{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-compound-interest-01", "mko_id": "mko-compound-interest", "title": { "zh-Hant": "複利模型：本金 1000、每年 2%、三年的餘額與單利比較", "en": "Compound-interest model: principal 1000 at 2% per year for three years, compared with simple interest" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-compound-interest-01", "source": { "mko_id": "mko-compound-interest", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "5e3f823637101eaaf3ba73ed1b448b858ef9d83163c30769859bfcd7bbb22bb8", "english_sha256": "adc5d56b7b60ab65bc2136aba773f4e2ce33bd70235a7c811eaed0b7e1bcb065" }, "mode": "worked_example", "audience_prerequisites": [ "nat_induction", "eq_replace", "order_basics", "square_nonneg", "distributive" ], "goal": { "zh-Hant": "在期別為一年、每期利率 i=1/50（2%）、期間內無存提款與費用的模型中，以歸納證明 B_n=B_0(1+i)^n，算出 B_0=1000 時三年後的餘額 132651/125（即 1061.208）與利息；與單利 B_0(1+ni)=1060 比較，並以 Bernoulli 不等式證明複利餘額對每個 n 都不小於單利；說明期別混用與期間現金流如何使單一冪次公式失效。", "en": "In the model with one-year periods, rate i=1/50 (2%) per period and no deposits, withdrawals or fees, prove B_n=B_0(1+i)^n by induction, compute the balance after three years from B_0=1000, namely 132651/125 (that is 1061.208), and the interest; compare with simple interest B_0(1+ni)=1060 and prove by Bernoulli's inequality that the compound balance is never below the simple one for any n; explain how mixing period units and cash flows during the period break the single-power formula." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "期別為一年，利息在每年年末計入；每期利率 i 固定；模型期間內沒有存款、提款、費用、稅或四捨五入（MKO 假設 a1）。", "en": "The period is one year and interest is credited at the end of each year; the rate i per period is fixed; there are no deposits, withdrawals, fees, taxes or rounding during the modeled interval (MKO assumption a1)." } }, { "id": "G2", "statement": { "zh-Hant": "B_0=1000，i=1/50（每年 2%），期數 n=3；更新規則 B_(n+1)=(1+i)B_n（假設 a2：本金非負、1+i>0）。", "en": "B_0=1000, i=1/50 (2% per year), number of periods n=3; update rule B_(n+1)=(1+i)B_n (assumption a2: nonnegative principal and 1+i>0)." } }, { "id": "G3", "statement": { "zh-Hant": "比較用的單利模型：S_n=B_0(1+ni)。", "en": "Simple-interest model for comparison: S_n=B_0(1+ni)." } }, { "id": "G4", "statement": { "zh-Hant": "兩個對照情境：(a) 若 2% 其實是每季利率，三年是 12 季；(b) 第一年末計息後存入 100。", "en": "Two contrast scenarios: (a) if the 2% were a quarterly rate, three years would be 12 quarters; (b) a deposit of 100 is made at the end of year 1, after interest is credited." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "G2" ], "before_tex": "\\text{fixed period, rate, timing; no cash flows}", "after_tex": "B_{n+1}=(1+i)B_n,\\ B_0=1000,\\ i=\\tfrac{1}{50}", "relation": "definition", "rule": "mko_ci_def", "conditions": [ "1+i>0" ], "explanation": { "zh-Hant": "依 MKO 寫出模型：每期先把目前餘額乘以同一個累積因子 1+i。模型的適用前提是期別、計息時點與利率都已固定，且期間內沒有任何現金流、費用或四捨五入（G1）。", "en": "State the model as in the MKO: each period multiplies the current balance by the same accumulation factor 1+i. The model applies only once the period, interest timing and rate are fixed, with no cash flows, fees or rounding during the interval (G1)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "G2", "mko_ci_def" ], "explanation": { "zh-Hant": "此步只寫出模型，不推出新事實；任何違反 G1 的情況都不屬於此模型（見 S11、S12）。", "en": "This step only states the model and derives no new fact; any situation violating G1 is outside the model (see S11 and S12)." } }, "leaf_basis": "mko_ci_def + def_intro" }, { "id": "S02", "uses": [ "G2" ], "before_tex": "1+i=1+\\tfrac{1}{50}", "after_tex": "1+i=\\tfrac{51}{50}>0", "relation": "equivalent", "rule": "frac_rules", "conditions": [], "explanation": { "zh-Hant": "累積因子 1+1/50=51/50，是正數，符合 MKO 假設 a2 的 1+i>0；本金 1000 也非負。", "en": "The accumulation factor 1+1/50=51/50 is positive, meeting 1+i>0 from MKO assumption a2; the principal 1000 is nonnegative too." