Question 1542: Forward Invariance versus Pointwise Fixing

Original question and givens

Use the complete finite self-map on Fin(6)={0,…,5} given by T(x)=(a*x+b) mod 6, a=17, b=13. Choose the least nonnegative residue, including for negative a*x+b. Time is discrete, indexed from 0; a finite orbit is not a general stability result. Candidate subset=[0, 1, 4], normalized as a set. Give (modulus, a, b, normalized subset, sorted distinct image, is the image contained in the subset?).

• G001 — Use the complete finite self-map on Fin(6)={0,…,5} given by T(x)=(a*x+b) mod 6, a=17, b=13. Choose the least nonnegative residue, including for negative a*x+b. Time is discrete, indexed from 0; a finite orbit is not a general stability result. Candidate subset=[0, 1, 4], normalized as a set. Give (modulus, a, b, normalized subset, sorted distinct image, is the image contained in the subset?).

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-discrete-dynamical-system

Step-by-step derivation

X={0 through 5}; T(x)=(17x+(13)) mod 6

remainder representative always lies in X

listed subset=[0, 1, 4]

set S=[0, 1, 4]

member0: a·x := mul(17, 0)

member0: a·x = 0

member0: a·x+b := sum(0, 13)

member0: a·x+b = 13

reduce 13 modulo 6

13=2·6+1

member1: a·x := mul(17, 1)

member1: a·x = 17

member1: a·x+b := sum(17, 13)

member1: a·x+b = 30

reduce 30 modulo 6

30=5·6+0

member4: a·x := mul(17, 4)

member4: a·x = 68

member4: a·x+b := sum(68, 13)

member4: a·x+b = 81

reduce 81 modulo 6

81=13·6+3

T(S)=[0, 1, 3]

T(S)⊆S: No; exiting witnesses=[[4, 3]]

x₀∈S and T(S)⊆S

induction keeps every later iterate in S

Reverse checks

• S013 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S012 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S011 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S010 · premise_and_result_audit — Recompute member4: a·x+b with exact rational arithmetic, not rounded decimals.
• S009 · premise_and_result_audit — Recompute member4: a·x with exact rational arithmetic, not rounded decimals.
• S008 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S007 · premise_and_result_audit — Recompute member1: a·x+b with exact rational arithmetic, not rounded decimals.
• S006 · premise_and_result_audit — Recompute member1: a·x with exact rational arithmetic, not rounded decimals.
• S005 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S004 · premise_and_result_audit — Recompute member0: a·x+b with exact rational arithmetic, not rounded decimals.
• S003 · premise_and_result_audit — Recompute member0: a·x with exact rational arithmetic, not rounded decimals.
• 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

(6, 17, 13, [0, 1, 4], [0, 1, 3], No)

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.