Proof: Sum 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:

The area of the orange square is

a ⋅ a

The area of the blue rectangle is:

a ⋅ b

The area of the green square is

b ⋅ b

Then, the total area of 1 orange square, 2 blue rectangles, and 1 green square is:

(a ⋅ a) + 2 ⋅ (a ⋅ b) + (b ⋅ b)

The graph shows that we can arrange the 4 rectangles into a larger square. The side of this larger square is a + b. Therefore, the area of the larger square is:

(a + b) ⋅ (a + b)

The two areas are equal, so we claim that:

(a ⋅ a) + 2 ⋅ (a ⋅ b) + (b ⋅ b) = (a + b) ⋅ (a + b)

This theorem can also be proved fairly easily using the distributive property, as shown below.

Proof:

Proof Table

#	Claim	Reason
1	(a + b) ⋅ (a + b) = ((a + b) ⋅ a) + ((a + b) ⋅ b)	(a + b) ⋅ (a + b) = ((a + b) ⋅ a) + ((a + b) ⋅ b) (Distributive Property)
2	(a + b) ⋅ a = (a ⋅ a) + (a ⋅ b)	(a + b) ⋅ a = (a ⋅ a) + (a ⋅ b) (Distributive Property Variation 2)
3	(a + b) ⋅ b = (a ⋅ b) + (b ⋅ b)	(a + b) ⋅ b = (a ⋅ b) + (b ⋅ b) (Distributive Property Variation 3)
4	((a + b) ⋅ a) + ((a + b) ⋅ b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b))	if (a + b) ⋅ a = (a ⋅ a) + (a ⋅ b) and (a + b) ⋅ b = (a ⋅ b) + (b ⋅ b), then ((a + b) ⋅ a) + ((a + b) ⋅ b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) (Equality Example)
5	(a + b) ⋅ (a + b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b))	if (a + b) ⋅ (a + b) = ((a + b) ⋅ a) + ((a + b) ⋅ b) and ((a + b) ⋅ a) + ((a + b) ⋅ b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)), then (a + b) ⋅ (a + b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) (Transitive Property of Equality)
6	(((a ⋅ a) + (a ⋅ b)) + (a ⋅ b)) + (b ⋅ b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b))	(((a ⋅ a) + (a ⋅ b)) + (a ⋅ b)) + (b ⋅ b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) (Associative Property of Addition)
7	(a ⋅ b) + (a ⋅ b) = (a ⋅ b) ⋅ 2	(a ⋅ b) + (a ⋅ b) = (a ⋅ b) ⋅ 2 (Doubling a Number)
8	(((a ⋅ a) + (a ⋅ b)) + (a ⋅ b)) + (b ⋅ b) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b)	if (a ⋅ b) + (a ⋅ b) = (a ⋅ b) ⋅ 2, then (((a ⋅ a) + (a ⋅ b)) + (a ⋅ b)) + (b ⋅ b) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b) (Substitution 16)
9	((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b)	if (((a ⋅ a) + (a ⋅ b)) + (a ⋅ b)) + (b ⋅ b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) and (((a ⋅ a) + (a ⋅ b)) + (a ⋅ b)) + (b ⋅ b) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b), then ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b) (Transitive Property of Equality Variation 2)
10	(a + b) ⋅ (a + b) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b)	if (a + b) ⋅ (a + b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) and ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b), then (a + b) ⋅ (a + b) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b) (Transitive Property of Equality)

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