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "frac_rules", "numeric_evaluation" ], "explanation": { "zh-Hant": "精確有理數計算，雙向相同；正負由分子分母皆正得出。", "en": "Exact rational arithmetic, the same both ways; positivity follows from a positive numerator and denominator." } }, "leaf_basis": "frac_rules + numeric_evaluation + order_basics" }, { "id": "S03", "uses": [ "S01", "L_balance" ], "before_tex": "B_{n+1}=(1+i)B_n", "after_tex": "\\forall n\\in\\mathbb{N}:\\ B_n=B_0(1+i)^n", "relation": "implies", "rule": "L_balance", "conditions": [], "explanation": { "zh-Hant": "依引理 L_balance（在 S03.1 到 S03.3 以數學歸納法證明）：反覆乘同一個因子 n 次得到冪次公式。", "en": "By Lemma L_balance (proved by induction in S03.1 to S03.3): multiplying by the same factor n times gives the power formula." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_balance" ], "explanation": { "zh-Hant": "依賴該引理；引理用到利率不隨期改變。", "en": "It depends on the lemma, which uses that the rate does not change from period to period." } }, "leaf_basis": "nat_induction + nat_power_def + mul_assoc + mul_comm" }, { "id": "S03.1", "uses": [ "nat_power_def" ], "before_tex": "n=0", "after_tex": "B_0=B_0\\cdot 1=B_0(1+i)^0", "relation": "induction_base", "rule": "nat_induction", "conditions": [], "explanation": { "zh-Hant": "基底：(1+i)^0=1，所以 B_0(1+i)^0=B_0。", "en": "Base: (1+i)^0=1, so B_0(1+i)^0=B_0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "nat_power_def", "mul_one" ], "explanation": { "zh-Hant": "只用到零次冪與乘法單位元。", "en": "It uses only the zeroth power and the multiplicative identity." } }, "leaf_basis": "nat_power_def + mul_one" }, { "id": "S03.2", "uses": [ "S01", "nat_power_def" ], "before_tex": "B_n=B_0(1+i)^n", "after_tex": "B_{n+1}=(1+i)B_0(1+i)^n=B_0(1+i)^{n+1}", "relation": "induction_step", "rule": "eq_replace", "conditions": [], "explanation": { "zh-Hant": "歸納步：假設 B_n=B_0(1+i)^n。由更新規則 B_(n+1)=(1+i)B_n，代入歸納假設並以乘法交換結合律重排為 B_0((1+i)^n(1+i))，依次冪定義即 B_0(1+i)^(n+1)。", "en": "Inductive step: assume B_n=B_0(1+i)^n. By the update rule B_(n+1)=(1+i)B_n; substitute the hypothesis and rearrange by commutativity and associativity to B_0((1+i)^n(1+i)), which by the definition of powers is B_0(1+i)^(n+1)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G2", "eq_replace", "mul_assoc", "mul_comm", "nat_power_def" ], "explanation": { "zh-Hant": "需要每期使用同一個 i；若某期有存款或利率改變，第一個等號就不成立。", "en": "It needs the same i in every period; with a deposit or a rate change in some period, the first equality fails." } }, "leaf_basis": "eq_replace + mul_assoc + mul_comm + nat_power_def" }, { "id": "S03.3", "uses": [ "S03.1", "S03.2" ], "before_tex": "P(0),\\ \\forall n\\,(P(n)\\Rightarrow P(n+1))", "after_tex": "\\forall n:\\ B_n=B_0(1+i)^n", "relation": "implies", "rule": "nat_induction", "conditions": [], "explanation": { "zh-Hant": "依數學歸納原理對所有 n 成立。S03.1 到 S03.3 合起來證得引理 L_balance。", "en": "By the principle of induction it holds for all n. S03.1 to S03.3 together prove Lemma L_balance." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03.1", "S03.2", "nat_induction" ], "explanation": { "zh-Hant": "兩個前提由 S03.1、S03.2 提供。", "en": "The two premises come from S03.1 and S03.2." } }, "leaf_basis": "nat_induction" }, { "id": "S04", "uses": [ "S01", "S02" ], "before_tex": "B_0=1000", "after_tex": "B_1=1020,\\ B_2=\\tfrac{5202}{5},\\ B_3=\\tfrac{132651}{125}", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "依更新規則逐期計算：1000·51/50=1020；1020·51/50=5202/5（即 1040.4）；(5202/5)·51/50=132651/125（即 1061.208）。全程精確，沒有四捨五入。", "en": "Compute period by period from the update rule: 1000·51/50=1020; 1020·51/50=5202/5 (that is 1040.4); (5202/5)·51/50=132651/125 (that is 1061.208). Everything is exact, with no rounding." