Proof: Algebra Square Sum
Let's prove the following theorem:
if the following are true:
- x = (a ⋅ a) + (b ⋅ b)
- m = a
- n = b
then x = (n ⋅ n) + (m ⋅ m)
Proof:
Given
| 1 | x = (a ⋅ a) + (b ⋅ b) |
|---|---|
| 2 | m = a |
| 3 | n = b |
| # | Claim | Reason |
|---|---|---|
| 1 | m ⋅ m = a ⋅ a | if m = a, then m ⋅ m = a ⋅ a |
| 2 | x = (m ⋅ m) + (b ⋅ b) | if x = (a ⋅ a) + (b ⋅ b) and m ⋅ m = a ⋅ a, then x = (m ⋅ m) + (b ⋅ b) |
| 3 | n ⋅ n = b ⋅ b | if n = b, then n ⋅ n = b ⋅ b |
| 4 | x = (m ⋅ m) + (n ⋅ n) | if x = (m ⋅ m) + (b ⋅ b) and n ⋅ n = b ⋅ b, then x = (m ⋅ m) + (n ⋅ n) |
| 5 | (m ⋅ m) + (n ⋅ n) = (n ⋅ n) + (m ⋅ m) | (m ⋅ m) + (n ⋅ n) = (n ⋅ n) + (m ⋅ m) |
| 6 | x = (n ⋅ n) + (m ⋅ m) | if x = (m ⋅ m) + (n ⋅ n) and (m ⋅ m) + (n ⋅ n) = (n ⋅ n) + (m ⋅ m), then x = (n ⋅ n) + (m ⋅ m) |
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