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R0031-16488 · q-20261014-advance-20261101-r01-vector-subspace-0038
Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[13, -6] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).
G001 — Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[13, -6] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).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.
Before
same usual real R2;fixed normal=[13, -6];U consists of ALL x with(13)*x0+(-6)*x1=0
After
zero vector:( 13)*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.
Rule: subspace_homogeneous_zero_vector_universal_parent · Uses: G001
Before
ANY u,v in this U;(13)*u0+(-6)*u1=0 and(13)*v0+(-6)*v1=0
After
(13)*(u0+v0)+(-6)*(u1+v1)=((13)*u0+(-6)*u1)+((13)*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.
Rule: subspace_addition_linearity_for_all_solutions · Uses: G001, S001
Before
ANY real scalar r and ANY u in U;(13)*u0+(-6)*u1=0
After
(13)*(r*u0)+(-6)*(r*u1)=r*((13)*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.
Rule: subspace_every_real_scalar_not_finite_sample_scalars · Uses: G001, S002
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.([13, -6], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)
The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.
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.
input commit: 0f7be61e2c22317a4075561dc822c87d881a50dboriginal question SHA-256: fed409a670237cde4501c37ec835d2feabb001f23a396263c20b3aab108b7cf5candidate record SHA-256: e31feab559bc704b7de0425fb4156af8f53face3beae4eb17e3bee2543b3ef0esemantic review receipt SHA-256: 55bb21783e2125ecf1e67e42f83fd25bec0ccd493f911f4bb937a5e9d6f9369aquestion-first witness SHA-256: 754ddf2a6039c5ef4aae78a538c6e9a1e9a58d4825aee194e869a27b8a4c95bbLean: NOT_RUN · human review: NOT_PERFORMED