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S01", "numeric_evaluation" ], "explanation": { "zh-Hant": "每期只由前一期決定；精確有理數計算可獨立重算。", "en": "Each period depends only on the previous one; the exact rational arithmetic can be recomputed independently." } }, "leaf_basis": "numeric_evaluation + frac_rules" }, { "id": "S05", "uses": [ "S03", "S04" ], "before_tex": "B_0(1+i)^3=1000\\left(\\tfrac{51}{50}\\right)^3", "after_tex": "1000\\cdot\\tfrac{132651}{125000}=\\tfrac{132651}{125}=B_3", "relation": "equivalent", "rule": "power_laws", "conditions": [], "explanation": { "zh-Hant": "依閉式：51³=132651，50³=125000，1000·132651/125000=132651/125，與 S04 的逐期計算一致。", "en": "By the closed form: 51³=132651, 50³=125000, and 1000·132651/125000=132651/125, agreeing with the period-by-period computation of S04." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "power_laws", "frac_rules", "numeric_evaluation" ], "explanation": { "zh-Hant": "精確計算，雙向是同一個數。", "en": "Exact computation; both sides are the same number." } }, "leaf_basis": "power_laws + frac_rules + numeric_evaluation" }, { "id": "S06", "uses": [ "S05", "mko_ci_notes" ], "before_tex": "B_3-B_0", "after_tex": "\\tfrac{132651}{125}-1000=\\tfrac{7651}{125}", "relation": "equivalent", "rule": "mko_ci_notes", "conditions": [], "explanation": { "zh-Hant": "依 MKO，利息金額是期末餘額減本金：132651/125−125000/125=7651/125，即 61.208。", "en": "By the MKO, the interest amount is the final balance minus the principal: 132651/125−125000/125=7651/125, that is 61.208." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "frac_rules", "numeric_evaluation" ], "explanation": { "zh-Hant": "加回 1000 即得 B_3，雙向成立。", "en": "Adding back 1000 gives B_3, so both directions hold." } }, "leaf_basis": "frac_rules + numeric_evaluation" }, { "id": "S07", "uses": [ "G3", "G2" ], "before_tex": "S_3=B_0(1+3i)", "after_tex": "S_3=1000\\left(1+\\tfrac{3}{50}\\right)=1060", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "單利只對本金計息：1000·(53/50)=1060，利息 60。單利是另一個模型，不是複利模型的精確等式改寫。固定 n 時，(1+i)^n 的一次項為 1+ni，因此單利式可作小 i 的一階近似；本題仍比較兩個模型的精確數值，不把它們視為相等。", "en": "Simple interest pays only on the principal: 1000·(53/50)=1060, with interest 60. Simple interest is a different model, not an exact rewriting of compound interest. For fixed n the linear part of (1+i)^n is 1+ni, so the simple-interest expression can serve as a first-order approximation for small i. This exercise still compares the exact values of the two models rather than identifying them." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "numeric_evaluation", "frac_rules" ], "explanation": { "zh-Hant": "精確計算，雙向相同。", "en": "Exact computation, the same both ways." } }, "leaf_basis": "numeric_evaluation + frac_rules" }, { "id": "S08", "uses": [ "S05", "S07" ], "before_tex": "B_3-S_3=B_0\\left[(1+i)^3-(1+3i)\\right]", "after_tex": "B_0(3i^2+i^3)=\\tfrac{151}{125}", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "展開 (1+i)³=1+3i+3i²+i³（N10 為多項式恆等式），減去 1+3i 得 3i²+i³，這就是「利息再生利息」的部分。代入：1000·(3/2500+1/125000)=6/5+1/125=151/125（即 1.208），與 132651/125−1060 一致。", "en": "Expand (1+i)³=1+3i+3i²+i³ (N10 checks the polynomial identity) and subtract 1+3i to get 3i²+i³, the 'interest on interest' part. Substituting: 1000·(3/2500+1/125000)=6/5+1/125=151/125 (that is 1.208), agreeing with 132651/125−1060." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "numeric_evaluation" ], "explanation": { "zh-Hant": "展開是恆等式，代值是精確計算，雙向成立。", "en": "The expansion is an identity and the substitution is exact, so both directions hold." } }, "leaf_basis": "distributive + power_laws + numeric_evaluation" }, { "id": "S09", "uses": [ "L_bernoulli" ], "before_tex": "i\\ge -1", "after_tex": "\\forall n\\in\\mathbb{N}:\\ (1+i)^n\\ge 1+ni", "relation": "implies", "rule": "L_bernoulli", "conditions": [ "i\\ge -1" ], "explanation": { "zh-Hant": "依引理 L_bernoulli（在 S09.1 到 S09.3 以歸納證明）。