Question 12154: A Strict Reciprocal Bound for the Selected Epsilon

Original question and givens

Keep the quantifier order: for every epsilon>0 there exists N such that for all m,n≥N, d(x_m,x_n)<epsilon. No limit or completeness premise is assumed. A finite recorded tail alone does not prove this infinite statement. The infinite model is x_n=1/(n+shift), with shift=4, epsilon=1/6. Give (shift,epsilon,least nonnegative N with 1/(N+shift)<epsilon,bound,m,n>=N implies distance<=1/(N+shift)<epsilon,cauchy_for_epsilon?). Use exact reduced rational outputs with positive denominators; omit /1 for integers.

• G001 — Keep the quantifier order: for every epsilon>0 there exists N such that for all m,n≥N, d(x_m,x_n)<epsilon. No limit or completeness premise is assumed. A finite recorded tail alone does not prove this infinite statement. The infinite model is x_n=1/(n+shift), with shift=4, epsilon=1/6. Give (shift,epsilon,least nonnegative N with 1/(N+shift)<epsilon,bound,m,n>=N implies distance<=1/(N+shift)<epsilon,cauchy_for_epsilon?). Use exact reduced rational outputs with positive denominators; omit /1 for integers.

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-cauchy-sequence

Step-by-step derivation

epsilon reciprocal threshold := div(6, 1)

epsilon reciprocal threshold = 6

positive epsilon=1/6;reciprocal threshold=6

floor threshold=6;6*1<=6<(6+1)*1

unclamped reciprocal cutoff := sum(6, -4, 1)

unclamped reciprocal cutoff = 3

reciprocal tail bound denominator := sum(3, 4)

reciprocal tail bound denominator = 7

reciprocal tail bound := div(1, 7)

reciprocal tail bound = 1/7

previous cutoff denominator := sum(2, 4)

previous cutoff denominator = 6

previous cutoff bound := div(1, 6)

previous cutoff bound = 1/6

N=3;bound=1/7;previous bound=1/6;epsilon=1/6

current bound is strict;previous cutoff fails;N is least nonnegative

explicit x_n=1/(n+4);N=3;bound=1/7<epsilon=1/6

all m,n>=N:both terms in[0,bound],so distance<=bound<epsilon

Reverse checks

• S009 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• 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 previous cutoff bound with exact rational arithmetic, not rounded decimals.
• S006 · premise_and_result_audit — Recompute previous cutoff denominator with exact rational arithmetic, not rounded decimals.
• S005 · premise_and_result_audit — Recompute reciprocal tail bound with exact rational arithmetic, not rounded decimals.
• S004 · premise_and_result_audit — Recompute reciprocal tail bound denominator with exact rational arithmetic, not rounded decimals.
• S003 · premise_and_result_audit — Recompute unclamped reciprocal cutoff 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 — Recompute epsilon reciprocal threshold with exact rational arithmetic, not rounded decimals.

Result

(4, 1/6, 3, 1/7, m,n>=N implies distance<=1/(N+shift)<epsilon, Yes)

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.