Plane vectors in a fixed basis: componentwise operations, equality and the direction of subtraction

Goal and scope

In the fixed standard basis let u=(3,−1), v=(−2,4), w=(1,3) and k=−2. Compute u+v, u−v, v−u, ku, 0u and 2u+v componentwise; find the vector x with u+x=v; check k(u+v)=ku+kv; and decide whether u=w and whether u=(−1,3).

Works inside the MKO's scope: a fixed standard basis, real components and scalars, with exact integer values in the task. Shows componentwise operations and componentwise equality, and proves three small general lemmas (u−v=−(v−u), the unique solution of u+x=v, and distributivity of a scalar over vector addition). Does not infer lengths, angles, unit vectors or changes of basis from components; does not treat general vector-space axioms or physical units; does not change the canonical MKO or its English edition.

Givens and symbols

• cartesian_product_def
• add_inverse
• distributive
• neg_mul
• numeric_evaluation
• neg_vec_def

• G1 — The basis is the fixed standard basis; a vector is a component pair (z₁,z₂)∈ℝ², addition, subtraction and scalar multiplication act componentwise, two vectors are equal exactly when both corresponding components are equal, and the zero vector is (0,0).
• G2 — u=(3,−1), v=(−2,4), w=(1,3), k=−2 (integer components and an integer scalar, MKO assumption a1).
• G3 — Ordered pairs here represent vector components; geometric points, position vectors and arbitrary coordinate systems are not identified as one object (MKO assumption a2).

• u=(u_1,u_2) — a plane vector given by its components in the fixed standard basis
• u+v,\ u-v — componentwise sum and difference
• ku — the scalar k times the vector u, again a vector
• -z — (−1)z=(−z_1,−z_2)
• (0,0) — the zero vector

Rules and sources

• cartesian_product_def · prerequisite — Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d.
• add_inverse · prerequisite — Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v).
• distributive · prerequisite — Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu.
• neg_mul · prerequisite — Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv.
• numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable.
• add_comm · prerequisite — Commutativity of addition: u+v=v+u.
• eq_congruence · prerequisite — Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule.
• add_assoc · prerequisite — Associativity of addition: (u+v)+w=u+(v+w).
• add_zero · prerequisite — Additive identity: u+0=u.
• mul_zero · prerequisite — Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0.
• mul_comm · prerequisite — Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ).
• and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q.
• mko_vector_ops · cited — In the fixed standard basis, a plane vector is u=(u₁,u₂)∈R². Define u+v=(u₁+v₁,u₂+v₂), k u=(k u₁,k u₂), and u−v=(u₁−v₁,u₂−v₂). Two vectors are equal exactly when both corresponding components are equal; the zero vector is (0,0). · mko-vector
• neg_vec_def · prerequisite — Negative of a vector: −z means (−1)z=(−z₁,−z₂).
• L_sub_reverse · lemma — For all u, v∈ℝ²: u−v=−(v−u).
• L_solve_vec · lemma — For all u, v, x∈ℝ²: u+x=v ⇔ x=v−u.
• L_scalar_distrib · lemma — For every real k and all u, v∈ℝ²: k(u+v)=ku+kv.

Main derivation

u=(3,-1),\ v=(-2,4),\ w=(1,3),\ k=-2

u+v=(u_1+v_1,u_2+v_2),\ ku=(ku_1,ku_2),\ u-v=(u_1-v_1,u_2-v_2)

u+v

(3+(-2),\,-1+4)=(1,3)

u-v

(3-(-2),\,-1-4)=(5,-5)

v-u

(-2-3,\,4-(-1))=(-5,5)

u-v=(5,-5),\ v-u=(-5,5)

u-v=-(v-u)

u-v=(5,-5),\ v-u=(-5,5)

u-v\neq v-u

u+x=v

x=v-u=(-5,5)

ku,\ k=-2

(-2\cdot 3,\,-2\cdot(-1))=(-6,2)

0u

(0\cdot 3,\,0\cdot(-1))=(0,0)

2u+v

(6,-2)+(-2,4)=(4,2)

k(u+v),\ ku+kv

-2(1,3)=(-2,-6)=(-6,2)+(4,-8)

u=(3,-1),\ w=(1,3),\ u_1^2+u_2^2=w_1^2+w_2^2=10

u\neq w

u=(3,-1),\ (-1,3)

u\neq(-1,3)

\text{S02 to S13}

u+v=(1,3),\ u-v=(5,-5),\ v-u=(-5,5),\ ku=(-6,2),\ 0u=(0,0),\ 2u+v=(4,2),\ x=(-5,5)

Substeps and lemmas

u-v,\quad -(v-u)

(u_1-v_1,\,u_2-v_2),\quad (-(v_1-u_1),\,-(v_2-u_2))

-(v_i-u_i)

u_i-v_i\quad(i=1,2)

u+x=v

x=v-u

x=v-u

u+x=v

k(u+v)