這把 S08 的單一比較推廣到所有期數。", "en": "By Lemma L_bernoulli (proved by induction in S09.1 to S09.3). It extends the single comparison of S08 to every number of periods." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_bernoulli" ], "explanation": { "zh-Hant": "依賴該引理；本例 i=1/50≥−1。", "en": "It depends on the lemma; here i=1/50≥−1." } }, "leaf_basis": "nat_induction + order_basics + square_nonneg" }, { "id": "S09.1", "uses": [ "nat_power_def" ], "before_tex": "n=0", "after_tex": "(1+i)^0=1\\ge 1=1+0\\cdot i", "relation": "induction_base", "rule": "nat_induction", "conditions": [], "explanation": { "zh-Hant": "基底：兩邊都等於 1，不等式成立（等號）。", "en": "Base: both sides equal 1, so the inequality holds (with equality)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "nat_power_def", "mul_zero" ], "explanation": { "zh-Hant": "只用到零次冪與 0·i=0。", "en": "It uses only the zeroth power and 0·i=0." } }, "leaf_basis": "nat_power_def + mul_zero" }, { "id": "S09.2", "uses": [ "nat_power_def", "order_basics", "square_nonneg" ], "before_tex": "(1+i)^n\\ge 1+ni", "after_tex": "(1+i)^{n+1}\\ge(1+ni)(1+i)=1+(n+1)i+ni^2\\ge 1+(n+1)i", "relation": "induction_step", "rule": "order_basics", "conditions": [ "1+i\\ge 0", "n\\ge 0" ], "explanation": { "zh-Hant": "歸納步：假設 (1+i)^n≥1+ni。兩邊同乘非負數 1+i（因 i≥−1）保持不等號方向：(1+i)^(n+1)=(1+i)^n(1+i)≥(1+ni)(1+i)。展開得 1+(n+1)i+ni²（N11），而 ni²≥0（n≥0 且平方非負），所以 ≥1+(n+1)i。", "en": "Inductive step: assume (1+i)^n≥1+ni. Multiplying both sides by the nonnegative number 1+i (since i≥−1) keeps the direction: (1+i)^(n+1)=(1+i)^n(1+i)≥(1+ni)(1+i). Expanding gives 1+(n+1)i+ni² (N11), and ni²≥0 (n≥0 and squares are nonnegative), so it is ≥1+(n+1)i." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "order_basics", "square_nonneg", "distributive", "nat_power_def" ], "explanation": { "zh-Hant": "關鍵條件是 1+i≥0：乘以負數會反轉不等號。這正是引理要求 i≥−1 的原因。", "en": "The key condition is 1+i≥0: multiplying by a negative number would reverse the inequality. That is exactly why the lemma requires i≥−1." } }, "leaf_basis": "order_basics + square_nonneg + distributive" }, { "id": "S09.3", "uses": [ "S09.1", "S09.2" ], "before_tex": "P(0),\\ \\forall n\\,(P(n)\\Rightarrow P(n+1))", "after_tex": "\\forall n:\\ (1+i)^n\\ge 1+ni", "relation": "implies", "rule": "nat_induction", "conditions": [ "i\\ge -1" ], "explanation": { "zh-Hant": "依數學歸納原理對所有 n 成立。S09.1 到 S09.3 合起來證得引理 L_bernoulli。", "en": "By the principle of induction it holds for all n. S09.1 to S09.3 together prove Lemma L_bernoulli." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S09.1", "S09.2", "nat_induction" ], "explanation": { "zh-Hant": "兩個前提由 S09.1、S09.2 提供。", "en": "The two premises come from S09.1 and S09.2." } }, "leaf_basis": "nat_induction" }, { "id": "S10", "uses": [ "S03", "S09", "G2", "G3" ], "before_tex": "(1+i)^n\\ge 1+ni,\\ B_0\\ge 0", "after_tex": "\\forall n:\\ B_n\\ge S_n;\\ B_1=S_1=1020,\\ B_3-S_3=\\tfrac{151}{125}>0", "relation": "implies", "rule": "order_basics", "conditions": [ "B_0\\ge 0" ], "explanation": { "zh-Hant": "兩邊同乘非負數 B_0=1000：B_n=B_0(1+i)^n≥B_0(1+ni)=S_n 對每個 n 成立。n=0 與 n=1 時兩者相等（B_1=S_1=1020），n=3 時差 151/125>0。所以「複利不少於單利」對所有 n 成立，但不是每期都嚴格較多。", "en": "Multiply both sides by the nonnegative B_0=1000: B_n=B_0(1+i)^n≥B_0(1+ni)=S_n for every n. For n=0 and n=1 the two are equal (B_1=S_1=1020), and for n=3 the difference is 151/125>0. So 'compound is at least simple' holds for all n, but not strictly in every period." