Deduction, contradiction and counterexamples · 演绎法、反证法和反例
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| counterexample/ˈkaʊntəreɡzæmpl/ | 反例 | fǎn lì |
When does a pattern become a proof?
- Checking several integers can suggest a pattern. What turns the pattern into a proof for every integer?
- This lesson studies counterexample 反例: A single valid case that disproves a universal claim.
何时模式成为证明？
- 检验多个整数可能暗示某种规律。是什么将这种规律转化为对所有整数的证明？
- 本课学习counterexample 反例：一个能推翻普遍断言的有效案例。
Choose the mathematical structure
- A deductive proof starts from stated definitions or assumptions. A counterexample refutes an all-values claim. For contradiction, assume the contrary and derive an impossibility. For exhaustion, check every case in a finite complete list.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
选择数学结构
- 演绎证明始于既定定义或假设；反例用于反驳全称命题；反证法需假设对立面并推导出矛盾；穷举法则需逐一检验有限完整列表中的每种情况。
- 计算前请先明确允许的输入项和单位。方程应表达关系本身，而不仅仅是记录计算器按键过程。
Which description correctly defines counterexample? · 以下哪项描述正确定义了反例？
A single valid case that disproves a universal claim. · 一个有效的反例，用以推翻全称命题。
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
Odd integers 2a+1 and 2b+1 sum to 2(a+b+1), hence even. To disprove that n²+n+41 is always prime, use n=41: the value is 41×43=1763. In a contradiction proof that √2 is irrational, assume √2=p/q in lowest terms; p²=2q² forces both p and q even, contradicting lowest terms.
通过验证案例进行推导
- 将结果与初始量进行比对。代入原始关系式，或在适当情况下对比图表与数值答案。
奇整数 2a+1 和 2b+1 之和为 2(a+b+1)，因此是偶数。要证伪 n²+n+41 恒为质数的结论，取 n=41：此时值为 41×43=1763。在 √2 是无理数的反证法中，假设 √2=p/q（最简分数形式）；则 p²=2q² 迫使 p 和 q 均为偶数，与最简分数形式矛盾。
Deduction, contradiction and counterexamples · 演绎法、反证法和反例
A deductive proof starts from stated definitions or assumptions · 演绎证明从给定的定义或假设出发
Distinguish a proof for all integers from one counterexample. · 区分针对所有整数的证明与单个反例。
Evaluate n²+n+41 at n=41. · 当 n=41 时，计算 n²+n+41 的值。
At n=41 the value is 41²+41+41=41×43=1763. · 当n=41时，其值为41²+41+41=41×43=1763。
Test a tempting shortcut
- Do not assume the conclusion while proving it. In contradiction, identify the impossible consequence and reject the initial contrary assumption. Induction is not compulsory in AQA 7357; it belongs in its own appropriate further-mathematics scope.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Checking the first ten integers proves a claim for every positive integer. This claim is false. Explain which definition or assumption it violates.
检验一个诱人的捷径
- 证明过程中不要预设结论。在反证法中，识别出必然导致的不可能结果并否定初始的相反假设。AQA 7357 不强制要求使用数学归纳法，它应归属于专门的进一步数学范畴。
- 当捷径失效时，找出其违背的假设。保留精确值直到题目要求的最终舍入步骤。
验证前十个整数只能证明该命题对每一个正整数成立。此命题是错误的。请解释它违反了哪一定义或假设。
For 2a+1 and 2b+1 with a=3,b=4, find their sum. · 对于2a+1和2b+1，其中a=3, b=4，求它们的和。
The two odd integers are 7 and 9, whose sum is 16. · 这两个奇整数是7和9，它们的和为16。
Checking the first ten integers proves a claim for every positive integer. · 检验前十个整数可证明适用于所有正整数的命题。
Do not assume the conclusion while proving it. In contradiction, identify the impossible consequence and reject the initial contrary assumption. Induction is not compulsory in AQA 7357; it belongs in its own appropriate further-mathematics scope. · 证明时不要预设结论。在反证法中，找出不可能出现的后果并否定初始的相反假设。归纳法并非AQA 7357所必需；它属于独立的进一步数学范畴。
Interpret a new situation
- AQA proof includes deduction, exhaustion and contradiction. IAL P2 introduces exhaustion and counterexample, while P4 introduces contradiction. Keep the method matched to the named unit.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
解读新情境
- AQA 的证明包括演绎、穷举法和反证法。IAL P2 引入穷举法和反例，而 P4 引入反证法。请确保所使用方法与指定单元名称相符。
- 完整解答需给出数学结果并解释其含义。检查其在所述背景下是否可行。
Find 1+2+...+20. · 求1+2+...+20。
Pair the endpoints: 20 terms have average 10.5, giving 20×10.5=210. · 配对端点：20项的平均值为10.5，得出20×10.5=210。
Match each part of a complete solution to its purpose. · 将完整解答的每个部分与其目的相匹配。
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · 假设用于论证模型的合理性；检查用于验证结果；解释用于将其与问题建立联系。
Use this in your course
- 7357 · A-level · A. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
A single valid case that disproves a universal claim. Choose the relationship, show the method, check its assumptions and interpret the result.
将其应用于你的课程
- 7357 · A-level · A. 在布置拓展任务前，先匹配目标层级及考试大纲。
- 在给出最终答案前先展示解题方法，并遵循试卷的计算器及公式使用规则。通过定位第一个无效步骤来复盘错误答案。
能推翻全称命题的一个有效特例。选择关系式，展示方法，检验其假设并解读结果。