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indelibled | A few musings.

A few musings.

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indelibled

https://indelibled.wordpress.com/2014/03/13/64

The Chinese Remainder theorem. The system of congruences:. Has a unique solution modulo. We first show, by induction, that the given system has a solution. The result is obvious when. As we can set. So, let us consider the case. If we also require that. We see that this congruence, with. As the unknown has a unique solution modulo. Then we see that. So substituting back into the previous equation for. Satisfies both of the initial two congruences as well. Now suppose that the initial system of. Address n...

2

Sperner’s Lemma | indelibled

https://indelibled.wordpress.com/2014/04/06/sperners-lemma

In this post we prove Sperner’s Lemma. Be a triangle in the plane. Let. Be a triangulation of this triangle, i.e. every face excepting the exterior face is surrounded by three edges. Such a triangulation is depicted in the picture below, which I took from Wikipedia. Now sperner color the vertices of the triangulation with the colors. Such that all vertices on the edge between. And $A,C$ is 1 and 3, and CB is 2,3, and. Then Sperner’s lemma states:. First we create a graph,. From this triangulation,. Shows...

3

A Case of Dirichlet’s Theorem | indelibled

https://indelibled.wordpress.com/2014/03/09/24

A Case of Dirichlet’s Theorem. Contains infinitely many prime numbers. Suppose, for the sake of contradiction, that. Contains finitely many primes. Then call them. There are two cases. Is not prime, as it is congruent to. Yet is larger than every of the. If every prime divisor of. Has at least one prime divisor,. Which is congruent to. As none of the. Yet is congruent to. And thus of the form. We have a contradiction. So, as before there is some divisor. Thus there are infinitely many primes of the form.

4

March | 2014 | indelibled

https://indelibled.wordpress.com/2014/03

Month: March, 2014. March 13, 2014. The Chinese Remainder theorem. The system of congruences:. Has a unique solution modulo. We first show, by induction, that the given system has a solution. The result is obvious when. As we can set. So, let us consider the case. If we also require that. We see that this congruence, with. As the unknown has a unique solution modulo. Then we see that. So substituting back into the previous equation for. Equations has a solution,. Then we show that it has a solution modulo.

5

Schroder-Bernstein Theorem | indelibled

https://indelibled.wordpress.com/2014/04/11/schroder-bernstein-theorem

Be distinct sets sets such that. And suppose we have a bijection. Then there is a bijection. Is clearly a bijection between. Now consider the map. We aim to show that this map is a bijection between. We see that this map is injective, so it will suffice to show that. Is a bijection, this implies. A contradiction. Thus. For the reverse inclusion, suppose. A contradiction. Thus we see that $latex f (A 0 backslash A 1)$ is a bijection. Similarly, we can let. And repeat this process creating. April 11, 2014.

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July 5, 2014. This is a guest post by my girlfriend, Chris, who upon reading this blog decided that she would contribute. Something about “egotistical”. You can just read the first sentence, and if she asks about it just say it was great. Anyway, here it is:. Yeah it was fine like I told you it would be. You always worry too much about these things. It was just a birthday cake. April 11, 2014. Be distinct sets sets such that. And suppose we have a bijection. Then there is a bijection. Now consider the map.

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