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S09", "order_basics" ], "explanation": { "zh-Hant": "需要 B_0≥0；若本金為負，不等號方向會反轉，但那不在 MKO 的範圍內。", "en": "It needs B_0≥0; with a negative principal the inequality would reverse, but that is outside the MKO's scope." } }, "leaf_basis": "order_basics + L_bernoulli" }, { "id": "S11", "uses": [ "G4", "S03", "S09", "mko_ci_notes" ], "before_tex": "2\\%\\ \\text{per quarter},\\ 3\\ \\text{years}=12\\ \\text{quarters}", "after_tex": "1000\\left(\\tfrac{51}{50}\\right)^{12}\\ge 1000\\left(1+\\tfrac{12}{50}\\right)=1240>\\tfrac{132651}{125}", "relation": "implies", "rule": "mko_ci_notes", "conditions": [ "\\text{period}=\\text{quarter}" ], "explanation": { "zh-Hant": "依 MKO，期別與期數必須一致。若 2% 是每季利率，三年有 12 期，餘額是 1000·(51/50)^12；依 Bernoulli 不等式它至少是 1240，遠大於年利率解讀下的 1061.208。同樣的「2%、三年」因期別不同而給出不同答案，所以必須先固定期別。", "en": "By the MKO, the period and the number of periods must agree. If 2% is a quarterly rate, three years are 12 periods and the balance is 1000·(51/50)^12; by Bernoulli's inequality it is at least 1240, far above 1061.208 under the annual reading. The same '2%, three years' gives different answers for different periods, so the period must be fixed first." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S09", "mko_ci_notes" ], "explanation": { "zh-Hant": "只需要冪次公式與 Bernoulli 下界；此步只說明期別錯置的後果，不換算名目與有效利率。", "en": "It needs only the power formula and the Bernoulli lower bound; it shows the effect of a mismatched period and does not convert between nominal and effective rates." } }, "leaf_basis": "mko_ci_notes + L_bernoulli + numeric_evaluation" }, { "id": "S12", "uses": [ "G4", "S04", "mko_ci_notes" ], "before_tex": "\\text{deposit }100\\ \\text{at end of year 1}", "after_tex": "(1020+100)\\left(\\tfrac{51}{50}\\right)^2=\\tfrac{145656}{125}\\neq\\tfrac{1459161}{1250}=1100\\left(\\tfrac{51}{50}\\right)^3", "relation": "implies", "rule": "mko_ci_notes", "conditions": [], "explanation": { "zh-Hant": "第一年末計息後存入 100：餘額變成 1120，再以 51/50 複利兩年，得 145656/125（即 1165.248）。若誤把本金當作 1100 套用單一冪次公式，得 1459161/1250（即 1167.3288），兩者不同。有期間現金流時已不是 MKO 的單一冪次模型，需要逐筆追蹤現金流。", "en": "Depositing 100 at the end of year 1 after interest makes the balance 1120, which then compounds at 51/50 for two years to 145656/125 (that is 1165.248). Wrongly treating the principal as 1100 in the single-power formula gives 1459161/1250 (that is 1167.3288); the two differ. With cash flows during the interval the MKO's single-power model no longer applies, and each cash flow must be tracked." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "mko_ci_notes", "numeric_evaluation" ], "explanation": { "zh-Hant": "需要第一年末的餘額 1020 與存款時點；結論只說明單一冪次公式不適用，不提出現金流模型。", "en": "It needs the year-1 balance 1020 and the timing of the deposit; the conclusion only shows that the single-power formula does not apply and does not propose a cash-flow model." } }, "leaf_basis": "mko_ci_notes + numeric_evaluation + power_laws" }, { "id": "S13", "uses": [ "S03", "S05", "S06", "S07", "S10", "S11", "S12" ], "before_tex": "\\text{proved facts}", "after_tex": "B_3=\\tfrac{132651}{125},\\ \\text{interest }\\tfrac{7651}{125},\\ B_n\\ge S_n\\ \\forall n", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "合併結論：B_n=B_0(1+i)^n；三年後餘額 132651/125，利息 7651/125；單利為 1060；對所有 n 複利不少於單利；期別混用與期間存款都使單一冪次公式不適用。這些是模型計算，不是投資建議或市場預測。", "en": "Combine the conclusions: B_n=B_0(1+i)^n; after three years the balance is 132651/125 and the interest 7651/125; simple interest gives 1060; compound is never below simple for any n; mixing periods or depositing during the interval makes the single-power formula inapplicable. These are model calculations, not investment advice or market predictions." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S03", "S05", "S06", "S07", "S10", "S11", "S12", "and_intro_elim" ], "explanation": { "zh-Hant": "目標 G：本稿目標的各項判定。足夠條件 H：所列各步已證（H⇒G 由合取引入）。H 由給定條件正向推出，沒有把 G 當作前提。", "en": "Goal G: the determinations in the goal. Sufficient condition H: the listed steps, each already proved (H⇒G by conjunction introduction). H is derived forwards from the givens; G is never used as a premise." