PROOF 155

Prove.

Vector addition is commutative

On plane vectors.

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.

What you may use

A1a + b = b + a
commutativity of real addition
A2(a+b)+c = a+(b+c)
associativity of real addition
A3(x + y) − x = y
additive inverse (Proof 7)
A4PQ = √((Δx)²+(Δy)²)
distance between two points (Proof 21)
A5Q−P = S−R ⟹ PQ ∥ RS
arrows with equal components are parallel — translation carries a line to a parallel line
Ea=b, b=c ⟹ a=c
properties of equality — substitution and transitivity

D1vector = ordered pair (a₁, a₂)
vectors and equality
D2a+b = (a₁+b₁, a₂+b₂)
vector addition — componentwise
D3placing a at P gives P+a
placement of arrows
L1placing twice gives P+(a+b)
Lemma 1 — two placements
T1a + b = b + a
Theorem 1 — commutativity in components
T2the two paths end at the same point
Theorem 2 — the parallelogram closes

Nothing outside this list may be cited.

Definition D1

Define a vector of the plane to be an ordered pair of real numbers a = (a₁, a₂). Two vectors are equal when, and only when, their components are respectively equal.

Proof.

(1)Call a₁ the x-component and a₂ the y-component.convention

(2)(a₁, a₂) = (b₁, b₂) means a₁ = b₁ and a₂ = b₂.convention

∎

Definition D2

Define a + b = (a₁ + b₁, a₂ + b₂).

Proof.

(1)Addition is real addition performed component by component.convention

(2)Subtraction is defined the same way: a − b = (a₁ − b₁, a₂ − b₂).convention

∎

Definition D3

For a point P = (p₁, p₂) and a vector a, set P + a = (p₁ + a₁, p₂ + a₂), and call the arrow from P to P + a the result of placing a at P.

Proof.

(1)Define the components of an arrow PQ to be Q − P.convention

(2)The components of the arrow obtained by placing a at P are (P + a) − P = (a₁, a₂) = a(1), A3, D2, D1

∎

Lemma L1

Let Q be the endpoint of the arrow obtained by placing a at P, and let R be the endpoint of the arrow obtained by placing b at that Q. Then R = P + (a + b)

Proof.

(1)Q = P + a = (p₁ + a₁, p₂ + a₂)D3

(2)R = Q + b = ((p₁ + a₁) + b₁, (p₂ + a₂) + b₂)(1), D3

(3)(p₁ + a₁) + b₁ = p₁ + (a₁ + b₁) and (p₂ + a₂) + b₂ = p₂ + (a₂ + b₂)A2

(4)R = (p₁ + (a₁ + b₁), p₂ + (a₂ + b₂)) = P + (a + b)(2), (3), D1, D2, D3, E

∎

Theorem T1

For all vectors a, b, a + b = b + a

Proof.

(1)a + b = (a₁ + b₁, a₂ + b₂)D2

(2)a₁ + b₁ = b₁ + a₁ and a₂ + b₂ = b₂ + a₂A1

(3)The components are respectively equal, so (a₁ + b₁, a₂ + b₂) = (b₁ + a₁, b₂ + a₂)(2), D1

(4)(b₁ + a₁, b₂ + a₂) = b + aD2

(5)a + b = b + a(1), (3), (4), E

∎

Theorem T2

The point reached from O by placing a and then b equals the point reached by placing b and then a.

Proof.

(1)Let C be the endpoint of the first path. Then C = O + (a + b)L1

(2)Let C′ be the endpoint of the second path. Then C′ = O + (b + a)L1

(3)a + b = b + aT1

(4)O + (a + b) = O + (b + a), so C = C′(1), (2), (3), D3, E

∎

Corollary C

Let A = O + a, B = O + b. If the three points O, A, B are not collinear, then the quadrilateral OACB is a parallelogram.

Proof.

(1)The two paths reach the same point C, so C = A + b and C = B + aT2, L1, D3

(2)The arrows OA and BC are both placements of a, so both have components a.(1), D3

(3)Equal components give OA ∥ BC, and both lengths equal √(a₁² + a₂²).(2), A5, A4

(4)The arrows AC and OB are both placements of b, so by the same argument AC ∥ OB with equal lengths.(1), D3, A5, A4

(5)O, A, B are not collinear, so the four points form the quadrilateral OACB, whose two pairs of opposite sides are respectively parallel.assumption, (3), (4), convention

∎

Remark

Can A1 be removed from this proof?

The endpoints of the two paths are each determined even without A1. Find the line that says the two are the same point.