Deduction, contradiction and induction · 演绎、反证法和归纳法
| 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.
Choose the mathematical structure
- A deductive proof starts from stated definitions or assumptions. A counterexample refutes an all-values claim. Induction requires a base case and an implication from n=k to n=k+1; checking successive cases alone is not induction.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
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.
For odd integers 2a+1 and 2b+1, their sum is 2(a+b+1), hence even. The claim n²+n+41 is always prime fails at n=41: the value is 41×43=1763. For 1+...+n=n(n+1)/2, the induction step adds k+1 to the assumed sum.
Deduction, contradiction and induction · 演绎、反证法和归纳法
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.
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.
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
- edexcel IAL mathematics; official unit P2. Other-unit enrichment is identified in the scope review; it is not an extra cash-in requirement.
- 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.