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Daily Insights from Magic Internet Math courses. Learn at https://mathacademy-cyan.vercel.app
πŸ“ Multiplicative Group is Cyclic The multiplicative group $\\mathbb{F}_q^* = \\mathbb{F}_q \\setminus \\{0\\}$ is cyclic of order $q-1$. From: Algebraic Number Theory Learn more: Explore all courses:
πŸ“– Individualism The political tradition that recognizes the individual as the ultimate judge of their own ends, respects their opinions and tastes as supreme within their sphere, and believes society should be organized to allow the maximum scope for individual initiative and choice. From: The Road to Serfdom Learn more: Explore all courses:
πŸ“– Primitive Element A primitive element (generator) $g$ of $\\mathbb{F}_q$ satisfies $\\mathbb{F}_q^* = \\{g^0, g^1, \\ldots, g^{q-2}\\}$. From: Algebraic Number Theory Learn more: Explore all courses:
πŸ“ Cofactor Expansion The determinant can be computed by expansion along any row $i$: $\\det(A) = \\sum_{j=1}^n a_{ij} C_{ij}$ where $C_{ij} = (-1)^{i+j} \\det(M_{ij})$ is the cofactor. From: Advanced Linear Algebra Learn more: Explore all courses:
πŸ“ Best Low-Rank Approximation The best rank-$k$ approximation to $A$ (minimizing $\\|A - B\\|$ over all rank-$k$ matrices $B$) is $A_k = \\sum_{i=1}^k \\sigma_i \\mathbf{u}_i \\mathbf{v}_i^T$. From: Linear Algebra Learn more: Explore all courses:
πŸ“ Uniqueness of the Zero Element In any linear space there is one and only one zero element. Proof: Axiom 5 tells us there is at least one zero element. Suppose there were two, say $O_1$ and $O_2$. Taking $x = O_1$ and $0 = O_2$ in Axiom 5, we obtain $O_1 + O_2 = O_1$. Similarly, taking $x = O_2$ and $0 = O_1$, we find $O_2 + O_1 = O_2$. But $O_1 + O_2 = O_2 + O_1$ by the commutative law, so $O... From: calc2 Learn more: Explore all courses:
πŸ“ Sample Theorem If $A \\subseteq B$ and $B \\subseteq A$, then $A = B$ Proof: Let $x \\in A$. Since $A \\subseteq B$, we have $x \\in B$ by definition of subset. Therefore, every element of $A$ is in $B$. Now, let $y \\in B$. Since $B \\subseteq A$, we have $y \\in A$ by definition. Therefore, every element of $B$ is in $A$. Since $A \\subseteq B$ and $B \\subseteq A... From: coding-theory Learn more: Explore all courses:
πŸ“– Man-in-the-Middle Attack (MITM) A cyberattack where an attacker secretly relays and possibly alters the communications between two parties who believe they are directly communicating. The attacker can eavesdrop, modify data, or inject malicious content. From: branta Learn more: Explore all courses:
πŸ“ Derivatives and Integrals are Inverses Integration and differentiation are inverse operations: $\\frac{d}{dx}\\int_a^x f(t)\\,dt = f(x)$ and $\\int_a^x f\ From: Calculus: A Liberal Art Learn more: Explore all courses:
πŸ“ Theorem 5.10 (Mean Value Theorem) If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists $x \\in (a, b)$ with $f(b) - f(a) = f\ Proof: Apply Theorem 5.9 with $g(x) = x$. Then $g(b) - g(a) = b - a$ and $g'(x) = 1$. We get $f(b) - f(a) = f'(x)(b - a)$ for some $x \\in (a, b)$. ∎ From: rudin Learn more: Explore all courses:
πŸ“ Economic Calculation Problem Without market prices, central planners cannot calculate whether resources are being used efficiently. This is the fundamental problem that makes rational socialist planning impossible. Proof: To allocate resources rationally, one must know the relative values of different uses. Prices emerge only from voluntary exchange. Socialism abolishes private ownership of means of production. Without private ownership, there is no exchange of capital goods. Without exchange, there are no... From: Human Action Learn more: Explore all courses:
πŸ“– Prometheus Unchained John Galt is Prometheus who changed his mindβ€”after centuries of being torn by vultures for bringing fire to men, he withdrew his fire until men withdraw their vultures. From: Atlas Shrugged Learn more: Explore all courses:
πŸ“ Euclid Given any finite list of primes $p_1, p_2, \\ldots, p_k$, there exists a prime not on this list. Proof: Consider $n = p_1 p_2 \\cdots p_k + 1$. By the Lemma, $n$ has some prime divisor $p$. But $n \\equiv 1 \\pmod{p_i}$ for each $i$, so $p$ cannot equal any $p_i$. Therefore there exists a prime not on the original list. Since any finite list can be extended, there are infinitely many primes. From: Thales to Euclid Learn more: Explore all courses:
πŸ“– Vertex and Edge Connectivity The vertex connectivity $\\kappa(G)$ is the minimum size of a vertex cut (or $n-1$ if $G = K_n$). The edge connectivity $\\kappa\ From: Introduction to Graph Theory Learn more: Explore all courses:
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