Theorem Examples

Published

Last modified: 2026-05-19 10:59:38 (PDT)

This chapter demonstrates all the theorem-like environments available through the callouty-theorem and custom-callout extensions.

1 Theorems

Theorem 1 (Pythagorean Theorem) In a right triangle with legs of length $a$ and $b$ and hypotenuse of length $c$, we have: \[a^2 + b^2 = c^2\]

See Theorem 1 for the famous Pythagorean theorem.

2 Lemmas

Lemma 1 (Fundamental Lemma) Every non-empty subset of $\mathbb{N}$ has a smallest element.

The result in Lemma 1 is used to prove many theorems in number theory.

3 Corollaries

Corollary 1 (Rational Root Test Corollary) If $p/q$ is a rational root of the polynomial $a_n x^n + \cdots + a_1 x + a_0$ with integer coefficients, then $p$ divides $a_0$ and $q$ divides $a_n$.

Corollary 1 follows directly from the Rational Root Theorem.

4 Propositions

Proposition 1 (Triangle Inequality) For any real numbers $x$ and $y$: \[|x + y| \leq |x| + |y|\]

The triangle inequality (Proposition 1) is fundamental in analysis.

5 Conjectures

Conjecture 1 (Goldbach’s Conjecture) Every even integer greater than 2 can be expressed as the sum of two prime numbers.

Conjecture 1 remains one of the oldest unsolved problems in number theory.

6 Definitions

Definition 1 (Derivative) The derivative of a function $f$ at a point $x$ is defined as: \[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\] provided this limit exists.

See Definition 1 for the formal definition of a derivative.

7 Examples

Example 1 (Solving a Quadratic Equation) Solve $x^2 - 5x + 6 = 0$.

We can factor this as $(x-2)(x-3) = 0$, so the solutions are $x = 2$ and $x = 3$.

Example 1 shows how to solve a simple quadratic equation.

8 Exercises

Exercise 1 (Integration Practice) Compute the following integral: \[\int_0^1 x^2 \, dx\]

Try Exercise 1 to practice basic integration.

9 Proofs

Proof (Proof of the Pythagorean Theorem)

Proof (Proof of the Pythagorean Theorem). Consider a square with side length $a+b$. We can divide this square into four right triangles and a central square with side length $c$.

The area equations show: \[a^2 + b^2 = c^2\]

Proof (Proof that the derivative of a constant is zero)

Proof (Proof that the derivative of a constant is zero). Let $f(x) = c$ for some constant $c$. Then $f'(x) = 0$.

10 Solutions

Solution (Solution to Exercise on Integration)

Solution 1 (Solution to Exercise on Integration). We use the power rule for integration: \[\int_0^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}\]

Solution (Alternative Solution to Quadratic Equation)

Solution (Alternative Solution to Quadratic Equation). For the quadratic equation, we can use the quadratic formula: \[x = \frac{5 \pm 1}{2}\]

This gives us $x = 3$ or $x = 2$.

11 Remarks

Remark (Convergence Note)

Remark 1 (Convergence Note). The limit in Definition 1 may not exist for all functions at all points. For example, the absolute value function $|x|$ is not differentiable at $x=0$.

Remark 1 highlights an important limitation of the derivative definition.

12 Mixed Example with Theorem and Solution

Here’s an example that combines multiple theorem types:

Theorem 2 (Mean Value Theorem) If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a point $c \in (a,b)$ such that: \[f'(c) = \frac{f(b) - f(a)}{b - a}\]

Exercise 2 (Applying the Mean Value Theorem) Let $f(x) = x^2$ on $[0, 2]$. Find the value of $c$ guaranteed by the Mean Value Theorem (Theorem 2).

Solution (Solution to Mean Value Theorem Exercise)

Solution 2 (Solution to Mean Value Theorem Exercise). We have $f(x) = x^2$, so $f'(x) = 2x$.

By the Mean Value Theorem: \[f'(c) = \frac{f(2) - f(0)}{2 - 0} = \frac{4 - 0}{2} = 2\]

Since $f'(c) = 2c$, we need: \[2c = 2\] \[c = 1\]

Therefore, $c = 1$ is the value guaranteed by the Mean Value Theorem.

