Euclidean Proof:

1. Consider any triangle ABC.
2. Draw a line through point A parallel to side BC.
3. This creates three angles at point A: the original angle A from the triangle, and two alternate angles that are equal to angles B and C (due to parallel line properties).
4. These three angles form a straight line, which equals 180 degrees.
5. Therefore, angle A + angle B + angle C = 180 degrees.

Method:

1. Start with a regular hexagon inscribed in a circle of radius 1.
2. Calculate the perimeter of this hexagon: 6 × side length.
3. Double the number of sides to get a 12-sided polygon, then 24, 48, 96.
4. As the number of sides increases, the perimeter approaches 2π.
5. Archimedes used a 96-sided polygon to determine that π is between 3.1408 and 3.1429.

This method gave one of the first accurate approximations of π in history.

Example: Find the hypotenuse of a right triangle with sides 3 and 4 units.

1. Apply the Pythagorean theorem: c² = 3² + 4²
2. c² = 9 + 16 = 25
3. c = √25 = 5 units

Example: Find the square root of 5 using Newton's method.

1. Let f(x) = x² - 5, we want to find where f(x) = 0
2. f'(x) = 2x
3. The iteration formula becomes: x_n+1 = x_n - (x_n² - 5)/(2x_n)
4. Simplify: x_n+1 = (x_n + 5/x_n)/2
5. Start with x₀ = 2:
   • x₁ = (2 + 5/2)/2 = 2.25
   • x₂ = (2.25 + 5/2.25)/2 ≈ 2.2361
   • x₃ = (2.2361 + 5/2.2361)/2 ≈ 2.2361
6. The method converges to √5 ≈ 2.2361

Example: Find the derivative of f(x) = x³ using Leibniz's approach.

1. Leibniz viewed the derivative as a ratio of infinitesimal differences.
2. For f(x) = x³, we compute:
   • f(x + dx) = (x + dx)³
   • f(x + dx) = x³ + 3x²dx + 3x(dx)² + (dx)³
   • df = f(x + dx) - f(x) = 3x²dx + 3x(dx)² + (dx)³
   • For infinitesimal dx, the higher-order terms become negligible
   • df/dx ≈ 3x²

Example: Representing a circle in the Cartesian plane.

1. A circle with radius r centered at the origin can be defined as all points (x,y) where the distance from (0,0) is exactly r.
2. Using the distance formula: √(x² + y²) = r
3. Square both sides: x² + y² = r²
4. This gives us the equation of a circle: x² + y² = r²
5. For a circle with radius 5: x² + y² = 25

Proof sketch:

1. Start with Euler's formula: e^ix = cos(x) + i·sin(x)
2. Substitute x = π: e^iπ = cos(π) + i·sin(π)
3. We know that cos(π) = -1 and sin(π) = 0
4. Therefore: e^iπ = -1
5. Rearranging: e^iπ + 1 = 0

Problem: Is it possible to walk through the city of Königsberg, crossing each of its seven bridges exactly once?

1. Euler represented land masses as nodes and bridges as edges in a graph.
2. He proved that such a path (later called an Eulerian path) exists if and only if the graph has either zero or two nodes with an odd number of edges.
3. In the Königsberg graph, all four nodes had an odd number of edges.
4. Therefore, Euler concluded that the walk was impossible.

Example: Solve the system of equations:

2x + y - z = 8

-3x - y + 2z = -11

-2x + y + 2z = -3

1. Write as an augmented matrix:

   [ 2  1 -1 |  8]
   [-3 -1  2 | -11]
   [-2  1  2 | -3]

2. Use row operations to transform into row echelon form:

   [ 2  1 -1 |  8]
   [ 0  0.5 0.5 | 1]
   [ 0  0  1 | -1]

3. Back-substitute to find: z = -1, y = 3, x = 3

Proof sketch:

1. Assume, for contradiction, that the real numbers between 0 and 1 are countable.
2. This means we could list them all: r₁, r₂, r₃, ...
3. Write each number in decimal form:

   r₁ = 0.a₁₁a₁₂a₁₃...
   r₂ = 0.a₂₁a₂₂a₂₃...
   r₃ = 0.a₃₁a₃₂a₃₃...
   ...

4. Construct a new number s = 0.b₁b₂b₃... where b₁ ≠ a₁₁, b₂ ≠ a₂₂, b₃ ≠ a₃₃, etc.
5. The number s differs from every number in our list in at least one decimal place.
6. Therefore, s is a real number between 0 and 1 that is not in our list.
7. This contradicts our assumption that we listed all real numbers between 0 and 1.

Components of a Turing Machine:

1. A tape divided into cells, each containing a symbol from a finite alphabet.
2. A head that can read and write symbols on the tape and move left and right.
3. A state register that stores the state of the Turing machine.
4. A finite table of instructions that tells the machine what to do based on the current state and the symbol it's reading.

Example: A simple Turing machine that converts all 0s to 1s and vice versa:

• If in state A and reading 0: write 1, move right, stay in state A.
• If in state A and reading 1: write 0, move right, stay in state A.
• If in state A and reading blank: halt.

Example: The Prisoner's Dilemma

Two suspects are interrogated separately with the following payoffs (years in prison):

• Both remain silent: Each gets 1 year
• One betrays, one remains silent: Betrayer goes free, silent gets 3 years
• Both betray: Each gets 2 years

Analysis:

1. From Prisoner A's perspective:
   • If B remains silent: Better to betray (0 years) than remain silent (1 year)
   • If B betrays: Better to betray (2 years) than remain silent (3 years)
2. Prisoner B has the exact same reasoning.
3. The Nash equilibrium is for both to betray, resulting in each serving 2 years.
4. This is stable because neither can unilaterally improve their situation by changing strategy.
5. Paradoxically, mutual cooperation would be better for both, but is not stable.

History and Solution:

1. Pierre de Fermat claimed in 1637 to have a proof but never wrote it down ("the margin was too small").
2. For 350+ years, mathematicians tried unsuccessfully to prove it.
3. Wiles worked in secret for 7 years, connecting two seemingly unrelated areas of mathematics:
   • Elliptic curves: y² = x³ + ax + b
   • Modular forms: complex functions with special symmetry properties
4. He proved the Taniyama-Shimura conjecture for semistable elliptic curves, which implied Fermat's Last Theorem.
5. After fixing a gap in his original proof, Wiles published the complete proof in 1995.