PROOF 92

Prove.

Identities and equations are different

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

When P(x) = Q(x) holds for every real x, the expression is called an identity.

Proof.

(1)An identity is a single claim, and the scope of that claim is all of the reals.convention

∎

Definition D2

When P(x) = Q(x) is viewed as the question of finding the x that make it hold, the expression is called an equation.

Proof.

(1)An equation is not a claim but a condition, and its answer is the solution set.convention

(2)The two names refer not to the shape of the expression but to the range of x.convention, D1

∎

Theorem T1

If P(x) = Q(x) is an identity, all corresponding coefficients are equal. That is, with P = aₙxⁿ + … + a₀ and Q = bₙxⁿ + … + b₀, aₖ = bₖ for every k.

Proof.

(1)Align the degrees of both polynomials at n, taking the coefficients of missing terms to be 0.convention

(2)Suppose P(x) = Q(x) for every real x.assumption, D1

(3)Setting R = P − Q, R has degree at most n and its coefficients are aₖ − bₖ.(1), A2

(4)At each real a, P(a) = Q(a).(2), A4

(5)R(a) = P(a) − Q(a) = 0(3), (4), A3

(6)Therefore every real is a root of R.(5)

(7)There are more than n + 1 distinct reals.A5

(8)All coefficients of R are 0.(6), (7), A1

(9)aₖ − bₖ = 0, so aₖ = bₖ(3), (8), A6

∎

Theorem T2

2x + 1 = 7 is not an identity, so T1 cannot be applied to it.

Proof.

(1)Substituting x = 0 gives 1 = 7N

(2)1 ≠ 7, so it fails at x = 0.(1), A7

(3)Therefore this expression is not an identity.(2), D1

(4)Substituting x = 3 gives 7 = 7N

(5)Some x makes it hold, so the expression is meaningful as an equation.(4), D2

(6)The hypothesis of T1 fails by (3), so one may not compare coefficients and write 2 = 0.(3), T1

∎

Corollary C

If ax² + bx + c = 3x² − 5 is an identity, then a = 3, b = 0, c = −5

Proof.

(1)Suppose it holds for every real x.assumption, D1

(2)Write the right side as 3x² + 0x + (−5).convention

(3)a = 3, b = 0, c = −5(1), (2), T1

(4)The three equalities were obtained at once, without checking values of x one by one.(3)

∎

Remark

Can A1 be removed from this proof?

Check whether comparing coefficients comes from '=' or from somewhere else.