Mathaematics.

For my exam time absence~.

(Photos, videos, concepts, or interesting problems are well appreciated)

Formal definition.

Let \(\phi :[0,1]\to T\) (where \( T\) is a topological space) be a continous function (i.e. \(\phi ^{-1}\) of any open set \( O\) of \( T\) is an open set of the unit interval) from the unit interval to a topological space. Such map is said to be an space-filling curve if \(\phi\) is onto (i.e. if \phi passes through every element of \( T\) ).

Note

Generally, space filling curves are represented on spaces homeomorphic to \( E^{n}\) euclidian spaces. So to say, the plane, or any geometric solid, since these are the easiest ways to visualize a space filling curve. The most popular example, perhaps, is the two-dimensional Hilbert curve.

Images

A Hilbert curve from the unit interval to the unit cube

Peano Curve from the unit interval to the unit cube.

Flow Snake sapce filling curve.

Last three school weeks. Final projects, exams, and kind-of complicated topics.

When I end my Group Theory, advanced linear algebra courses and the overwhelming amounts of multivariable calc and mechanics I’ll be back. In the meantime, I’ll ony be posting pretty images.

So, this is something that constantly comes into my mind.

During elementary school, one is thaught arithmethics, the base for every other areas of mathematics, since all of them rely, in any amount, on arithmetics. One is taught the basic operations with natural numbers, and, eventually, the set of real numbers.

At least in my country, middle school is where one starts with basic algebra, I mean, solving linear equations, coloquially called “finding x”. This is still no problem, since showing how equations work is pretty self-explanatory.

The messy part on education in mathematics starts when one reaches high school, that is the part when students ask “Why does that happen?” when they are thaught trigonometric functions, identities, limits, derivatives, integrals, et cetera. However, it is not precissely easy to answer those questions. One cannot explain the linearity of calculus operators to their students if they have no idea of linear algebra concepts such as linear transformation. Whan cannot explain the reason of integrals to a group of students if they have no idea of mathematical analysis. Then, what is the point?

I personally believe that, in mathemathics, theory is first. This means, high school should first teach formal logic and set theory (instead of having a recap in middle school topics), and notions on geometry and abstract algebra, at least, defining functions, binary operations, notions on group theory and analysis, since one can construct every topic in mathematics by knowing that, and then, reasons for concepts seen later would be known. And then, math wouldn’t be seen as “the hardest subject in school” (well, perhaps at the beginning).

Yet, again, it’s just me drabbling and ranting on a topic that will never be concluded.

In absract algebra, mainly group theory, given a group \( G\) aconjugateof an element \( x\in G\)  is another element \( y\in G\) of the form \( y=zxz’\) with \( z\in G\) (and obviously \( z’\) , since as a group all element must have its inverse included.

Theconjugacy problemis a decision problem, which consists in determining if an element \( y\in G\) can be represented as a conjugate of \( x\in G\) .

An easy example is the group \(\left( Z,+\right)\) where \( Z\) is the set of integer numbers.

¿Is it possible to express \( y=z+x-z\) for \( y\neq x\) and \( z\in Z\) ?

For example, for:

\( 3=z+1-z\)

¿Is there a value of \( z\) for which this equation is valid?

I’ll post something when I’m back home or when I’m not drunk.

Which will be… pretty much at the same time.

In abstract algebra, apermutation groupis a group \( G\) that consists in all possible permutations of a set. So to say, the elements of a permutation group are the possible permutations of the elements of a different, \( H\) group. Then we get a group \(\left( G,+\right)\) where \( +\) is the composition of permutations. Also, each permutation \(\phi_{i}\left(H\right)\) . We can assume that permutations are bijective functions, since as a group under composition of permutations, an inverse permutation that yields the identity permutation when composed with one specific permutation exists within \( G\) .

But how is the Rubik’s 3x3 cube a permutation group?

Let’s talk about the cube itself, disregarding the colors from now. The cube has six fixed piecesand 20 permutable pieces, therefore, every piece itself can be placed in \( 20!\) different positions. Hence, the Rubik’s Cube pieces form an order 20! permutation group. But why would we even care on a rubiks cube without distinction of sides. Even a rock would solve one of those.

The cube WITH face distinction is another permutation group, over a greater set \(\bar{H}\) , this set is not formed by the pieces, but by the stickers on the pieces. The mathematics yielding this are slightly more complex, but the number of possible permutations without disassembling the cube is \( 8!\times 3^{7}\times 12!/2\times 2^{11}\). So we have a permutation group on this order. The fact scrambling and solvingn the cube shows the existence of inverse permutations, let’s say that the solved cube is the identity permutation, the resut of that scramble is \(\psi\left( \bar{H}\right)\) and, the permutation that solves the cube would be \(\psi^{-1}\left( \bar{H}\right)\). But \(\psi^{-1}\) can be expressed as the composition of multiple, elemental permutations, which would be, face rotations. Which means that\(\psi^{-1}\left(\bar{H}\right) =\sum_{i=1}^{12}\alpha_{j}\varphi_{i} (\bar{H} )\) where \(\varphi_{i} (\bar{H} )\) are the single step face rotations and \(\alpha\in \{ 1,2,3\}\) since one-face rotations form cyclic subgroups of our permutation group \( G\) and the fourth rotation is equal to the identity rotation which is, not rotating the face at all. The single face rotations that \(\varphi_{i}\) represent are the steps in the solution of the cube.

