@continuum_many

I talk about math!

Joined August 2026
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If p is a prime number, the only group of order p is, up to isomorphism, the cyclic group with p elements.
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The converse of this statement is false: the derivative of f(z) = eᶻ never vanishes, but f is not one-to-one.
Continuum Many@continuum_manySep 26
In complex analysis, the opposite implication holds: if f : Ω → ℂ is an holomorphic, one-to-one function (Ω ⊂ ℂ open), then f' ≠ 0 on Ω. Indeed, f has an holomorphic inverse g : f(Ω) → Ω. Since (g ∘ f)(z) = z for z ∈ Ω, 1 = (g ∘ f)'(z) = g'(f(z)) f'(z), so f'(z) ≠ 0.
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In complex analysis, the opposite implication holds: if f : Ω → ℂ is an holomorphic, one-to-one function (Ω ⊂ ℂ open), then f' ≠ 0 on Ω. Indeed, f has an holomorphic inverse g : f(Ω) → Ω. Since (g ∘ f)(z) = z for z ∈ Ω, 1 = (g ∘ f)'(z) = g'(f(z)) f'(z), so f'(z) ≠ 0.
Continuum Many@continuum_manySep 24
If f : I → ℝ is differentiable on an open interval I ⊂ ℝ and f' ≠ 0 on I, then f is one-to-one on I. Indeed, given a, b ∈ I such that f(a) = f(b), by the mean value theorem there exists c ∈ I such that 0 = f(a) − f(b) = f'(c) (a − b). Since f'(c) ≠ 0, we have a = b.
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The converse is false! For instance, f(x) = x³ is one-to-one but f'(0) = 0.
Continuum Many@continuum_manySep 24
If f : I → ℝ is differentiable on an open interval I ⊂ ℝ and f' ≠ 0 on I, then f is one-to-one on I. Indeed, given a, b ∈ I such that f(a) = f(b), by the mean value theorem there exists c ∈ I such that 0 = f(a) − f(b) = f'(c) (a − b). Since f'(c) ≠ 0, we have a = b.
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If f : I → ℝ is differentiable on an open interval I ⊂ ℝ and f' ≠ 0 on I, then f is one-to-one on I. Indeed, given a, b ∈ I such that f(a) = f(b), by the mean value theorem there exists c ∈ I such that 0 = f(a) − f(b) = f'(c) (a − b). Since f'(c) ≠ 0, we have a = b.
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Let (Ω, 𝒜, P) be a probability space. The expected value of a random variable X : Ω → ℝ is defined as E[X] = ∫_Ω X dP. If X is (respectively) discrete or continuous, we recover the usual formulas E[X] = ∑ x P(X=x) and E[X] = ∫ x f(x) dx.
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The real projective plane ℝℙ² has an area of 2π.
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6 is the only perfect number that is the product of two primes.
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Indeed, if a number of the form pq is perfect, then 1+p+q = pq (its proper divisors are 1, p, q). After rearranging, 2 = (p-1)(q-1), so either p-1 = 1 and q-1 = 2 or the other way around. Thus, p and q are 2 and 3, and the number is 6.
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The power set 𝒫(X) of a set X is in bijection with the set 2^X of functions X → {0,1}. A particular bijection is 𝒫(X) → 2^X, A ↦ χ_A where χ_A is the characteristic function of A (given x ∈ X, χ_A(x) = 1 if x ∈ A and 0 otherwise).
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This means that a finite set with n elements has 2ⁿ subsets. It also works for infinite sets. For example, |ℕ| = ℵ₀, so |𝒫(ℕ)| = 2^ℵ₀, i.e. the natural numbers have continuum many subsets.
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Contrary to popular belief, it is unknown whether the digits of π contain every finite sequence of numbers 0-9.
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Together with the CLT, this basic result is probably the most important theorem on statistics: Let X₁,...,Xₙ be i.i.d. random variables with finite mean and variance μ and σ². Then the sample mean and variance X̅ and S² satisfy: a) E[X̅] = μ b) Var[X̅] = σ²/n c) E[S²] = σ²
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Some of the first examples of groups that one comes accross with are the real numbers ℝ as an additive group and the nonzero real numbers ℝ* as a multiplicative group, but the strong relation between the two is often omitted: ℝ*/{1, −1} ≃ ℝ ℝ* ≃ ℝ × ℤ/2ℤ Thread! 🧵
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the absolute value and the projection are surjective). For injectivity, assume x ∈ ker(f), i.e. |x| = 1 and xH = H. From the first equality, x = ± 1. From the second, x ∈ H, i.e. x > 0. Thus x = 1, so the kernel is trivial and f is injective. (...)
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Therefore, we have ℝ* ≃ H × ℝ*/H ≃ ℝ × ℤ/2ℤ, as promised!
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Thread on factoring real polynomials! We know from the Fundamental Theorem of Algebra that any complex polynomial can be written as a product of linear factors in an essentially unique way, but what about polynomials with real coefficients? I claim that a real polynomial (...)
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some polynomial g. But writing u = a + bi, we have (X − u) (X − u̅) = X² − 2aX + a² + b², so this is a factor with real coefficients. It remains to justify that g also has real coefficients. To do this, dividing f by X² − 2aX + a² + b² in ℝ[X], we obtain polynomials (...)
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q,r ∈ ℝ[X] such that f(X) = (X² − 2aX + a² + b²) q(X) + r(X). By the unicity of the quotient and remainder of division in ℂ[X], it must be q = g and r = 0, so g has real coefficients. Therefore f(X) = (X² − 2aX + a² + b²) g(X) and deg(g) ≥ 1, so f is reducible in ℝ[X].
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