PROOF 91

Prove.

Substitution is valid

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

For a polynomial P and a real number a, P(a) is the result of putting a in every place of x and carrying out the prescribed additions and multiplications.

Proof.

(1)That process consists of finitely many additions and multiplications.convention

∎

Definition D2

If P(x) = Q(x) holds for every real x, it is called an identity; viewed as the question of finding the x for which it holds, it is called an equation.

Proof.

(1)Let S be the set of all reals x for which P(x) = Q(x) holds.convention

(2)An identity is the case S = ℝ.convention

∎

Lemma L1

For each real a, P(a) is determined as a single real number.

Proof.

(1)The process of D1 is finitely many additions and multiplications.D1

(2)Addition and multiplication are functions, so they give the same value on the same input.A1

(3)Each step gives a single value, so P(a) is determined as a single value.(1), (2)

∎

Theorem T1

If P(x) = Q(x) is an identity, then for every real a, P(a) = Q(a)

Proof.

(1)Suppose P(x) = Q(x) for every real x.assumption, D2

(2)a is real, so P(a) = Q(a).(1), A2

(3)P(a) and Q(a) are each determined as single real numbers.L1

(4)Therefore (2) is an equality between two real numbers.(2), (3)

∎

Theorem T2

If P(x) = Q(x) is not an identity, there exists a real a with P(a) ≠ Q(a).

Proof.

(1)Let S be the set of D2.D2

(2)It is not an identity, so S ≠ ℝ.(1), D2

(3)There is a real a not belonging to S.(2)

(4)At that a, P(a) ≠ Q(a).(1), (3)

(5)For 2x + 1 = 7, x = 3 belongs to S and x = 0 does not.N, (1)

(6)The equality remains true only when a belongs to S.(4), (5)

∎

Corollary C

If ax + b = 3x + 5 is an identity, then a = 3, b = 5

Proof.

(1)Suppose ax + b = 3x + 5 for every real x.assumption, D2

(2)Substituting x = 0 gives a × 0 + b = 3 × 0 + 5T1

(3)b = 5(2), A3, A4

(4)Substituting x = 1 gives a × 1 + b = 3 × 1 + 5T1

(5)a + b = 3 + 5(4), A5

(6)a + 5 = 3 + 5(3), (5)

(7)a = 3(6), A6

∎

Remark

Can A1 be removed from this proof?

Substitution giving a value and substitution preserving an equality are different things. Check which side rests on A1.