No, you’d reserve that for a proposition that you prove to be true (or cite a source for). Depending on the context I’d suggest calling it a conjecture which you subsequently disprove. Alternatively, you could pose the negation as the proposition of interest; once proven, this negation would be a theorem. @Li357
For example:
Conjecture: For any nonnegative integer $n$, the integer $2^{2^n}+1$ is prime.
Proposition: The integer 2^{2^5}+1=4294967297=641*6700417 is not prime.
Theorem: There exists a nonnegative integer $n$ such that $2^{2^n}+1$ is not prime.
@Semiclassical Hm, maybe. For the one it simply depends upon the convergence radius and behaviour of the perturbation series in general. But moreover the form of the equations simply differ for the degenerate and the non-degenerate case of the eigenfunction space.
in the standard linearity $f:E\to R$ we say that it is linear off for all $\lambda,\mu\in \mathbb{R},\forall x,y\in E, f(\lambda x+\mu y)= \lambda f(x)+ \mu f(y)$
$f:U \times V \to W$ est bilinear si 1. $U \to W : u \mapsto f(u,v)$ est linear pour tous $v \in V$, et 2. $V \to W : v \mapsto f(u,v)$ est linear pour tous $u \in U$
also, that's an interesting way to prove the uniqueness of projections in U perp that does not require too much "let x be an element in a set, show that stuff lies in the intersection"
@loch this paper's tendency to treat empty or zero objects as a special case is beginning to be annoying
in particular, they talk about "the zero vector space and free modules", "maximally linearly independent set exists because the torsion module is zero", ...
@LeakyNun I am very particular about the empty set and zero and stuff like that. When I write theorems and proofs, I make sure that they take care of the empty set and zero appropriately.
I don't read Bourbaki cover to cover. The level of generality is not compatible with other books, and they only cover a small fraction of math branches.
I think the most readable Bourbaki book though would be General Topology I and II.
How I would write it: We assume that $M_{tors} = \{0\}$, and find an [injecitve] linear map from $M$ to a free module, so that Prop 2(1) will imply that $M$ is free. We let $(x_1, \cdots, x_n)$ be a generating set of $M$. We let $(y_1, \cdots, y_m)$ with $m \le n$ be a family of elements taken from $\{x_1, \cdots, x_n\}$ with $m$ maximal such that $(y_1, \cdots, y_m)$ is $A$-linearly independent. That such a family exists is because $\{\}$ is linearly independent.
@loch to be rude, it's annoying because it feels like the author is incompetent
How I would write it: We assume that $M_{tors} = \{0\}$, and find an [injecitve] linear map from $M$ to a free module, so that Prop 2(1) will imply that $M$ is free. We let $(x_1, \cdots, x_n)$ be a generating set of $M$. We let $(y_1, \cdots, y_m)$ with $m \le n$ be a family of elements taken from $\{x_1, \cdots, x_n\}$ with $m$ maximal such that $(y_1, \cdots, y_m)$ is $A$-linearly independent. That such a family exists is because $\{\}$ is linearly independent.
I found a very nice, simple word processor recently. It doesn't have too many features but will suffice for most people. It is called Able Word, downloadable from ableword.net.
If you don't need the full features of Microsoft Word and think that LibreOffice Writer is too ugly and not polished, then try Able Word.
"if on closer inspection the conversation is all towards the shallow end of the pool, with moderately difficult questions going unanswered, then a Help Vampire infestation is likely."
While going through Probability: Theory and Examples by Rick Durrett (4th edition, p.9), I came across the familiar definition of $\sigma$-algebras where, if $A_i \in \mathcal{F}$ is a countable sequence of sets for some $\sigma$-algebra $\mathcal{F}$ and $\cup_i A_i \in \mathcal{F}$ by definitio...
Actually no, that $M_D$ is all wrong. The brackets and what x,y are matters, so we still need to run through all 27 associator base cases to completely characterise the magma
Commutators and associators sure are useful to keep track of nonassociativity and noncommutativity
My guess is that one way to set the parameters will give us an idempotent nonassociative magma
As we can see here, the main reason associativity in division by zero magma is broken is because there are associators that maps one element to another directly, instead of simply appending elements
So yes, you can divide by zero, and one possibility is a unital magma with a unique identity
typo: replace $*q$ with id
It actually surprised me that the nonassociativity is quite localised at q
This seemed to give us some hopes that division by zero ringnoids will be more interesting
All division by zero magma has this period 2 orbit formed by the commutator [0,q] which is the reason of the nonassociativity. Likewise, 3 associators (0,0,q), (0,q,q),(q,0,q) cycles between the 3 elements in this particular unital magma example
I think basically, it's how negation interacts with universal quantifier, thus the negation of "there exists x such that P(x) is true" becomes "for all x, P(x) is false"
thus cardinality does not matter
x is in the union of something if it is in at least one member of the union
$S=\{a_1,a_2,\dots,a_n,b_1,b_2,\dots,b_n\}$, $\bigcup\limits_i A_i=\{b_1,b_2,\dots,b_n, c_1,c_2,\dots,c_n\}$ olsun. $S-\bigcup\limits_i A_i=\{a_1,a_2,\dots,a_n\}$'dir. Biliyoruz ki, $A_i\subseteq\bigcup\limits_i A_i$. Yani $S-A_i\supseteq S-\bigcup\limits_i A_i$. Ve ayrıca $(\forall k\in\{1,2,\dots,n\})\, (\exists i)\, b_k\in A_i$. Bu demektir ki, $\bigcap\limits_i (S-A_i)=\{a_1,a_2,\dots,a_n\}$.
