1h
comment What is a zero morphism in an abelian category
@Tom: Yep. $A \to 0$ is always a zero morphism too.
2h
comment Abuse of site with homework assignments
... simply reading others' solutions until it sinks in through osmosis. The bad questions will drown out the good questions, further discouraging those who wish to ask them from participating, especially so when they receive bad answers from the people who have trained to engage in the "blunt question - submittable solution" pattern. It will drive away those who wish to answer questions that get turned off by these practices.
2h
comment Abuse of site with homework assignments
The problem is that you can't just look at individual questions in isolation: you have to look at them as a whole, and what effect our reactions to them have. And you can't just look at one side of the issue either. The more obvious, serious drawbacks are that if "bluntly stated" questions are rewarded with solution manual entries, that will encourage more ill-intentioned people to participate. That will encourage well-intentioned people not to learn how to ask good questions. It will even convey the impression that one "learns" mathematics not by thinking about problems, but...
2h
comment Proof that the coefficients of a polynomial are real
@Amr: By computing its imaginary component $\Im\{z\} = \frac{1}{2}(z - \bar{z})$ to be zero. Or just showing it's equal to its complex conjugate.
14h
comment Real solution of $\sin (\cos (\sin x)) = \cos(\sin (\cos x))$.
Have you looked at the graph yet?
15h
comment Are there contradictions in math?
@Marc: You only need the sign bit if you want to interpret a number as being positive or negative; except for inexact division, the actual arithmetic is just mod $2^n$ arithmetic, and the most significant bit has no special meaning. If you're encoding ordinary integers as $2$-adic numbers, you don't need just a single sign bit: you're negative if and only if your number is $\overline{1}xxxx$ and positive if it is $\overline{0}xxxx$.
15h
comment Abuse of site with homework assignments
+1. Not because I like what you have to say, but because I feel it's accurate. I came back to MSE once I realized that the community had finally settled upon being antagonistic towards these questions, but I've increasingly been feeling that MSE is simply a lost cause.
15h
answered How does negating a matrix affect its eigenvalues?
15h
comment Relation between continuity of $f$ and analyticity of $f(z)^8$
Your reasoning is sound! The only obstacle to making it a rigorous argument is finding a technical condition that captures your intuition: the eight branches of $z \mapsto z^{1/8}$ are disjoint and so any continuous function must stay "within" a branch and thus be analytic. I want to say something about "Riemann surfaces" or "sheaves" or "local homeomorphisms", but I'm not sure if you're familiar with any of these.
16h
comment Category of Presheaves on a small category $C$ is locally cartesian closed
If you truly believe $F$ to be one half of an equivalence, then it is going to be fully faithful and have an adjoint. So if you boldly charge ahead and try to write down what its adjoint would be, that would give you the $G$ you're looking for.
16h
comment Category of Presheaves on a small category $C$ is locally cartesian closed
The adjoint of an equivalence functor is the other half of the equivalence.
16h
comment Category of Presheaves on a small category $C$ is locally cartesian closed
Are you sure you have everything straight? So many $C$'s! And the two appearances of $q$ have quite different types. Your problem simply might be that you have confusing notation or actually actually confused something about the particulars of the problem.
1d
comment How to prove $\frac{\partial z}{\partial \bar{z}} = 0$ if and only if $\frac{\partial \bar{z}}{\partial z} = 0$
Anyways, you can complexify $\mathbf{R}^2$ to $\mathbf{C}^2$, and $z$ and $\bar{z}$ (given the same definitions relating them to $x$ and $y$) are now functionally independent as well; maybe you'll find the interplay between $\mathbf{R}^2 \subseteq \mathbf{C}^2$ a useful way to think about things.
1d
comment How to prove $\frac{\partial z}{\partial \bar{z}} = 0$ if and only if $\frac{\partial \bar{z}}{\partial z} = 0$
@Georges: Wait a minute, I'm confused by the point of your example: your function relates $x$ and $x^2$, and thus proves them dependent in a variety of senses (e.g. functionally, algebraically, smoothly), not independent.
1d
comment How to prove $\frac{\partial z}{\partial \bar{z}} = 0$ if and only if $\frac{\partial \bar{z}}{\partial z} = 0$
er, I meant "relevant" not "relative". And yes: $z$ and $\bar{z}$ are differentiably independent. They are clearly not functionally independent, but that's not the only sense that matters. Even in your example, we have examples of different kinds of dependence: $x$ and $x^2$ are linearly independent, for example.
1d
comment How to prove $\frac{\partial z}{\partial \bar{z}} = 0$ if and only if $\frac{\partial \bar{z}}{\partial z} = 0$
@JLA: They are independent in many important senses; e.g. there is no complex differentiable function such that $\bar{z} = f(z)$, or even $f(z,\bar{z}) = 0$. The differentials $\mathrm{d}z$ and $\mathrm{d}\overline{z}$ are $\mathbf{C}$-linearly independent. And so forth. The goal of working with the variables $z$ and $\bar{z}$ is that they are independent in enough of the ways that actually matter, and calculus is much simpler when you can leverage complex differentiation. The formulation of your homework problem just doesn't make sense, at least not without being familiar with the book.
1d
comment Find $T(1)$, $T(x)$ and $T(x^{2})$ and $T(ax^2+bx+c)$
If all else fails, turn it into a problem about matrices.
1d
comment How to prove $\frac{\partial z}{\partial \bar{z}} = 0$ if and only if $\frac{\partial \bar{z}}{\partial z} = 0$
@Georges: the relative notion of dependence here is almost surely "there exists a nontrivial differentiable function such that $f(z, \bar{z})$ is identically zero".
1d
answered How to prove $\frac{\partial z}{\partial \bar{z}} = 0$ if and only if $\frac{\partial \bar{z}}{\partial z} = 0$
1d
revised valuation ring is a field?
added 157 characters in body
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