📐 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: fermat-minkowski Learn more:

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📐 Axiom of Infinity There exists a set $\\Omega$ containing $\\emptyset$ and closed under the successor operation $X \\mapsto X \\cup \\{X\\}$. From: numbers-geometry Learn more:

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📖 Definition 7.1.1 (Finite Series) Let $m, n$ be integers, and let $(a_i)_{i=m}^n$ be a finite sequence. We define $\\sum_{i=m}^n a_i$ recursively by $\\sum_{i=m}^m a_i := a_m$ and $\\sum_{i=m}^{n+1} a_i := (\\sum_{i=m}^n a_i) + a_{n+1}$. From: tao-analysis-1 Learn more:

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📐 RSA Correctness For any message $m$ with $\\gcd(m, n) = 1$: $(m^e)^d \\equiv m \\pmod{n}$. Proof: Since $ed \\equiv 1 \\pmod{\\varphi(n)}$, write $ed = 1 + k\\varphi(n)$ for some integer $k$. By Euler's theorem, $m^{\\varphi(n)} \\equiv 1 \\pmod{n}$ when $\\gcd(m,n) = 1$. Therefore: $m^{ed} = m^{1 + k\\varphi(n)} = m \\cdot (m^{\\varphi(n)})^k \\equiv m \\cdot 1^k = m \\pmod{n}$. From: Algebraic Number Theory Learn more:

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📖 Definition 2.32 (Compact Set) A set $K$ is **compact** if every open cover of $K$ has a finite subcover. From: rudin Learn more:

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📐 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:

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🎯 Corollary 8.3.3 (2^N is Uncountable) The power set $2^{\\mathbb{N}}$ is uncountable. From: tao-analysis-1 Learn more:

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📖 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:

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An interactive course exploring F.A. Hayek

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💡 Proposition 8.1.5 (Q is Countable) The set $\\mathbb{Q}$ of rational numbers is countable. From: tao-analysis-1 Learn more:

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📖 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:

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📐 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:

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📐 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:

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📐 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:

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📐 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:

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📖 Subgroup A subset $H$ of group $G$ is a subgroup if $H$ is itself a group under the same operation. We write $H \\leq G$. From: aa-pinter Learn more:

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📖 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:

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📐 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:

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📐 Mobius Transformations Form a Group The Mobius transformations form a group under composition, isomorphic to $\\text{PGL}(2, \\mathbb{R})$. From: Four Pillars of Geometry Learn more:

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Explore geometry from four perspectives: Euclidean constructions, coordinates, projective geometry, and transformations. Based on John Stillwell

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📐 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:

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📐 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:

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Ludwig von Mises

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