References

--- title: "Theorem Examples" preview-changed: true --- This chapter demonstrates all the theorem-like environments available through the `callouty-theorem` and `custom-callout` extensions. # Theorems :::{#thm-pythagorean} ## Pythagorean Theorem In a right triangle with legs of length $a$ and $b$ and hypotenuse of length $c$, we have: $$a^2 + b^2 = c^2$$ ::: See @thm-pythagorean for the famous Pythagorean theorem. # Lemmas :::{#lem-fundamental} ## Fundamental Lemma Every non-empty subset of $\mathbb{N}$ has a smallest element. ::: The result in @lem-fundamental is used to prove many theorems in number theory. # Corollaries :::{#cor-rational-root} ## Rational Root Test Corollary If $p/q$ is a rational root of the polynomial $a_n x^n + \cdots + a_1 x + a_0$ with integer coefficients, then $p$ divides $a_0$ and $q$ divides $a_n$. ::: @cor-rational-root follows directly from the Rational Root Theorem. # Propositions :::{#prp-triangle-inequality} ## Triangle Inequality For any real numbers $x$ and $y$: $$|x + y| \leq |x| + |y|$$ ::: The triangle inequality (@prp-triangle-inequality) is fundamental in analysis. # Conjectures :::{#cnj-goldbach} ## Goldbach's Conjecture Every even integer greater than 2 can be expressed as the sum of two prime numbers. ::: @cnj-goldbach remains one of the oldest unsolved problems in number theory. # Definitions :::{#def-derivative} ## Derivative The derivative of a function $f$ at a point $x$ is defined as: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ provided this limit exists. ::: See @def-derivative for the formal definition of a derivative. # Examples :::{#exm-quadratic} ## Solving a Quadratic Equation Solve $x^2 - 5x + 6 = 0$. We can factor this as $(x-2)(x-3) = 0$, so the solutions are $x = 2$ and $x = 3$. ::: @exm-quadratic shows how to solve a simple quadratic equation. # Exercises :::{#exr-integration} ## Integration Practice Compute the following integral: $$\int_0^1 x^2 \, dx$$ ::: Try @exr-integration to practice basic integration. # Proofs :::{.proof} ## Proof of the Pythagorean Theorem Consider a square with side length $a+b$. We can divide this square into four right triangles and a central square with side length $c$. The area equations show: $$a^2 + b^2 = c^2$$ ::: :::{.proof} ## Proof that the derivative of a constant is zero Let $f(x) = c$ for some constant $c$. Then $f'(x) = 0$. ::: # Solutions :::{#sol-integration .solution} ## Solution to Exercise on Integration We use the power rule for integration: $$\int_0^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}$$ ::: :::{.solution} ## Alternative Solution to Quadratic Equation For the quadratic equation, we can use the quadratic formula: $$x = \frac{5 \pm 1}{2}$$ This gives us $x = 3$ or $x = 2$. ::: # Remarks :::{#rem-convergence .remark} ## Convergence Note The limit in @def-derivative may not exist for all functions at all points. For example, the absolute value function $|x|$ is not differentiable at $x=0$. ::: @rem-convergence highlights an important limitation of the derivative definition. # Mixed Example with Theorem and Solution Here's an example that combines multiple theorem types: :::{#thm-mean-value} ## Mean Value Theorem If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a point $c \in (a,b)$ such that: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ ::: :::{#exr-mvt-application} ## Applying the Mean Value Theorem Let $f(x) = x^2$ on $[0, 2]$. Find the value of $c$ guaranteed by the Mean Value Theorem (@thm-mean-value). ::: :::{#sol-mvt-application .solution} ## Solution to Mean Value Theorem Exercise We have $f(x) = x^2$, so $f'(x) = 2x$. By the Mean Value Theorem: $$f'(c) = \frac{f(2) - f(0)}{2 - 0} = \frac{4 - 0}{2} = 2$$ Since $f'(c) = 2c$, we need: $$2c = 2$$ $$c = 1$$ Therefore, $c = 1$ is the value guaranteed by the Mean Value Theorem. ::: # References {.unnumbered}