n]$ where not all of the coordinates are zero - that represents all the points of the form $(\lambda x_0, \lambda x_1, \cdots \lambda x_n)$ in $\Bbb R^{n+1} - 0$
I'd wager they're more just a product of the axioms of a projective space (any two points have a unique line between them) than a concrete visualizable thing
so if I look at the subset $Z$ of $\Bbb R^{n+1} - 0$ where $P = 0$, then that can be realized as a union of lines through the origin in $\Bbb R^{n+1} - 0$
I'll get to something concrete in a bit, but let me talk about a further level of generality here. please bear with me :p i'm sure a better person would have explained things in a better way
So points in $\Bbb P^n$ (abbreviation smabbreviation) correspond to lines-through-origin in $\Bbb R^{n+1}$. I want to construct a space points of which correspond to hypersurfaces in $\Bbb P^n$ :p
Actually, this is not too hard. Let $P = 0$ be a hypersurface in $\Bbb P^n$. Since $P$ is homogenenous write it as $P = \sum a_{i_0 i_1 \cdots i_n} X_0^{i_0} \cdots X_n^{i_n}$
So identify (a_blah) with (lambda a_blah). The relevant space which classifyies degree d hypersurface in $\Bbb P^n$ is $\Bbb P^{\binom{n+d - 1}{n} - 1}$
This is the "moduli space of degree d hypersurfaces in Pn" (TM)
Why do I care? Because this construction has super-cool applications. Remember the geometry fact that for any 5 points on R^2, a conic passes through them?
just try proving it for 5 points in P^2 (any conic on R^2 can be homogenized anyway; x^2 + y^2= 1 becomes (x/z)^2 + (y/z)^2 = 1, ie x^2 + y^2 = z^2 which is homogenenous)
@Daminark yeah, equivalently, a conic on $\Bbb P^2$ is $P = 0$ where $P$ is a degree 2 homogenenous polynomial
So suppose $p_1, p_2, p_3, p_4, p_5$ be 5 points on $\Bbb P^2$. Look at the space $S_i$ of conics passing through $p_i$ for each $i$. These are hyperplanes (copies of $\Bbb P^4$) in $\Bbb P^5$, because if you plug in the coordinates of $p_i$ in a conic $P$, you end up with a linear equation in $a, b, c, d, e, f$ = 0
Or if I'm worried that the people will press I'll just say "If I explain it, things will be confusing, but the more you look at it, the more trivial it becomes" (re: my analysis prof :P)
So you get 5 hyperplanes $S_1, S_2, \cdots, S_5$ in $\Bbb P^5$. Those intersect at a point (like in Euclidean spaces)
generically at least. unless $p_1, p_2, \cdots, p_5$ lie on a line or lie one above another or something, in which case they'll intersect in more than a point, but at least a point
room topic changed to Number Theory Study Group: Actually Number Theory Study Integral Domain but whatever (Weil's Number Theory for Beginners) (no tags)
Sounds good. It's a pretty comprehensive book on the basics of number theory. For what it's worth we'll probably only really start picking up steam later once finals are through.
And I mean, depending on when we get through this we may go deeper into other books
Very Secret Number Theory Study Group
Very Secret Number Theory Study Group