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "在每年 2%、年末計息、無現金流的模型中，歸納證明 B_n=1000·(51/50)^n；三年後 B_3=132651/125=1061.208，利息 7651/125=61.208。單利 1000(1+3/50)=1060，差額 1000(3i²+i³)=151/125。Bernoulli 不等式證明對所有 n 有 B_n≥S_n，n=0、1 時相等。若 2% 是季利率則三年為 12 期，餘額至少 1240；若第一年末存入 100，餘額為 145656/125，不能以單一冪次公式計算。以上不是投資建議。", "en": "In the model with 2% per year credited at year end and no cash flows, induction proves B_n=1000·(51/50)^n; after three years B_3=132651/125=1061.208 and the interest is 7651/125=61.208. Simple interest gives 1000(1+3/50)=1060, a difference of 1000(3i²+i³)=151/125. Bernoulli's inequality proves B_n≥S_n for all n, with equality at n=0 and 1. If the 2% were quarterly, three years would be 12 periods and the balance at least 1240; with a deposit of 100 at the end of year 1 the balance is 145656/125, which is not given by the original no-cash-flow single-power model. None of this is investment advice." }, "negative_controls": [ { "claim": { "zh-Hant": "本金 1000、年複利 2%、三年後的餘額是 1000(1+3·0.02)=1060", "en": "With principal 1000 at 2% compounded yearly, the balance after three years is 1000(1+3·0.02)=1060" }, "expected": "reject", "reason": { "zh-Hant": "那是單利；複利為 1000(51/50)³=132651/125=1061.208（S05）。", "en": "That is simple interest; compound gives 1000(51/50)³=132651/125=1061.208 (S05)." } }, { "claim": { "zh-Hant": "若 2% 是季利率，三年後餘額仍是 1000(1.02)³", "en": "If 2% is a quarterly rate, the balance after three years is still 1000(1.02)³" }, "expected": "reject", "reason": { "zh-Hant": "三年是 12 季，應為 1000(1.02)^12，且至少 1240（S11）。", "en": "Three years are 12 quarters, so it is 1000(1.02)^12, at least 1240 (S11)." } }, { "claim": { "zh-Hant": "第一年末存入 100 後，三年後餘額是 1100·(1.02)³", "en": "After a deposit of 100 at the end of year 1, the balance after three years is 1100·(1.02)³" }, "expected": "reject", "reason": { "zh-Hant": "存款只複利兩年：(1020+100)(1.02)²=145656/125，與 1459161/1250 不同（S12）。", "en": "The deposit compounds for two years only: (1020+100)(1.02)²=145656/125, unlike 1459161/1250 (S12)." } }, { "claim": { "zh-Hant": "對每個 n，複利餘額都嚴格大於單利餘額", "en": "For every n the compound balance is strictly greater than the simple balance" }, "expected": "reject", "reason": { "zh-Hant": "n=0 與 n=1 時兩者相等，例如 B_1=S_1=1020（S10）。", "en": "At n=0 and n=1 they are equal, for example B_1=S_1=1020 (S10)." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "nat_induction", "kind": "prerequisite", "statement": { "zh-Hant": "數學歸納原理：對從 n₀ 起的命題族 P(n)，若 P(n₀) 成立，且對每個 n≥n₀ 都有 P(n)⇒P(n+1)，則所有 n≥n₀ 都有 P(n)。", "en": "Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀." } }, { "id": "eq_replace", "kind": "prerequisite", "statement": { "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。", "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true." } }, { "id": "order_basics", "kind": "prerequisite", "statement": { "zh-Hant": "實數的序：任兩數恰有 uv 之一；u0 推出 uwvw。", "en": "Order of the reals: exactly one of uv holds; u0 give uwvw." } }, { "id": "square_nonneg", "kind": "prerequisite", "statement": { "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。", "en": "Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=0." } }, { "id": "distributive", "kind": "prerequisite", "statement": { "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。", "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu." } }, { "id": "def_intro", "kind": "prerequisite", "statement": { "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。", "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable." } }, { "id": "frac_rules", "kind": "prerequisite", "statement": { "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。", "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0." } }, { "id": "numeric_evaluation", "kind": "prerequisite", "statement": { "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。", "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable." } }, { "id": "mul_assoc", "kind": "prerequisite", "statement": { "zh-Hant": "乘法結合律：(uv)w=u(vw)。", "en": "Associativity