Question 11066: An Exact Point from Qualified Convex Weights

Original question and givens

Use a finite nonempty ordered vector list and aligned exact rational weights. Convex weights may be zero but must be nonnegative and sum exactly to 1, not merely approximately. A convex-combination witness establishes its weighted point, not unique coordinates without separate affine independence. vectors=[[3, 0], [5, 2]], weights=[16/18, 2/18]. Require qualified weights. Give (original vectors,normalized aligned weights,weighted point). Use exact rational arithmetic, reduce fractions to positive denominators and omit /1 for integer outputs.

• G001 — Use a finite nonempty ordered vector list and aligned exact rational weights. Convex weights may be zero but must be nonnegative and sum exactly to 1, not merely approximately. A convex-combination witness establishes its weighted point, not unique coordinates without separate affine independence. vectors=[[3, 0], [5, 2]], weights=[16/18, 2/18]. Require qualified weights. Give (original vectors,normalized aligned weights,weighted point). Use exact rational arithmetic, reduce fractions to positive denominators and omit /1 for integer outputs.

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.

• mko-finite-convex-combination

Step-by-step derivation

convex weights:raw rational pairs=[[16, 18], [2, 18]]

convex weights:normalized=[8/9, 1/9]

convex weights total := sum(8/9, 1/9)

convex weights total = 1

convex weights:weights=[8/9, 1/9];sum=1

nonnegative=Yes;total_one=Yes;qualified=Yes

convex point:ordered vectors=[[3, 0], [5, 2]];aligned weights=[8/9, 1/9]

dimension=2;preserve duplicates and order

convex point coordinate0 term0 := mul(8/9, 3)

convex point coordinate0 term0 = 8/3

convex point coordinate0 term1 := mul(1/9, 5)

convex point coordinate0 term1 = 5/9

convex point coordinate0 sum := sum(8/3, 5/9)

convex point coordinate0 sum = 29/9

convex point coordinate1 term0 := mul(8/9, 0)

convex point coordinate1 term0 = 0

convex point coordinate1 term1 := mul(1/9, 2)

convex point coordinate1 term1 = 2/9

convex point coordinate1 sum := sum(0, 2/9)

convex point coordinate1 sum = 2/9

Reverse checks

• S010 · premise_and_result_audit — Recompute convex point coordinate1 sum with exact rational arithmetic, not rounded decimals.
• S009 · premise_and_result_audit — Recompute convex point coordinate1 term1 with exact rational arithmetic, not rounded decimals.
• S008 · premise_and_result_audit — Recompute convex point coordinate1 term0 with exact rational arithmetic, not rounded decimals.
• S007 · premise_and_result_audit — Recompute convex point coordinate0 sum with exact rational arithmetic, not rounded decimals.
• S006 · premise_and_result_audit — Recompute convex point coordinate0 term1 with exact rational arithmetic, not rounded decimals.
• S005 · premise_and_result_audit — Recompute convex point coordinate0 term0 with exact rational arithmetic, not rounded decimals.
• S004 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• 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 — Recompute convex weights total with exact rational arithmetic, not rounded decimals.
• S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.

Result

([[3, 0], [5, 2]], [8/9, 1/9], [29/9, 2/9])

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.