(k(u_1+v_1),k(u_2+v_2))=(ku_1+kv_1,ku_2+kv_2)=ku+kv

Reverse audits

• S14 · sufficient_goal_reduction · S02, S03, S04, S07, S08, S09, S10, S11, S12, S13, and_intro_elim — Goal G: the computations and decisions the task asks for. Sufficient condition H: the conclusions of the listed steps; H⇒G is just conjunction, H is proved forwards from G1 and G2, and G is never used as a premise.
• S13 · premise_audit · cartesian_product_def — Ordered-pair equality needs agreement position by position; one differing position suffices to refute it.
• S12 · premise_audit · cartesian_product_def, numeric_evaluation — One differing component suffices for inequality (N20); no relation between the sum of squares and any geometric quantity is claimed.
• S11.1 · inverse_equivalence · mko_vector_ops, distributive, cartesian_product_def — Each equality is an identity and both components agree, so it holds both ways.
• S11 · premise_audit · L_scalar_distrib, S02, S08 — The lemma is proved for all k, u, v in S11.1; the numbers here only check it and do not replace the proof.
• S10 · inverse_equivalence · mko_vector_ops, numeric_evaluation — Recompute componentwise (N14, N15).
• S09 · inverse_equivalence · mul_zero, mul_comm, mko_vector_ops — The components agree (N12, N13), so it holds both ways.
• S08 · inverse_equivalence · mko_vector_ops, numeric_evaluation, neg_mul — Recompute componentwise (N10, N11); both sides are the same vector.
• S07.2 · premise_audit · add_inverse, add_assoc, add_comm, add_zero — Only the group properties of real addition are used, valid for every u and v.
• S07.1 · premise_audit · cartesian_product_def, eq_congruence, add_inverse — This is the '⇒' direction of the lemma, the other being proved in S07.2; adding the same number to both sides is always allowed.
• S07 · inverse_equivalence · L_solve_vec — Both directions of the lemma are proved in S07.1 and S07.2; the substitution check is N8 and N9.
• S06 · premise_audit · cartesian_product_def, S03, S04 — One differing component suffices for inequality; the closing general remark is not proved here and serves only as comment, the verdict resting on this example's components.
• S05.2 · inverse_equivalence · distributive, neg_mul, add_comm, cartesian_product_def — These are identities valid for all reals, so they hold both ways; vector equality needs both components to agree, and both do.
• S05.1 · inverse_equivalence · mko_vector_ops, neg_vec_def — Unfolding definitions is two-way.
• S05 · premise_audit · L_sub_reverse, S03, S04 — The lemma is proved for all vectors in S05.1 and S05.2; the numbers here are only a check.
• S04 · inverse_equivalence · mko_vector_ops, numeric_evaluation — Recompute componentwise (N5, N6).
• S03 · inverse_equivalence · mko_vector_ops, numeric_evaluation — Recompute componentwise (N3, N4); both sides are the same component pair.
• S02 · inverse_equivalence · mko_vector_ops, numeric_evaluation — Both sides are the same component pair; recompute componentwise (N1, N2).
• S01 · premise_audit · G1, G2, G3, mko_vector_ops — This step only states the definitions and data and derives no new fact; without a fixed basis, the same components could represent different vectors.

Error controls and negative controls

Components have positions; subtraction has a direction

One misconception named by the MKO is treating vector coordinates as interchangeable. (3,−1) and (−1,3) differ (S13); u−v=(5,−5) and v−u=(−5,5) are opposite vectors (S05, S06). To get from u to v one adds v−u (S07); writing u−v gives the opposite direction.

Equality is componentwise, not by length or component sum

The second misconception identified in the MKO is that equal length is enough to establish vector equality. u and w both have sum of squares 10, but 3≠1, so u≠w (S12); u and (1,1) both have component sum 2 and are not equal either (S13). The only criterion is that both corresponding components agree.

Scalar multiplication is not an inner product

The third misconception identified in the MKO is mistaking scalar multiplication for an inner product. ku=(−6,2) is a vector in ℝ² (S08), not a number; adding its components to get −4 already leaves the operations of this entry. The inner product is a different operation, which this entry neither defines nor uses.

• u−v=v−u, since both are the difference of u and v → reject — u−v=(5,−5) and v−u=(−5,5); the first components are 5≠−5.
• u=w, because u₁²+u₂²=w₁²+w₂²=10 → reject — Equality needs every component to agree; the first components are 3≠1.
• (3,−1)=(−1,3), since both use the same two numbers → reject — Ordered pairs are compared position by position; 3≠−1.
• ku is a number: (−2)·3+(−2)·(−1)=−4 → reject — ku=(−6,2) is a vector; adding its components is not part of scalar multiplication.

Result and limitations

In the fixed standard basis: u+v=(1,3), u−v=(5,−5), v−u=(−5,5)=−(u−v), so u−v≠v−u; ku=(−6,2), 0u=(0,0), 2u+v=(4,2); the unique vector with u+x=v is x=v−u=(−5,5); k(u+v)=ku+kv=(−2,−6). u≠w (although both have sum of squares 10), and u≠(−1,3) (components in swapped positions).

• N7 and N18 are grid checks of polynomial identities of degree at most one in each variable matching the algebra in S05.2 and S11.1; the argument does not depend on them.
• This guide does not involve length, angle, inner product or change of basis (MKO paragraph three says these carry separate obligations).

Verification and source binding

20 declared checks for this guide; only the listed expressions are covered.

• source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f
• adopted source SHA-256: bc749cb55fd9feab11176f4265e52ebf388cff1d90049efe15d0c6b29999d5c1
• original source SHA-256: bc749cb55fd9feab11176f4265e52ebf388cff1d90049efe15d0c6b29999d5c1
• Lean: NOT_RUN · independent AI: NOT_MEASURED

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