Complete cases and counterexamples
導入済み| English | 意味 |
|---|---|
| proof by exhaustion/pruːf baɪ eɡˈzɔːstʃn/ | 帰納的証明 |
When do three cases cover every integer?
- Testing a thousand integers may miss the next counterexample. A complete list of possible remainders can instead settle every integer.
- This lesson studies proof by exhaustion 穷举证明: A proof that checks every possibility in a finite, complete set of cases.
Choose the mathematical structure
- Write n=3k, 3k+1 or 3k+2 for integer k. These are all possible remainders on division by 3, including negative integers. Evaluate the expression in each case. One counterexample disproves a universal claim; examples supporting a claim do not prove it.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Why does checking n=0,1,2 not prove a claim about every integer’s remainder on division by 3?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The three remainder classes with a general k cover every integer, positive or negative, in one argument each. Checking sample values verifies only those values.
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 n=3k, n²=9k² is divisible by 3. For n=3k+1, n²=3(3k²+2k)+1. For n=3k+2, n²=3(3k²+4k+1)+1. Thus an integer square has remainder 0 or 1, never 2. The claim that n²+n+41 is always prime fails at n=41: its value is 1763=41×43.
Complete cases and counterexamples
Write n=3k, 3k+1 or 3k+2 for integer k
Classify each argument and explain the condition that makes it valid or incomplete.
How many remainder cases are needed for division by 3?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
The possible remainders are 0,1,2, giving three cases.
Test a tempting shortcut
- A finite sample is not exhaustion unless it contains every possible case. Checking n=0,1,2 alone is not enough: the expressions involving arbitrary integer k justify all integers.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
If a claim works for the first hundred integers, it has been proved for all integers. This claim is false. Explain which definition or assumption it violates.
What remainder does 8² have on division by 3?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
64=3×21+1.
If a claim works for the first hundred integers, it has been proved for all integers.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
A finite sample is not exhaustion unless it contains every possible case. Checking n=0,1,2 alone is not enough: the expressions involving arbitrary integer k justify all integers.
Interpret a new situation
- Choose a complete case split, such as parity or remainders. State why no case is missing. For a false universal claim, give a valid input and verify the failure directly.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Evaluate n²+n+41 at n=41.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
41²+41+41=1763=41×43.
Match each proof outcome to what establishes it.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
A finite complete case list proves by checking each case; a finite INCOMPLETE sample is just evidence.
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 proof that checks every possibility in a finite, complete set of cases. Choose the relationship, show the method, check its assumptions and interpret the result.