PROOF 110

Prove.

Factoring is reading the product formulas backwards

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

After first fixing the range from which coefficients are chosen, factoring a polynomial f means finding polynomials g, h with coefficients in that range satisfying f = gh, where g and h are not constants.

Proof.

Factoring does not create a new expression; it rewrites the existing f in product form. Without fixing the coefficient range, this question has no meaning.convention

∎

Definition D2

Expansion means repeatedly applying the distributive law to an expression in product form until it becomes a sum of terms.

Proof.

Expansion rewrites an equality, so the result has the same value as the original expression.convention

∎

Definition D3

a − b means a + (−b).

Proof.

Subtraction is notation, not a new operation.convention

∎

Theorem T1

The product of two polynomials always expands. But if coefficients are chosen from the reals, x² + 1 does not factor.

Proof.

(1)An expression in product form becomes a sum of terms by repeated application of the distributive law.D2, A1

(2)This process does not ask about the coefficient range, so expansion always terminates.(1)

(3)Now choose coefficients from the reals, and suppose there are g, h with x² + 1 = g(x)h(x).assumption, D1

(4)deg g + deg h = 2(3), A3

(5)g and h are not constants, so their degrees are at least 1, hence both are linear.(4), D1

(6)g(x) = ax + b with a ≠ 0, so there is a real α with g(α) = 0.(5), A4

(7)α² + 1 = g(α)h(α) = 0 × h(α) = 0(3), (6), A5, A6

(8)Hence α² = −1(7), A10

(9)But no real α has α² = −1.A7

(10)Contradiction. Over real coefficients, x² + 1 does not factor.(8), (9)

∎

Theorem T2

If f = gh is a factorization, then expanding gh yields f.

Proof.

(1)Suppose f = gh.assumption, D1

(2)Expanding gh gives a sum of terms, and the result equals gh.D2, A1

(3)Hence the result of the expansion equals f.(1), (2)

(4)Factoring reads f = gh from left to right; expansion reads it from right to left.(3), D1, D2

(5)The two operations are not two different rules but two directions of reading one and the same equality.(4)

∎

Corollary C

If coefficients are chosen from the complex numbers, then x² + 1 = (x + i)(x − i).

Proof.

(1)(x + i)(x − i) = x² − i²A2

(2)i² = −1A8

(3)x² − i² = x² + (−(−1)) = x² + 1(1), (2), D3, A9

(4)x + i and x − i are not constants, so this is a factorization.(3), D1

(5)The same polynomial factors or fails to factor depending on the coefficient range.(4), T1

∎

Remark

Can A7 be removed from this proof?

Find the place where the fact that the coefficient range is the reals enters this proof. There is only one such place.