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Question 16413: Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set

On this page
  1. Original question and givens
  2. Definition and scope
  3. Step-by-step derivation
  4. Reverse checks
  5. Result
  6. This question review and source binding

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R0031-16413 · q-20261014-advance-20261101-r01-vector-subspace-0088

Original question and givens

In R² with usual operations consider a*x+b*y=c with [a,b,c]=[2, 6, 0]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.

  • G001 — In R² with usual operations consider a*x+b*y=c with [a,b,c]=[2, 6, 0]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.

Definition and scope

Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.

Step-by-step derivation

S001 · implies

Before

same usual real R2;fixed normal=[2, 6];U consists of ALL x with(2)*x0+(6)*x1=0

After

zero vector:( 2)*0+(6)*0=0;zero belongs to U

Conditions: none

The equation defines the whole set,not a finite sample.First substitute the zero vector;all later vectors and scalars stay in the same usual real vector space.

Step identifiers and dependencies

Rule: subspace_homogeneous_zero_vector_universal_parent · Uses: G001

S002 · implies

Before

ANY u,v in this U;(2)*u0+(6)*u1=0 and(2)*v0+(6)*v1=0

After

(2)*(u0+v0)+(6)*(u1+v1)=((2)*u0+(6)*u1)+((2)*v0+(6)*v1)=0+0=0;u+v in U

Conditions: none

Distribute and regroup into the two equations already equal to zero.This proves closure for any two solutions,not just selected points.

Step identifiers and dependencies

Rule: subspace_addition_linearity_for_all_solutions · Uses: G001, S001

S003 · implies

Before

ANY real scalar r and ANY u in U;(2)*u0+(6)*u1=0

After

(2)*(r*u0)+(6)*(r*u1)=r*((2)*u0+(6)*u1)=r*0=0;r*u in U

Conditions: none

Commute the fixed coefficients with an arbitrary real r and factor it out.The result is zero,including zero,negative and irrational scalars.Finitely many integer tests are not a substitute for every-real-scalar closure.

Step identifiers and dependencies

Rule: subspace_every_real_scalar_not_finite_sample_scalars · Uses: G001, S002

Reverse checks

  • S003 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S002 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.

Result

([2, 6, 0], homogeneous_line, Yes, universal_homogeneous_equation_certificate)

The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.

This question review and source binding

Review date: 2026-10-26

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.

Source hashes and review receipts
  • input commit: 0f7be61e2c22317a4075561dc822c87d881a50db
  • original question SHA-256: eb346d252f6f832905133845244e558c5277336ad30431446186ce5a26cc88fe
  • candidate record SHA-256: ebe1362304fc6cfaf61420c11c26a08e64e359f1dbcddb8a70235807fac0df8d
  • semantic review receipt SHA-256: 55bb21783e2125ecf1e67e42f83fd25bec0ccd493f911f4bb937a5e9d6f9369a
  • question-first witness SHA-256: 754ddf2a6039c5ef4aae78a538c6e9a1e9a58d4825aee194e869a27b8a4c95bb
  • Lean: NOT_RUN · human review: NOT_PERFORMED
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