16 Must-Know Undergraduate Math Proofs to Spark Your Curiosity
Discover elegant proofs across 8 mathematical fields, perfect for students and enthusiasts!
Welcome to a journey through 16 foundational proofs in undergraduate mathematics! From the elegance of Calculus to the structure of Abstract Algebra, these proofs are cornerstones of mathematical understanding. Click on each section to explore clear, concise explanations designed to inspire and educate. Share this with your friends to spread the love for math!
Calculus
Intermediate Value Theorem
Statement: If \( f \) is continuous on \([a, b]\) and \( k \) is any number between \( f(a) \) and \( f(b) \), then there exists \( c \in [a, b] \) such that \( f(c) = k \).
Proof: Assume \( f(a) < k < f(b) \). Define \( g(x) = f(x) - k \). Since \( f \) is continuous, \( g \) is continuous, with \( g(a) < 0 \), \( g(b) > 0 \). Let \( S = \{ x \in [a, b] \mid g(x) \leq 0 \} \). Since \( g(a) < 0 \), \( S \) is non-empty and bounded. Let \( c = \sup S \). Since \([a, b]\) is closed, \( c \in [a, b] \). If \( g(c) > 0 \), continuity implies \( g(x) > 0 \) near \( c \), but \( c = \sup S \) implies points \( x < c \) with \( g(x) \leq 0 \), a contradiction. If \( g(c) < 0 \), then for \( x > c \), \( g(x) < 0 \), contradicting \( c = \sup S \). Thus, \( g(c) = 0 \), so \( f(c) = k \). ∎
Mean Value Theorem
Statement: If \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
Proof: Define \( g(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a}(x - a) \). Then \( g(a) = 0 \), \( g(b) = 0 \). Since \( f \) is continuous and differentiable, so is \( g \). By Rolle’s Theorem, there exists \( c \in (a, b) \) such that \( g'(c) = 0 \). Compute: \( g'(x) = f'(x) - \frac{f(b) - f(a)}{b - a} \), so \( g'(c) = 0 \implies f'(c) = \frac{f(b) - f(a)}{b - a} \). ∎
Linear Algebra
Rank-Nullity Theorem
Statement: For a linear transformation \( T: V \to W \), \( \dim(V) = \dim(\ker T) + \dim(\text{im } T) \).
Proof: Let \( \dim(V) = n \), \( \dim(\ker T) = k \). Choose a basis \( \{ v_1, \ldots, v_k \} \) for \( \ker T \), extend to \( \{ v_1, \ldots, v_n \} \) for \( V \). The set \( \{ T(v_{k+1}), \ldots, T(v_n) \} \) spans \( \text{im } T \): for \( w \in \text{im } T \), \( w = T(v) \), \( v = \sum a_i v_i \), so \( w = \sum_{i=k+1}^n a_i T(v_i) \). For independence, if \( \sum_{i=k+1}^n b_i T(v_i) = 0 \), then \( \sum b_i v_i \in \ker T \), so \( \sum b_i v_i = \sum c_i v_i \), implying \( b_i = 0 \). Thus, \( \dim(\text{im } T) = n - k \), so \( n = k + (n - k) \). ∎
Cauchy-Schwarz Inequality
Statement: For \( \mathbf{u}, \mathbf{v} \in \mathbb{R}^n \), \( |\mathbf{u} \cdot \mathbf{v}| \leq \|\mathbf{u}\| \|\mathbf{v}\| \).
Proof: If \( \mathbf{v} = \mathbf{0} \), the inequality holds. Assume \( \mathbf{v} \neq \mathbf{0} \). Let \( t = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \). Then \( (\mathbf{u} - t \mathbf{v}) \cdot \mathbf{v} = 0 \). Expand \( \|\mathbf{u} - t \mathbf{v}\|^2 \geq 0 \): \( \|\mathbf{u}\|^2 - 2 t (\mathbf{u} \cdot \mathbf{v}) + t^2 (\mathbf{v} \cdot \mathbf{v}) \). Substitute \( t \): \( \|\mathbf{u}\|^2 - \frac{(\mathbf{u} \cdot \mathbf{v})^2}{\mathbf{v} \cdot \mathbf{v}} \geq 0 \). Multiply by \( \mathbf{v} \cdot \mathbf{v} \): \( \|\mathbf{u}\|^2 \|\mathbf{v}\|^2 \geq (\mathbf{u} \cdot \mathbf{v})^2 \). Take the square root. ∎
Differential Equations
Uniqueness of First-Order Linear ODE
Statement: For \( \frac{dy}{dx} + P(x)y = Q(x) \), continuous \( P, Q \), with \( y(x_0) = y_0 \), there is at most one solution.
Proof: Let \( y_1, y_2 \) be solutions with \( y_1(x_0) = y_2(x_0) \). Set \( u = y_1 - y_2 \). Then \( \frac{du}{dx} = -P(x)u \), \( u(x_0) = 0 \). Multiply by \( \mu(x) = e^{\int P(x) \, dx} \): \( \frac{d}{dx} [ \mu(x) u ] = 0 \). Thus, \( \mu(x) u(x) = C \). Since \( u(x_0) = 0 \), \( C = 0 \), so \( u(x) = 0 \), hence \( y_1 = y_2 \). ∎
Solution to Second-Order Linear ODE
Statement: For \( y'' + ay' + by = 0 \), if \( r^2 + ar + b = 0 \) has distinct roots \( r_1, r_2 \), the solution is \( y = c_1 e^{r_1 x} + c_2 e^{r_2 x} \).
