Mathematics desk
< March 20	<< Feb \| March \| Apr >>	Current desk >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

March 21

Variation of the definition of a subspace of a vector space

Let $V$ be a vector space over the real numbers. Let $B\subset V$ be a subset (think of it as a basis). The subspace generated by $S$ is the set of all finite linear combinations with scalar coefficients from the real numbers.

Now change this to allow only non-negative coefficients. Does this thing have a name or any well known properties?

Thanks. 95.115.169.61 (talk) 10:04, 21 March 2015 (UTC)[reply]

A cone (linear algebra). Sławomir Biały (talk) 13:10, 21 March 2015 (UTC)[reply]

That's it! Thank you! 95.115.169.61 (talk) 13:49, 21 March 2015 (UTC)[reply]

Well, almost. More precisely, AFAICT, it is a convex cone. —Quondum 14:55, 21 March 2015 (UTC)[reply]

Well, to be even more precise, it is a Conical_combination, but I was interested in the general thing, just missing the magical word to start searching on my own. 95.115.169.61 (talk) 15:28, 21 March 2015 (UTC)[reply]

Right triangle, sides a + 2a, can the hypotenuse be a natural number?

Given a right triangle with two legs of the length 'a' and '2a', both integers, can the hypotenuse be an integer too? Or we other words, can the square root √(a + 2a) be an integer?--Fend 83 (talk) 14:30, 21 March 2015 (UTC)[reply]

No. For there to be a solution, the square root of 5 has to be rational. Georgia guy (talk) 14:48, 21 March 2015 (UTC)[reply]

Surely you mean "can √(a^2 + 4a^2) be an integer, for integer a?", which, as Georgia Guy implies, is the same as asking for "can √5a^2 be an integer, for integer a?" answering "yes" to which would imply that √5 was rational. Since it isn't (see here for a proof) we can conclude the answer to the first question is also "no", by reductio ad absurdam.-- Impsswoon (talk) 15:07, 21 March 2015 (UTC)[reply]

Yes, I meant √(a^2 + 4a^2). Thanks for the proof of √5 being irrational too. --Fend 83 (talk) 15:27, 21 March 2015 (UTC)[reply]

"Grothendieck logic"?

This page refers to "Grothendieck logic". I've found this explanation of the topic, but I don't find it even remotely enlightening. Does anyone know what this is? Is this related to topos theory? -- Impsswoon (talk) 14:54, 21 March 2015 (UTC)[reply]

See Institution for a def of institutions. Then, From [1] (which is in a part Google doesn't let you preview...sadly) define a logic as:

Definition

The following discussion has been closed. Please do not modify it.

A logic is an institution equipped with an entailment system consisting of an entailment relation $\vdash$ _Σ⊆ |Sen(Σ)| ×Sen(Σ), for each Σ ∈ |Sign|, such that the following conditions are satisfied:
1. reflexivity: for any ϕ ∈ Sen(Σ), {ϕ} $\vdash$ _Σ ϕ
2. monotonicity: if Γ $\vdash$ _Σ ϕ and Γ' ⊇ Γ then Γ' $\vdash$ _Σ ϕ,
3. transitivity: if Γ $\vdash$ _Σ ϕ_i, for i ∈ I, and Γ ∪ {ϕ_i| | i ∈ I} $\vdash$ _Σ ψ, then Γ $\vdash$ _Σ ψ,
4. $\vdash$ _Σ-translation: if Γ $\vdash$ _Σ ϕ, then for any σ: Σ → Σ' in Sign, σ[Γ] $\vdash$ _Σ' σ(ϕ),
5. soundness: for any Σ ∈ |Sign|, Γ ⊆ Sen(Sign) and ϕ ∈ Sen(Σ), Γ $\vdash$ _Σ ϕ implies Γ |=_Σ ϕ.

Then,

Def of Groth. Logic

The following discussion has been closed. Please do not modify it.

Definition 8. Given an indexed logic L: Ind^op −→Log, define the Grothendieck logic L^# as follows:
– signatures in L# are pairs (Σ, i), where i ∈ |Ind| and Σ a signature in the logic L(i),
– signature morphisms (σ, d): (Σ1, i)−→(Σ2, j) consist of a morphism d:i−→j ∈ Ind and a signature morphism σ: Σ₁ −→Φ^L(d)(Σ2) (here, L(d):L(j)−→ L(i) is the logic morphism corresponding to the arrow d:i−→j in the logic graph, and ΦL(d) is its signature translation component),
– the (Σ, i)-sentences are the Σ-sentences in L(i), and sentence translation along (σ, d) is the composition of sentence translation along σ with sentence translation along L(d),
– the (Σ, i)-models are the Σ-models in L(i), and model reduction along (σ, d) is the composition of model translation along L(d) with model reduction along σ, and
– satisfaction (resp. entailment) w.r.t. (Σ, i) is satisfaction (resp. entailment) w.r.t. Σ in L(i).

You may also want to have a look at [2] and [3]. I apologize if I am not clarifying anything...I'm very limited with anything symbol heavy in Wikipedia. A Google search for, or books/papers on, universal logic may be of use.Phoenixia1177 (talk) 15:27, 23 March 2015 (UTC)[reply]

My answer from the topos talk page: In the context you pointed out, I think it is more of an analogous construction than an actual topos. It is a construction over multiple logic systems in the form of Grothendeick institutions, for which we have an Institution (computer science) article. The construction is talked about in Foundations of Software Science and Computation Structures: 5th International Conference, FOSSACS 2002., p.334-335. --Mark viking (talk) 20:03, 23 March 2015 (UTC)[reply]

Many thanks to all of you. -- Impsswoon (talk) 11:01, 27 March 2015 (UTC)[reply]

Deciding if a point is on, inside or outside of a circle

On the net I found an elegant way to decide if a point d in the plane is inside of the circle that is given by a, b, and c. It is as simple as augmenting the vectors $(x_{1},x_{2})$ to $(x_{1},x_{2},x_{1}^{2}+x_{2}^{2},1)$ , and taking the sign of the determinant. This seems to work for any dimension.

