What we were talking about before, though: In $\mathbb Q(\sqrt2)$, $\pm\sqrt2$ are indistinguishable. In $\mathbb R$, they're not, since $\exists c(c^2=X)$ is true for one and not the other.
Right-the extra structure of "multiplication" means that we can talk about positives and negatives, because the rule positive times positive = positive is qualitatively different then negative times negative = positive
and not any business involving field extensions. (though i suspect i should know a bit more about that kind of thing, if only so i could properly understand wth differential galois theory is)
I should mention that the two families of solutions are very different: In the north, the distance you walk east/west doesn't matter; in the south, it does.
i think what i'm curious about are what conditions on the chart/surface are necessary to guarantee the existence of some closed coordinate curves on that surface, and maybe something about the distribution of their circumference on that surface
i figured, heh
obviously curvature plays a big part in the story, though i don't rightly know how to describe that
right. the trouble is that it no longer feels like a terribly 'natural' chart, though with that i'm definitely not able to talk about generic charts anymore
@DavidWheeler: the torus is interesting to me because, if one interprets it simply as a matter of walking in a certain direction
then there's two cases: either the trajectory eventually returns to itself (in which case the 'slope' of the line in the aforementioned square is rational) or it doesn't (irrational)
in the former case, that rational slope determines the number of times one winds around the two loops, and one can determine the length of the trajectory from that
in the latter case, though, the line never stops and indeed serves as a space-filling curve for the torus
though what's also interesting about the torus is that, for any given direction (in my sense) on the torus, the question of 'how long will i walk until i get back to where i started' is the same for all starting points
unlike the sphere, where walking east always gets me back to where i started, but takes a lot longer at the equator
Do invertible matrices have only one inverse? (I don't know much linear algebra, so I don't know.)
If so, then it's got to be real, because $i$ and $-i$ are indistinguishable in $\Bbb C$, so if $A^{-1}$ is one inverse, its complex conjugate has also got to be an inverse.
no, @DavidWheeler, that's not a terrible argument; let $A^*$ denote the complex conjugate of $A$. then, if $A$ is real, invertible, and $AB = I$, $$ I = I^* = (AB)^* = B^*A^* = B^* A.$$
since left and right inverses agree here, $B = B^*$ as desired.
@MikeMiller I don't know what you're on. Share plx.
Also, the term "complex conjugate" is a bit ambiguous in this context. Do you mean, for $A = (a_{ij})$ that $A^{\ast} = (\overline{a_{ij}})$ or: $A^{\ast} = (\overline{a_{ji}})$?
Typically, Id refers to the identity function. However, any matrix can be made into a function via matrix multiplication with the proper-sized column vectors: $A(v) = Av$.
It turns out the matrix for the identity function is the identity matrix. Not terribly surprising.
I'm reviewing for a test, and of course the last piece I get to review is the one I don't understand due to being sick and missing class (yay flu!).. Anyways, I've been working on this project for a little bit, but I'm stuck on how to find the values for the quadratic equation. I thought I had the correct method, but my graph for the lower part
was way off.. so that's wrong. Could anyone point me in the right direction to find a, b, and c so that I can then find the equation for the 2nd half of f(x)? This is what I have so far: i.imgur.com/RlnVZjb.png
A round membrane in space, is over the space $x^2+y^2 \leq a^2$.
The maximum coordinate $z$ of a point of the membrane is $b$.
We suppose that $(x, y, z)$ is a point of the inclined membrane.
Show that the respective point $(r , \theta , z)$ in cylindrical coordinates satisfies the conditio...
@ʙᴀᴅᴀᴛᴍᴀᴛʜ To prove that $d$ and $d'$ generate the same topology on $X$, you have to show that if a set is open in $(X, d)$, it's also open in $(X, d')$ and vice-versa.
That is, you have the same collection of open set in both spaces.
So if you have two collection of open sets $\tau$ and $\tau'$ on the underlying set $X$, then to show that these are same topologies really means you have to show that $\tau = \tau'$
@ʙᴀᴅᴀᴛᴍᴀᴛʜ You start with a collection of sets satisfying those properties. You call that collection a topology and you call those sets open, that is all.
A round membrane in space, is over the space $x^2+y^2 \leq a^2$.
The maximum coordinate $z$ of a point of the membrane is $b$.
We suppose that $(x, y, z)$ is a point of the inclined membrane.
Show that the respective point $(r , \theta , z)$ in cylindrical coordinates satisfies the conditio...
