PROOF 105

Prove.

Elimination does not change the solution set

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

The solution set of the system (E₁, E₂) is the set of all ordered pairs (x, y) that make both equations true.

Proof.

(1)Write E₁ as P = p and E₂ as Q = q. P, Q are linear expressions in x and y.convention

(2)The solution set is the intersection of the solution set of E₁ and that of E₂.convention

∎

Definition D2

E₂ + kE₁ means Q + kP = q + kp.

Proof.

(1)kE₁ is the equation kP = kp obtained by multiplying both sides of E₁ by k.convention

(2)Adding the two equations, left sides together and right sides together, gives E₂ + kE₁.convention

∎

Lemma L1

If (x, y) satisfies E₁ and E₂, then it also satisfies E₂ + kE₁.

Proof.

(1)Suppose P = p and Q = q.assumption, D1

(2)kP = kp(1), A1

(3)Q + kP = q + kp(1), (2), A2

(4)This equation is E₂ + kE₁.(3), D2

∎

Lemma L2

(E₂ + kE₁) + (−k)E₁ is the same equation as E₂.

Proof.

(1)kP + (−k)P = (k + (−k))PA6

(2)k + (−k) = 0A4

(3)(k + (−k))P = 0 × P = 0(2), A7

(4)(Q + kP) + (−k)P = Q + (kP + (−k)P)A3

(5)　　　　　　　　　 = Q + 0 = Q(1), (3), A5

(6)The right side likewise: (q + kp) + (−k)p = qA6, A4, A7, A3, A5

(7)That equation is Q = q, that is, E₂.(5), (6), D2

∎

Theorem T1

The solution set of (E₁, E₂) and that of (E₁, E₂ + kE₁) are equal.

Proof.

(1)A solution of the first system satisfies E₁, and also satisfies E₂ + kE₁.L1, D1

(2)Hence the first solution set is contained in the second.(1), D1

(3)Now take a solution of the second system. It satisfies E₁ and E₂ + kE₁.assumption, D1

(4)Apply L1 to the two equations E₁, E₂ + kE₁ with −k in place of k.(3), L1, A4

(5)That solution satisfies (E₂ + kE₁) + (−k)E₁.(4), D2

(6)That equation is E₂, so the second solution set is contained in the first as well.(5), L2, D1

(7)The two solution sets contain each other, hence they are equal.(2), (6), S

∎

Remark

Can A4 be removed from this proof?

Separate the lines that show only the forward direction from the lines that undo the step. Check whether one-sided containment lets you say the two systems ask the same question.