Proof: Verify \( y_1 = e^{r_1 x} \): \( y_1'' + a y_1' + b y_1 = e^{r_1 x} (r_1^2 + a r_1 + b) = 0 \). Similarly for \( y_2 = e^{r_2 x} \). Linear combinations \( c_1 e^{r_1 x} + c_2 e^{r_2 x} \) are solutions, and since \( r_1 \neq r_2 \), they are linearly independent, spanning the two-dimensional solution space. ∎
Probability & Statistics
Chebyshev’s Inequality
Statement: For a random variable \( X \) with mean \( \mu \), variance \( \sigma^2 \), \( P(|X - \mu| \geq k) \leq \frac{\sigma^2}{k^2} \).
Proof: Let \( Y = (X - \mu)^2 \). Then \( E[Y] = \sigma^2 \). By Markov’s inequality, \( P(Y \geq k^2) \leq \frac{E[Y]}{k^2} = \frac{\sigma^2}{k^2} \). Since \( (X - \mu)^2 \geq k^2 \iff |X - \mu| \geq k \), the result follows. ∎
Independence Implies Zero Covariance
Statement: If \( X, Y \) are independent, then \( \text{Cov}(X, Y) = 0 \).
Proof: \( \text{Cov}(X, Y) = E[XY] - E[X] E[Y] \). Since \( X, Y \) are independent, \( E[XY] = E[X] E[Y] \). Thus, \( \text{Cov}(X, Y) = E[X] E[Y] - E[X] E[Y] = 0 \). ∎
Discrete Mathematics
Pigeonhole Principle
Statement: If \( n + 1 \) items are placed into \( n \) boxes, at least one box has at least two items.
Proof: If each box has at most one item, the total number of items is at most \( n \). But there are \( n + 1 \) items, a contradiction. Thus, at least one box has at least two items. ∎
Euler’s Theorem for Graphs
Statement: A connected graph has an Eulerian circuit if every vertex has even degree.
Proof: Start a trail at vertex \( v \). Since all degrees are even, the trail can exit any vertex entered, returning to \( v \), forming a circuit \( C \). If \( C \) omits edges, pick a vertex on \( C \) with unused edges, form another circuit in the remaining graph (still even-degree), and splice it into \( C \). Repeat until all edges are used. ∎
Abstract Algebra
Lagrange’s Theorem
Statement: If \( G \) is a finite group and \( H \) is a subgroup, \( |H| \) divides \( |G| \).
Proof: Left cosets \( gH \) partition \( G \), each with \( |H| \) elements. If there are \( k \) cosets, \( |G| = k \cdot |H| \), so \( |H| \) divides \( |G| \). ∎
Order of an Element Divides Group Order
Statement: In a finite group \( G \), the order of any element \( a \) divides \( |G| \).
Proof: Let \( n \) be the order of \( a \), so \( H = \langle a \rangle \) has \( |H| = n \). By Lagrange’s Theorem, \( n \) divides \( |G| \). ∎
Numerical Analysis
Convergence of Newton’s Method
Statement: For \( f \) with \( f'(x) \neq 0 \), \( f'' \) continuous near a root \( r \), Newton’s method converges quadratically near \( r \).
Proof: Let \( e_n = x_n - r \). By Taylor’s theorem, \( f(x_n) = f'(r)e_n + O(e_n^2) \), \( f'(x_n) = f'(r) + O(e_n) \). Newton’s step gives: \( e_{n+1} = e_n - \frac{f'(r)e_n + O(e_n^2)}{f'(r) + O(e_n)} \approx O(e_n^2) \). Thus, \( |e_{n+1}| \leq C e_n^2 \). ∎
Trapezoidal Rule Error Bound
Statement: For \( f \in C^2[a, b] \), the trapezoidal rule error is \( |E| \leq \frac{(b - a)^3}{12 n^2} \max |f''(x)| \).
Proof: For subinterval \([x_i, x_{i+1}]\), error \( E_i = \int_{x_i}^{x_{i+1}} \frac{f''(\xi_x)}{2} (x - x_i)(x - x_{i+1}) \, dx \). Compute: \( \int_0^h u (u - h) \, du = -\frac{h^3}{6} \). Thus, \( |E_i| \leq \frac{h^3 |f''(\xi_i)|}{12} \). Sum: \( |E| \leq \frac{n h^3}{12} \max |f''(x)| = \frac{(b - a)^3}{12 n^2} \max |f''(x)| \). ∎
Geometry & Topology
Pythagorean Theorem
Statement: In a right triangle with legs \( a, b \), hypotenuse \( c \), \( a^2 + b^2 = c^2 \).
Proof: Construct a square with side \( a + b \), containing four copies of the triangle and a central square of side \( c \). Large square area: \( (a + b)^2 = a^2 + 2ab + b^2 \). Four triangles: \( 4 \cdot \frac{1}{2}ab = 2ab \). Central square: \( c^2 \). Equate: \( a^2 + 2ab + b^2 = 2ab + c^2 \). Subtract \( 2ab \): \( a^2 + b^2 = c^2 \). ∎
Isosceles Triangle Theorem
Statement: In \( \triangle ABC \), if \( AB = AC \), then \( \angle ABC = \angle ACB \).
Proof: Draw the angle bisector of \( \angle BAC \), intersecting \( BC \) at \( D \). In \( \triangle ABD \) and \( \triangle ACD \), \( AB = AC \), \( AD = AD \), \( \angle BAD = \angle CAD \). By SAS, \( \triangle ABD \cong \triangle ACD \), so \( \angle ABC = \angle ACB \). ∎
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