I can proof this for n=2 by computing the determinant and transforming it to an equation of a circle. I could do the same for n=3, it should be possible for n=4 but it surely would take the rest of my life to do it with n>4.

Is there a better way to proof this?

Thanks. 95.115.169.61 (talk) 15:39, 21 March 2015 (UTC)[reply]

In case it's unclear, here's what the OP is describing. You have four points

a=(a_{0},a_{1}),b=(b_{0},b_{1}),c=(c_{0},c_{1})

and

d=(d_{0},d_{1})

and you want to determine whether or not d is inside the circle running through a, b and c, where "inside" means "on the left as you travel from a to b to c". The method is to consider the sign of the determinant of the following matrix:

{\begin{bmatrix}d_{0}&d_{1}&|d|^{2}&1\\a_{0}&a_{1}&|a|^{2}&1\\b_{0}&b_{1}&|b|^{2}&1\\c_{0}&c_{1}&|c|^{2}&1\end{bmatrix}}

Positive means outside, negative means inside, and 0 means on the circle. I'm afraid I can't explain why this works, though.--80.109.80.31 (talk) 17:47, 21 March 2015 (UTC)[reply]

Thank you for providing the long version of my question ;-) . I'm still a bit uneasy about that "inside/outside" versus "left/right". n points in n dimensions fix a sphere, and from that, inside and outside are quite unambiguous. The translation from the sign of the determinant to the inside/outside depends, of course, on the order of the points in question. 95.115.169.61 (talk) 19:38, 21 March 2015 (UTC)[reply]

(ec) There are two parts to this; first that the equation of the circle/sphere is given by the determinant, and second that the inside vs. outside is determined by the sign. The first part is (or should be) found in any good text on analytic geometry or even linear algebra. Perhaps what you're missing is that you don't need to expand the determinant into monomial; you just use the cofactor expansion. For dim=4, say, the firs row of the determinant is

|1 x y z w x²+y²+z²+w²|.

When you expand using cofactors you get an expression of the form

A+Bx+Cy+Dz+Ew+F(x²+y²+z²+w²)

where A, B, C, D, E, and F are determinants, but except for F it doesn't matter what they are. F is a determinant which is 0 if the points are on the same hyperplane and non-zero otherwise. There is no sphere if the points are on the same hyperplane, but if not then divide through by F to get the equation

A'+B'x+C'y+D'z+E'w+x²+y²+z²+w²=0

which is the equation of a sphere. That this sphere contains all the points in question can be shown by plugging the coordinates into the determinant; you get two rows the same which implies that the determinant is 0, so all the points lie on the sphere with the equation given.

"Inside" and "Outside" are topological concepts, so the second part a bit trickier, at least conceptually. I would think that in order to even state a theorem which works in any dimension you would need to know something about the orientation of a set of point. But suppose you've already shown that the set of points is positively oriented if F is positive and negatively oriented if F is negative, (Perhaps you could take that as the definition of orientation.) F if is positive then the polynomial

A+Bx+Cy+Dz+Ew+F(x²+y²+z²+w²)

approaches infinity as the point goes to infinity. You also know that the polynomial has a unique minimum point, and since there are multiple points where the value is 0, the minimum must be negative. That means the sphere divides space into two regions, one where the determinant is positive and which contains a neighborhood of infinity, and the other where the determinant is negative and which is bounded. Since the determinant is continuous, you can't move from the negative region to the positive region without being zero at some point, so the boundary between the two regions is the sphere. I think this would be enough to satisfy your definition of "inside" and "outside". If F is negative then swap two points to make if positive and proceed from there.

When you talk about the topology of space you may get into some difficulties proving things that seem intuitively obvious; see Jordan Curve Theorem as an example. So there may be a few points in the above argument which require more work if you want to be completely rigorous about it, but I don't want to turn this into a course on topology so a certain amount of hand waving is needed I think. --RDBury (talk) 18:01, 21 March 2015 (UTC)[reply]

Thank you, that hint about cofactors did much help. I'm not quite sure yet if I underestimate the problem that the inside can be mapped to one sign and the outside to the other. Obviously, the order of points will change the sign, but once the order is fixed, I feel there should be no problem showing that one side sticks to one sign and the other to the other. 95.115.169.61 (talk) 19:38, 21 March 2015 (UTC)[reply]

If I recall correctly, there are much simpler ways to prove an analog to the JCT if you restrict to circles/spheres. Part of what makes it hard to prove and requires all the winding numbers, etc is that JCT applies to any simple closed loop. Having to consider only spheres should (I think) make it much easier to prove that they have one inside and one outside. SemanticMantis (talk) 15:07, 23 March 2015 (UTC)[reply]

Followup question. From what I've gathered right now, the same trick would not work for other metrics. In other words, it is not possible (at least not in an obvious extension) to uses some determinate like this to decide if the "point d" is inside the simplex or bounding box given by the other points. Is this true, or is there any trick around? 95.115.169.61 (talk) 19:57, 21 March 2015 (UTC)[reply]

I don't see how the same trick would work with the examples you're giving, but I would think that determinants might be useful in some way. It seems like the same trick might be used with other families of curves though. For example ellipses of the form x²+2y²+ax+by+c = 0. --RDBury (talk) 13:31, 22 March 2015 (UTC)[reply]