Proof: Difference of Two Squares

Let's prove the following theorem:

This theorem is equivalent to:

a² - b² = (a + b) ⋅ (a - b)

Here is a visualization of the theorem:

In this graph, the side of the largest square is a. The side of the orange square is b. Thus, the longer side of the blue rectangle is a - b.

The difference between the area of the largest square and the area of the orange square is:

a² - b²

Visually, if we remove the orange square from the largest square, then we are left with 2 blue rectangles and the green square.

The area of 2 blue rectangles and the green square is:

2 ⋅ ((a - b) ⋅ b) + (a - b)²

If we factor out (a - b) from both terms, we get:

(2 ⋅ 1 ⋅ b + (a - b)) ⋅ (a - b)

(2 ⋅ 1 ⋅ b + (a - b)) is a + b. Therefore, the area of 2 blue rectangles and the green square is:

(a + b) ⋅ (a - b)

We can also prove this theorem using the distributive property, as shown below.

Proof:

Proof Table

#	Claim	Reason
1	(a + b) ⋅ (a + (b ⋅ (-1))) = ((a + b) ⋅ a) + ((a + b) ⋅ (b ⋅ (-1)))	(a + b) ⋅ (a + (b ⋅ (-1))) = ((a + b) ⋅ a) + ((a + b) ⋅ (b ⋅ (-1))) (Distributive Property)
2	(a + b) ⋅ a = (a ⋅ a) + (b ⋅ a)	(a + b) ⋅ a = (a ⋅ a) + (b ⋅ a) (Distributive Property Variation 3)
3	(a + b) ⋅ (b ⋅ (-1)) = ((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1))	(a + b) ⋅ (b ⋅ (-1)) = ((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1)) (distributive_applied)
4	((a + b) ⋅ a) + ((a + b) ⋅ (b ⋅ (-1))) = ((a ⋅ a) + (b ⋅ a)) + (((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1)))	if (a + b) ⋅ a = (a ⋅ a) + (b ⋅ a) and (a + b) ⋅ (b ⋅ (-1)) = ((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1)), then ((a + b) ⋅ a) + ((a + b) ⋅ (b ⋅ (-1))) = ((a ⋅ a) + (b ⋅ a)) + (((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1))) (Equality Example)
5	(b ⋅ a) + ((a ⋅ b) ⋅ (-1)) = 0	(b ⋅ a) + ((a ⋅ b) ⋅ (-1)) = 0 (add_inverse)
6	((a ⋅ a) + (b ⋅ a)) + (((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1))) = (a ⋅ a) + ((b ⋅ b) ⋅ (-1))	if (b ⋅ a) + ((a ⋅ b) ⋅ (-1)) = 0, then ((a ⋅ a) + (b ⋅ a)) + (((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1))) = (a ⋅ a) + ((b ⋅ b) ⋅ (-1)) (equation)
7	(a + b) ⋅ (a + (b ⋅ (-1))) = (a ⋅ a) + ((b ⋅ b) ⋅ (-1))	if (a + b) ⋅ (a + (b ⋅ (-1))) = ((a + b) ⋅ a) + ((a + b) ⋅ (b ⋅ (-1))) and ((a + b) ⋅ a) + ((a + b) ⋅ (b ⋅ (-1))) = ((a ⋅ a) + (b ⋅ a)) + (((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1))) and ((a ⋅ a) + (b ⋅ a)) + (((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1))) = (a ⋅ a) + ((b ⋅ b) ⋅ (-1)), then (a + b) ⋅ (a + (b ⋅ (-1))) = (a ⋅ a) + ((b ⋅ b) ⋅ (-1)) (Transitive Property Application 1)
8	(a ⋅ a) + ((b ⋅ b) ⋅ (-1)) = (a + b) ⋅ (a + (b ⋅ (-1)))	if (a + b) ⋅ (a + (b ⋅ (-1))) = (a ⋅ a) + ((b ⋅ b) ⋅ (-1)), then (a ⋅ a) + ((b ⋅ b) ⋅ (-1)) = (a + b) ⋅ (a + (b ⋅ (-1))) (Symmetric Property of Equality)

Comments

Please log in to add comments