of multiplication: (uv)w=u(vw)." } }, { "id": "mul_comm", "kind": "prerequisite", "statement": { "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。", "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)." } }, { "id": "mul_one", "kind": "prerequisite", "statement": { "zh-Hant": "乘法單位元：1·u=u。", "en": "Multiplicative identity: 1·u=u." } }, { "id": "power_laws", "kind": "prerequisite", "statement": { "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。", "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1." } }, { "id": "mul_zero", "kind": "prerequisite", "statement": { "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。", "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0." } }, { "id": "and_intro_elim", "kind": "prerequisite", "statement": { "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。", "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q." } }, { "id": "mko_ci_def", "kind": "cited", "source": { "mko_id": "mko-compound-interest", "sha256": "5e3f823637101eaaf3ba73ed1b448b858ef9d83163c30769859bfcd7bbb22bb8", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "在期別、計息時點與每期利率 i 已固定，且期間內無存提款或費用時，複利餘額以 B_(n+1)=(1+i)B_n 更新。若累積因子 1+i>0，n 期後為 B_0(1+i)^n。", "en": "Fix periods, interest timing and rate i per period, with no deposits, withdrawals or fees during the modeled interval. Compound balance updates are B_(n+1)=(1+i)B_n. For 1+i>0, the balance after n periods is B_0(1+i)^n." } }, { "id": "mko_ci_notes", "kind": "cited", "source": { "mko_id": "mko-compound-interest", "sha256": "5e3f823637101eaaf3ba73ed1b448b858ef9d83163c30769859bfcd7bbb22bb8", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "利息金額是期末餘額減本金；若有定期存款、提款、費用或變動利率，就不是本條目的單一冪次模型。年利率、月利率與名目／有效利率不能混用；必須先固定期別與轉換慣例。", "en": "The interest amount is the final balance minus the principal; with regular deposits, withdrawals, fees or variable rates the single-power model of this entry no longer applies. Annual, monthly, nominal and effective rates must not be mixed; the period and conversion convention must be fixed first." } }, { "id": "nat_power_def", "kind": "prerequisite", "statement": { "zh-Hant": "自然數次冪的遞迴定義：a^0=1，a^(n+1)=a^n·a。", "en": "Recursive definition of natural-number powers: a^0=1 and a^(n+1)=a^n·a." } }, { "id": "L_balance", "kind": "lemma", "proved_in": [ "S03.1", "S03.2", "S03.3" ], "statement": { "zh-Hant": "在 G1、G2 的模型中，對每個自然數 n，B_n=B_0(1+i)^n。", "en": "In the model of G1 and G2, B_n=B_0(1+i)^n for every natural number n." } }, { "id": "L_bernoulli", "kind": "lemma", "proved_in": [ "S09.1", "S09.2", "S09.3" ], "statement": { "zh-Hant": "Bernoulli 不等式：若 i≥−1，則對每個自然數 n，(1+i)^n≥1+ni。", "en": "Bernoulli's inequality: if i≥−1 then (1+i)^n≥1+ni for every natural number n." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S02", "lhs": "1+1/50", "rhs": "51/50" }, { "id": "N2", "step": "S04", "lhs": "1000*51/50", "rhs": "1020" }, { "id": "N3", "step": "S04", "lhs": "1020*51/50", "rhs": "5202/5" }, { "id": "N4", "step": "S04", "lhs": "(5202/5)*(51/50)", "rhs": "132651/125" }, { "id": "N5", "step": "S05", "lhs": "1000*(51/50)**3", "rhs": "132651/125" }, { "id": "N6", "step": "S06", "lhs": "132651/125-1000", "rhs": "7651/125" }, { "id": "N7", "step": "S07", "lhs": "1000*(1+3/50)", "rhs": "1060" }, { "id": "N8", "step": "S08", "lhs": "132651/125-1060", "rhs": "151/125" }, { "id": "N9", "step": "S08", "lhs": "1000*(3*(1/50)**2+(1/50)**3)", "rhs": "151/125" }, { "id": "N10", "step": "S08", "vars": [ "i" ], "degree": 