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rjlipton.com | ||
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mattbaker.blog
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| | | | | I'm teaching Graduate Algebra this semester, and I wanted to record here the proof I gave in class of the (existence part of the) structure theorem for finitely generated modules over a PID. It's a standard argument, based on the existence of the Smith Normal Form for a matrix with entries in a PID, but... | |
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anuragbishnoi.wordpress.com
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| | | | | The Ramsey number $latex R(s, t)$ is the smallest $latex n$ such that every graph on $latex \geq n$ vertices either contains a clique of size $latex s$ or an independent set of size $latex t$. Ramsey's theorem implies that these numbers always exist, and determining them (precisely or asymptotically) has been a major challenge... | |
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www.jeremykun.com
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| | | | | Problem: Prove there are infinitely many prime numbers. Solution: First recall that an arithmetic progression with difference $ d$ is a sequence of integers $ a_n \subset \mathbb{Z}$ so that for every pair $ a_k, a_{k+1}$ the difference $ a_{k+1} - a_k = d$. We proceed be defining a topology on the set of integers by defining a basis $ B$ of unbounded (in both directions) arithmetic progressions. That is, an open set in this topology is an arbitrary union of arithmetic progressions from $ -\infty$ to $ \infty$. | |
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thatsmaths.com
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| | | The rational numbers $latex {\mathbb{Q}}&fg=000000$ are dense in the real numbers $latex {\mathbb{R}}&fg=000000$. The cardinality of rational numbers in the interval $latex {(0,1)}&fg=000000$ is $latex {\boldsymbol{\aleph}_0}&fg=000000$. We cannot list them in ascending order, because there is no least rational number greater than $latex {0}&fg=000000$. However, there are several ways of enumerating the rational numbers. The... | ||