What you've said finishes proof in the second last line. @Secret
How do I show that the sequence $f_n(x)=x^n$ does not converge to any other polynomial in the space of polynomials defined on $[0,1]$ with the norm given by $\sup\|\cdot\|$ (supremum of the absolute value)?
Given a person can answer 9 out of 10 questions correctly. If we give him 10 questions what is the probability that he will answer all 10 of them correct?
I think it should be (9/10)^10
But I was reading a question today and it solved it something like this.
9/10*8/9*7/8...... which will eventually be 0. So what will be the correct method?
Given a person can answer 9 out of 10 questions correctly. If we give him 10 questions what is the probability that he will answer all 10 of them correct?
if he answers 9 out of 10 questions correctly and you give him 10 questions, then he obviously answers 9 correctly :P
What is a smooth surface in terms of tangents and normals?
I read in a book that surfaces are smooth if its surface normals depend continuously on the points of that surface.
I did not understand this definition, could somebody simplify it for me?
Ways to prove it : 1)To be a smooth surface there must exist at tangent plane for every point.Why is does not exist a tangent plane in the peak ( taking arbitrary curves)
Can we talk it . I read the answers from 3 different posts
but that means that the limit of the surface normal as you approach the tip doesn't exist and therefore the surface normal doesn't vary continuously a the tip
@Semiclassical The sneaky way out of this mess is noting that $\dfrac{\hat{R}}{R^2} \cdot d\mathbf{a'}$ is the projection and normalization of the vectorial surface element $d\mathbf{a'}$ to the unit sphere, i.e. the differential solid angle $d\mathbf{a'}$ occupies seen from $\mathbf{r}$. So the integrand reduces to $d\Omega$ which gives us $E_z = \dfrac{\sigma \Omega}{4 \pi \epsilon_0}$
@Semiclassical Suppose $C:(x,y,z): z=\sqrt{x^2+y^2}$ suppose it is a smooth manifold meaning there exists a smooth parametrization $φ(u,v): U\subset R^2 \rightarrow C$ such that $φ(u,v)=(φ_1(u,v),φ_2(u,v),φ_3(u,v))$ such that $φ_3= \sqrt{ φ_2^2+φ_3^2} $ being continuously differentiable at every point but it fails at the peak since the limit at the peak of the derivative fails to be unique. (So which is the the derivative and what limiti to take to continue my proof.?
My ability to prove that is limited to "If you approach the vertex of a cone from some direction direction, the surface normals are constant along the way but depend on the direction of approach"
I'm not going to write down a more rigorous proof than that
@Lozansky going alone the lines I proposed above, I get $E_z=\frac{\sigma}{4\pi \epsilon_0}\int_0^{2\pi}\int_0^{\Theta(\phi)}\sin\theta\,d\theta\,d\phi$
and then $\int_0^{2\pi}\int_0^{\Theta(\phi)}\sin\theta\,d\theta\,d\phi = \Omega$ from identifying this with a surface integral in spherical coordinates
I have a problem... by convex set L I mean if points A and B are elements of L, then the line from A to B is its subset. now, if f is an isometric transform, I have to prove that f(L) is also convex. my intuition tells me that starting from the assumption that f(L) is not convex, i.e. that there exists a point f(C) on the line f(A) to f(B) which is not an element of f(L), I should prove that f is not isometric or something. any ideas?
figured it out. I had to think about L as an actual set of points and f as a bijection between L and f(L). then, every A has an f(A) in f(L), and if f(L) is not convex there would be an f(C) on the line between f(A) and f(B) which was not in f(L) which is a contradiction
What is a smooth surface in terms of tangents and normals?
I read in a book that surfaces are smooth if its surface normals depend continuously on the points of that surface.
I did not understand this definition, could somebody simplify it for me?
