PROOF 23
Prove.
The product formulas come from the distributive law
On the real numbers ℝ.
A way to read this proof
First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.
Definition D1
a − b means a + (−b), and a² means a × a.
Proof.
Subtraction and squaring are notation, not new operations.convention
∎
Lemma L1
For all a, b, c, d, (a + b)(c + d) = ac + ad + bc + bd
Proof.
(1)(a + b)(c + d) = (a + b)c + (a + b)dA1, A2
(2)(a + b)c = ac + bcA1, A2
(3)(a + b)d = ad + bdA1, A2
(4)(a + b)(c + d) = (ac + bc) + (ad + bd)(1), (2), (3)
(5) = ac + ad + bc + bd(4), A5, A6
∎
Theorem T1
For all a, b, (a + b)² = a² + 2ab + b²
Proof.
(1)(a + b)² = (a + b)(a + b)D1
(2)Setting c = a, d = b in L1 gives (a + b)(a + b) = aa + ab + ba + bbL1
(3)aa = a², bb = b²D1
(4)ba = abA2
(5)ab + ab = (1 + 1)ab = 2abA1, A2, A3
(6)(a + b)² = a² + 2ab + b²(1), (2), (3), (4), (5), A5, A6
∎
Corollary C
For all a, b, (a + b)(a − b) = a² − b²
Proof.
(1)(a + b)(a − b) = (a + b)(a + (−b))D1
(2)Setting c = a, d = −b in L1 gives = aa + a(−b) + ba + b(−b)(1), L1
(3)aa = a², a(−b) = −(ab)D1, A4
(4)ba = ab, b(−b) = −(bb) = −(b²)A2, A4, D1
(5)(−(ab)) + ab = 0A6, A7
(6)(a + b)(a − b) = a² + 0 + (−(b²)) = a² − b²(2), (3), (4), (5), A5, A6, A8, D1
∎
Remark
Can A1 be removed from this proof?
Check whether the two formulas collapse separately or together.