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Yes, this is pretty much something I will need to show my abstract Algebra I groub by the end of the semester.

Let \( V\) be a nonempty set, whose elements are called “vectors” , let \( K\) be a field, whose elements are called “scalars”, and define two operations:

\( +:V\times V\to V\)

\( ·:K\times V\to V \)

\( V\) is called a vector space over a field \( K\) if:

1. \(\left( V, +\right)\) is an abelian group.
2. Let \(\alpha ,\beta\in K, v,w\in V\) then:

• \(\alpha ·\left( v+w\right) =\alpha ·v+\alpha ·w\)
• \(\left(\alpha +\beta\right) ·v=\alpha ·v+\beta ·v\)
• \(\left(\alpha\beta\right) ·v=\alpha\left(\beta ·v\right)\)
• \(\dot{e} ·v=v\) where \(\dot{e}\) is the multiplicative identity on the field \( K\)

Continuation on algebraic structures:

Ring

A ring is a nonempty set along with two operations, denoted by \( \left( R,+,· \right) \) Where

• \( \left( R,+ \right) \) is an abelian group.
• \( \left( R,· \right) \) is a monoid.

Structures between rings and fields.

A commutative ring is a ring that is commutative under ·

A ring with identity is a ring that contains an identity element \( \dot{e} \) for the · operation.

An integral domain implies that \( e\) , the identity for \( +\) has no inverse under · in R.

An integral domain from which all elements except \( e\) have an inverse under · is called a semifield.

A commutative semifield is a field.

Injective, suprajective and bijective functions.

Let \( \phi :G\to H \) be an application from a set G to a set H . We say the application \( \phi \) is a function if for each element in the domain, there exists one and only one element in the image.

If \( \phi \) is injective, it means that every element in \( Image\left( \phi \right) \) is associated to only one element in the domain of the function. These are also called one-to-one functions

If \( \phi \) is suprajective, it means that every element in the H is mapped to an element of G.

When a function is biyective, it means  that a function is both inyective and suprajective, also, for bijective functions there exists a function\( \phi ^{-1} : H\to G \) that works as an inverse function for phi.

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Given these, now we can define a vector space.

So I decided to start my blog on linear algebra; however, before entering the topic, one must understand some basic concepts, such as algebraic structures, linear transformation, injective, suprajective and bijective functions.

To begin with…

Binary operation:

Let C be a non-empty set. A binary operation is a mapping

\( +:C\times C→C\)

That has the characteristic of being closed and determined for every ordered pair from CxC.

Where + is NOT the normal sum of numbers.+

For example: Let \( I_{k}\left( G,G \right) \) be the set of functions from G into G. The operation

\( +: I_{k}\left( G,G \right)\times I_{k}\left( G,G \right)\to I_{k}\left( G,G \right) \) being + the composition of functions

Is a binary operation. Since the referred set includes all the functions from G to G, the set is closed under composition, since \( \phi :G\to G\) and \( \psi :G\to G\), then \( \phi \circ \psi \in I_{k}\left( G,G \right) \). The same fore \( \psi \circ \phi \) .

Note that \( +(a,b)\) can also be written a+b.

Algebraic Structures:

Group

Let G be a nonempty set, the set G is a group under a binary operation + (denoted \( (G,+)\) ) if

• \( +(a,b)\) is associative: \( (a+b)+c=a+(b+c)\) ∀a,b,c∈G
• There exists an identity element \( e\in G\) that follows \( a+e=e+a=a \)
• ∀ a∈G ∃ a’∈G that follows a+a’=a’+a=e

It is important to note that + is not necessarily a commutative operation.

Indeed, if the operation is commutative, it is said that the group is an Abelian Group.

As an example to the previous sentence

• (\( \Re \) , +) forms an abelian group with the normal number sum.
• (ℚ, +) with \( +(a,b)=a-b \) forms a non-abelian group.

Lower order structures of a group:

Semigroup:

A (G, +) that is associative but lacks identity element and inverse elements is called a Semigroup

Monoid:

A (G, +) that is associative, has an identity element but lacks inverse elements is called a monoid.

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Also to be said, when working on algebras, one is not necessarily working with sets of numbers, only with non-empty sets, these might have numbers, sets, colors, shapes, functions, iterations, et al.

Later on I will finish explaining the preliminaries on one or two more posts and enter then to vector spaces.

This blog is a recent creation, with the objective of explaining the concepts of linear and abstract algebras, as well as other topics in calculus and mathematical analysis. Let’s see how it goes.