Continued Fractions; Loose Ends

In today' s blog, I will provide details that are used in the general proof for Continued Fractions.

Lemma 1: If x is a positive integer, then x² - 4 is not a square.

(1) First, I will show that for all positive integers greater than 2, the difference between one square and the next highest square is greater than 4 therefore no square x² - 4 can exist.

(2) For n=2, the smallest difference is 9 - 4 = 5 which is greater than 4.

(3) We assume that this is true up to n ≥ 2.

(4) So (n+1)² - n² = n² + 2n + 1 - n² = 2n + 1 ≥ 2(2) + 1 = 5.

(5) In other words, the minimal difference between any two successive squares is at least 5.

(6) Now, I will show that all differences 2 or less don't work either.

3² - 2² = 9 - 4 = 5
2¹ - 1² = 1
1² - 0² = 1

The only possible differences then are 1, 2, or a number ≥ 5.

QED

Quadratic Equation Solved

One of the most useful formulas of all time historically is the solution of the equation:
ax² + bx + c = 0.

The equation above is known as the quadratic equation.

Theorem: (-b ± √b² - 4ac)/2a is the solution to the quadratic equation.

(1) First, we multiply both sides by 4a and get:

4a²x² + 4abx + 4ac = 0

(2) Next, we add b² - b² to the equation:

4a²x² + 4abx + b² + 4ac - b² = 0

(3) Now, we add b² - 4ac to both sides which gives us:

4a²x² + 4abx + b² = b² - 4ac

(4) Further, we know that:

(2ax + b)² = 4a²x² + 4axb + b²

(5) Combining #4 and #3, gives us:

(2ax + b)² = b² - 4ac

(6) Now, taking the square root of both sides gives us:

2ax + b = ±√b² - 4ac

(7) Now, using basic algebra, we get to:

x = (-b ±√b² - 4ac)/2a.