Spivak's Calculus on Manifolds asks the reader to prove this:
If there is a basis $x_1, x_2, ..., x_n$ of $\mathbb{R}^n$ and numbers $\lambda_1, \lambda_2, ..., \lambda_n$ such that $T(x_i) = \lambda_i x_i$, $1 \leq i \leq n$, prove that $T$ is angle-preserving iff $\left| \lambda_i \right| =...
if the speed of light equal 186,ooo miles per second how fast will the speed of light be if 1000 years equal 1 day or a 24 hour period of time this question has stomp me for years plesae help me an can you shown how you got the answer
What I'm asking is basically: what are sufficient conditions on a basis for $\mathbb{R}^n$ in order that all "angle-preserving" linear maps are just flipping of some of the basis coordinates, and scaling them all up by the same number?
if I have y^3/y^-4 and y^-4 is 1/y^4. So I have y^3/1/y^4. Wouldnt you cross multiply with y^4 to cancel the bottom half out and end up with y^3 * y^4 on top? The example says thats wrong, that yuo end up with y^3 - y^4 on top..
@anon can you help explain this to me, (I am seriously lost with this, I am not asking you top do my homework, its just i have been at this problem for 2 hours now)
We say a subset $I\subset R$ is an ideal if it is an additive subgroup and closed under ambient multiplication, i.e. $RI\subseteq I$. The quotient ring $R/I$ is comprised of the additive cosets $a+I$, with addition defined as $(a+I)+(b+I)=(a+b)+I$ and $(a+I)(b+I)=ab+I$. At this point we have an exercise: show that these are well-defined operations on $R/I$.
With algebra, it is nice to carry examples in your head to test these general definitions and theorem against. Like you can think about what $Z/6Z$ should be or $Z/2Z$, etc.
If $a\in I$, then $a+I=0+I$ is zero in the quotient ring. Thus, if $I=(a)$ is the principal ideal generated by the element $a$, i.e. the set of all multiples of $a$, i.e. $I=aR$, then in $R/I$ we have that $ra+I=0+I$, hence is zero in $R/I$, for any $r\in R$.
In particular, adding 1 to itself four times in Z/4Z gives you zero. And any multiple of 2+2i is zero in the quotient ring Z[i]/(2+2i).
The fact that $a+I=0+I$ as additive cosets of $I$ when $a\in I$ is because $I$ is closed under addition. This is no different from $xH=H$ when $x\in H$ in group theory.
Any ring R is an abelian group under addition, which we denote (R,+). The additive order of x in R is the order of x in the group (R,+). Why, where did you see this?
in example 11: To demonstrate that there is not, we will show that these five cosets are distinct. It suf-fices to show that 1 + (2 + 2i) has additive order 5.
In my mind, the quotient ring R/I is what we get if we pretend everything in I is zero, and face up to the consequences. (For instance, if $a,b\in R$ differ by an element in $I$, then a consequence would be that they differ by zero hence $a=b$ in $R/I$.)
Note that, given $X\subseteq Y\subseteq Z$ where $Z$ is a topological space, if $Y$ is open in $Z$ and $X$ is open in $Y$ then $X$ is necessarily open in $Z$.
the open sets of [0,1] are obtained as the intersections of open sets in R against [0,1]. In particular, (0,1] is obtained as the intersection of the open interval (0,a) with [0,1], for any a>1.
there is a definition of interior that is specific to Euclidean space, and there is also a definition of interior point that is generic and applies to any space. which are you using?
well, any point in (0,1] has a ball around it contained in (0,1]. that should be clear for all x in (0,1), so it suffices to see that 1 has a ball around it contained in (0,1]. pick the open ball of radius 1/2 around 1 inside the topological space [0,1], and you will get (1/2,1], which is indeed contained in (0,1].
that is, B(1,1/2) = {x in [0,1] s.t. |x-1|<1/2} = (1/2,1] is contained in (0,1]
A neighborhood of a point x in a space S is a set containing x, which also contains some open set U which contains x. An open ball (defined in metric spaces) of radius r, denoted B(x,r) (notation varies), is the set of all points a distance less than r away from x. An open interval in R is a set of the form (a,b) (where a and/or b might be infinity), and other intervals include those of the form [a,b),(a,b] and [a,b].
But let's break things into pieces like I suggested @hyg17. (a) A point in a metric space is interior to the set S if there is a ball around that point contained in S. (b) Every element of (0,1) is an interior point of (0,1] in the space S=[0,1]. (c) x=1 is an interior point of the set (0,1] in the space [0,1]. Which part do you want to go over first?
To show (0,1] is open in [0,1], under your definition of open (as containing all interior points), we must show everything in (0,1] is an interior point of (0,1]. This breaks into two parts: showing everything in (0,1) is an interior point of (0,1], and showing 1 is an interior point of (0,1].
@hyg17 yes, but that's not relevant unless I misunderstand why you want that fact
to show 1 is interior inside the subset (0,1] of the metric space [0,1], you need to show there is an open ball around 1 that is contained within (0,1]. this does not mean we want to show every open ball around 1 has merely some point inside (0,1].
so you're asking why (0,1] is not open in R? Because in the metric space R, 1 is not an interior point of (0,1] (even though it is an interior point if the metric space is instead [0,1]).