3, "lhs": "(1+i)**3", "rhs": "1+3*i+3*i**2+i**3" }, { "id": "N11", "step": "S09.2", "vars": [ "n", "i" ], "degree": 2, "lhs": "(1+n*i)*(1+i)", "rhs": "1+(n+1)*i+n*i**2" }, { "id": "N12", "step": "S10", "vars": [ "n" ], "range": [ 0, 30 ], "lhs": "(51/50)**n", "relation": ">=", "rhs": "1+n/50" }, { "id": "N13", "step": "S11", "lhs": "1000*(1+12/50)", "rhs": "1240" }, { "id": "N14", "step": "S11", "lhs": "1000*(51/50)**12", "relation": ">=", "rhs": "1240" }, { "id": "N15", "step": "S12", "lhs": "(1020+100)*(51/50)**2", "rhs": "145656/125" }, { "id": "N16", "step": "S12", "lhs": "1100*(51/50)**3", "rhs": "1459161/1250" } ] }, "presentation": { "md_scope": { "zh-Hant": "只在 MKO 的模型假設下計算：期別、利率與期數單位一致，利率固定，無現金流、稅費或四捨五入，本金為非負精確有理數且 1+i>0（假設 a1、a2）。這些計算不是投資建議、報價或市場預測。不處理現值、名目與有效利率的換算或年金；不改動 canonical MKO 或英文 edition。", "en": "Computes only under the MKO's assumptions: period, rate and number of periods use the same unit, the rate is fixed, there are no cash flows, taxes, fees or rounding, the principal is a nonnegative exact rational and 1+i>0 (assumptions a1 and a2). These calculations are not investment advice, quotes or market predictions. Present value, conversion between nominal and effective rates, and annuities are not treated; the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "B_n", "第 n 期期末餘額", "balance at the end of period n" ], [ "B_0", "本金", "principal" ], [ "i", "每期利率", "rate per period" ], [ "1+i", "每期累積因子", "accumulation factor per period" ], [ "S_n=B_0(1+ni)", "單利模型的餘額", "balance under simple interest" ] ], "md_error_controls": [ { "title": "單利與複利不能混同 / Simple and compound interest must not be confused", "zh": "MKO 指出的誤解之一：B_0(1+ni) 是單利，B_0(1+i)^n 是複利。三年後兩者差 151/125，來自 3i²+i³ 這部分「利息的利息」；依 Bernoulli 不等式，複利對每個 n 都不少於單利，n=0、1 時相等。", "en": "One misconception named by the MKO: B_0(1+ni) is simple interest and B_0(1+i)^n is compound interest. After three years they differ by 151/125, coming from the 'interest on interest' part 3i²+i³; by Bernoulli's inequality compound is never below simple for any n, with equality at n=0 and 1." }, { "title": "期別與利率單位要一致 / The period and the rate unit must agree", "zh": "MKO 指出的另一個誤解是混用不同期別的利率。同樣的「2%、三年」，年利率給出 1061.208，季利率給出至少 1240（S11）。計算前必須確定利率是每什麼期的利率，以及總共有幾期。", "en": "Another misconception named by the MKO is mixing rates of different periods. The same '2%, three years' gives 1061.208 with an annual rate and at least 1240 with a quarterly rate (S11). Before computing, fix what period the rate refers to and how many periods there are." }, { "title": "有現金流就不是單一冪次模型 / With cash flows it is not the single-power model", "zh": "S12 中第一年末存入 100 後，正確餘額是 145656/125，而不是把本金當 1100 的 1459161/1250。存款、提款、費用、稅、變動利率或四捨五入都會破壞 S03.2 的歸納步，因為更新不再是單純乘以同一因子。", "en": "In S12, after a deposit of 100 at the end of year 1, the correct balance is 145656/125, not 1459161/1250 from treating the principal as 1100. Deposits, withdrawals, fees, taxes, variable rates or rounding all break the inductive step of S03.2, because an update is no longer a plain multiplication by the same factor." } ], "md_open_items": [ { "zh-Hant": "N12 是 n=0 到 30 的有限核對，只涵蓋這些期數；對所有 n 的結論由 S09 的歸納證明。", "en": "N12 is a finite check for n=0 to 30 only; the statement for all n is proved by the induction of S09." }, { "zh-Hant": "現值折現、名目與有效利率換算、年金與現金流時點屬於 MKO 的 formal next。", "en": "Present-value discounting, nominal and effective rate conversion, annuities and cash-flow timing belong to the MKO's formal next step." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "40a7814824c3f040be3cac68187574b75fa4bc4de6b05d465c54acd3cd9c8741", "original_source_sha256": "1484cf0ee66f4257f1254bcbb2872828ba4fb74c4c8367c12c4072a75908034b", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [ "BR-M044" ], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }