Proof: Difference Squared Theorem

Let's prove the following theorem:

This theorem is equivalent to:

(a - b)² = a² - 2⋅a⋅b + b²

Here is a visualization of the theorem:

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

The area of the orange square is (a - b)².

Let's now try to describe the area of the orange square in terms of the other rectangles.

The area of the orange square is:

(The area of the largest square) - 2 ⋅ (the area of the blue rectangle) - (the area of the green square)

The area of the largest square is

a²

The area of the blue rectangle is

b ⋅ (a - b)

The area of the green rectangle is

b²

Then using substitution, the area of the orange square is

a² - 2 ⋅ (b ⋅ (a - b)) - b²

2 ⋅ (b ⋅ (a - b)) is equal to

(2 ⋅ b ⋅ a) - (2 ⋅ b ⋅ b)

Which is

(2 ⋅ b ⋅ a) - (2 ⋅ b²)

After substitution, we get:

a² - ((2 ⋅ b ⋅ a) - (2 ⋅ b²)) - b²

Which reduces to

a² - (2 ⋅ b ⋅ a) + b²

We can also prove this theorem using the Sum Squared Theorem, which used the distributive property, as shown below.

Proof:

Proof Table

#	Claim	Reason
1	(a + (b ⋅ (-1))) ⋅ (a + (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ (b ⋅ (-1))) ⋅ 2)) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1)))	(a + (b ⋅ (-1))) ⋅ (a + (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ (b ⋅ (-1))) ⋅ 2)) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) (sum_squared_theorem)
2	(a ⋅ (b ⋅ (-1))) ⋅ 2 = (a ⋅ b) ⋅ (-2)	(a ⋅ (b ⋅ (-1))) ⋅ 2 = (a ⋅ b) ⋅ (-2) (manipulation_2)
3	(a ⋅ a) + ((a ⋅ (b ⋅ (-1))) ⋅ 2) = (a ⋅ a) + ((a ⋅ b) ⋅ (-2))	if (a ⋅ (b ⋅ (-1))) ⋅ 2 = (a ⋅ b) ⋅ (-2), then (a ⋅ a) + ((a ⋅ (b ⋅ (-1))) ⋅ 2) = (a ⋅ a) + ((a ⋅ b) ⋅ (-2)) (Substitution)
4	((a ⋅ a) + ((a ⋅ (b ⋅ (-1))) ⋅ 2)) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1)))	if (a ⋅ a) + ((a ⋅ (b ⋅ (-1))) ⋅ 2) = (a ⋅ a) + ((a ⋅ b) ⋅ (-2)), then ((a ⋅ a) + ((a ⋅ (b ⋅ (-1))) ⋅ 2)) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) (Additive Property of Equality)
5	(b ⋅ (-1)) ⋅ (b ⋅ (-1)) = b ⋅ b	(b ⋅ (-1)) ⋅ (b ⋅ (-1)) = b ⋅ b (manipulation_3)
6	((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + (b ⋅ b)	if (b ⋅ (-1)) ⋅ (b ⋅ (-1)) = b ⋅ b, then ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + (b ⋅ b) (Substitution)
7	((a ⋅ a) + ((a ⋅ (b ⋅ (-1))) ⋅ 2)) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + (b ⋅ b)	if ((a ⋅ a) + ((a ⋅ (b ⋅ (-1))) ⋅ 2)) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) and ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + (b ⋅ b), then ((a ⋅ a) + ((a ⋅ (b ⋅ (-1))) ⋅ 2)) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + (b ⋅ b) (Transitive Property of Equality)
8	(a + (b ⋅ (-1))) ⋅ (a + (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + (b ⋅ b)	if (a + (b ⋅ (-1))) ⋅ (a + (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ (b ⋅ (-1))) ⋅ 2)) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) and ((a ⋅ a) + ((a ⋅ (b ⋅ (-1))) ⋅ 2)) + ((b ⋅ (-1)) ⋅ (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + (b ⋅ b), then (a + (b ⋅ (-1))) ⋅ (a + (b ⋅ (-1))) = ((a ⋅ a) + ((a ⋅ b) ⋅ (-2))) + (b ⋅ b) (Transitive Property of Equality)

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