My book tells me that "A point p is interior of E if there is a neighborhood N of p such that N is a subset of E" Why is 1 the problem? isn't 1 inside (0,1] ?
Firstly, do you agree that that definition of interior is equivalent to "a point p is interior of E iff there exists an open ball around p that is a subset of E" (in the case of metric spaces)?
Also, you seem not to have read what I just said, which stated that 1 is interior to (0,1] if the metric space is [0,1] but is not interior if the metric space is all of R.
(0,1] is not open (in R) because 1 is not an interior point of (0,1] (in R) since any open ball around 1 (in R) will contain points strictly bigger than 1 whereas (0,1] does not contain any points strictly bigger than 1 hence none of these balls can be contained in (0,1].
@GustavoBandeira yes. do you know the law of the excluded middle? either something is true or it isn't. thus, either "x is a rational number" is true, or it's false. the logical negation of "x is a rational number" is by definition "x is an irrational number"
"x is either rational or irrational" is exactly the same as "x is either rational or it is not rational," by the very definition of irrational (= "not rational"), which is as obvious a thing as you can get ("something is either true or it isn't")
algebraic numbers, being of an algebraic persuasion of course, are roots of integer-coefficient polynomials (polynomials are algebraic thingies). those that transcend polynomials, those numbers that are not roots of any integer coefficient polynomial, are transcendental numbers.
@anon I possess a book in which the author says that algebraic numbers are numbers that can be expressed as solutions to a polynomial with integer coefficients. But wikipedia says about rational coefficients.
@GustavoBandeira Multiply a polynomial in $\mathbb Q[x]$ by a common denominator of the coefficients to get a polynomial in $\mathbb Z[x]$ with the same roots.
For instance, $0.1\overline{23}=0.1(0.23+0.23/10^2+0.23/10^4+\cdots)$. Moral of the story, any eventually recurring decimal expansion can be made into a rational using the geometric series formula.
TeXworks wouldn't parse my latex document because I think I had a & misplaced. I removed all &'s from my latex code (literally all of them), but it still cites "\insertsectionhead ...er analysis on ${\bf Q}_p$ & Tate's thesis" as the error (which is a version of the section title that is no longer present in my code). I have experienced this before, where I remove something TeXworks didn't like but then it still thinks it's there in the code. Anybody familiar with this?
I now literally copy/pasted the code to a new tex file in a different folder, and now it parses fine. same fix I used the other times too.
I have a file tensor.tex, and when compiled, it creates a file called tensor.aux which has all section names (for a toc even though I don't have one) along with other information.
Emil Artin showed in his 1955 papers$^1{}^2$ that groups of Lie type of characteristic $p$ have a large Sylow $p$-subgroup. That explains why the families $^2D_n,^2E_6,F_4,^2G_2$ and $E_n$ ($n=6,7,8$) have a "spike" at exactly their characteristic, and why the $2$-power in the orders of other gr...
yay last minute studying for Linear Algebra quiz tomorrow morning
>.<
The worst part about Linear Algebra is when you read a definition for something and you have no idea what they hell a letter is referring to (a basic/matrix/subspace/etc)
In a ring R, if ab=0 for nonzero elements a,b in R, then neither a nor b can have an inverse. For suppose (wlog) a has an inverse a', so aa'=1. Then a'(ab)=0 implies b=0, a contradiction.
since the set {0,1,2,3} is closed under subtraction, to show they are distinct (which is to show that a-b is not 0 in the quotient ring when a,b are distinct from that set) it suffices to show that none of 1,2,3 are 0
but if k=m(2+2i) in the gaussian integers Z[i] for some m, then taking norms we get 8|k^2, which does not hold for k=1,2,3. this shows that there are at least 4 elements in this quotient ring.
One could also appeal to Lagrange's theorem, since the additive subgroup--not ideal--generated by 1 has order 4, the ring has to have size 4r for some integer r.
now, the general form of elements will be a+bi. 2ki=-2k and (2k+1)i=-2k+i. hence every element in Z[i]/(2+2i) can be written in the form a+bi where a is in {0,1,2,3} and b is 0 or 1. this is precisely 8 elements.
@user1 for this we still have to rule out characteristic 2 (and the trivial ring) in order to know 1 has additive order 4 (which is still easier than using norms when one hasn't heard of them, I bet)
Spivak's Calculus on Manifolds asks the reader to prove this:
If there is a basis $x_1, x_2, ..., x_n$ of $\mathbb{R}^n$ and numbers $\lambda_1, \lambda_2, ..., \lambda_n$ such that $T(x_i) = \lambda_i x_i$, $1 \leq i \leq n$, prove that $T$ is angle-preserving iff $\left| \lambda_i \right| =...
if the speed of light equal 186,ooo miles per second how fast will the speed of light be if 1000 years equal 1 day or a 24 hour period of time this question has stomp me for years plesae help me an can you shown how you got the answer
