Mathematics: why does a proof establish more than examples?
Введено| English | Значение |
|---|---|
| conjecture/kənˈdʒektʃə/ | предположение |
| proof/pruːf/ | доказательство |
| induction/ɪnˈdʌkʃn/ | математическая индукция |
A pattern seems to continue
- The first odd number is 1; the first two sum to 4; the first three sum to 9. A student claims the first n positive odd numbers always sum to n².
- A conjecture 猜想 is a proposed statement awaiting justification. Checking three cases suggests a pattern but leaves infinitely many positive integers unchecked.
State the claim and its domain
- The claim concerns every positive integer n and the list 1, 3, 5, up to 2n−1. It does not concern any arbitrary collection of odd numbers.
- A proof 证明 gives a chain of valid reasons that covers the stated domain. One counterexample could defeat a universal claim, while many successful examples do not by themselves establish it.
What do the sums 1, 4 and 9 establish by themselves?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The finite observations do not cover the whole domain.
Explain how the next case follows
- Suppose the first n positive odd numbers sum to n². The next odd number is 2n+1, so the next sum is n²+2n+1=(n+1)². This step works for any positive integer n.
- The starting case is n=1, where the sum is 1=1². Starting there and applying the general step establishes each successive positive integer case.
What is the next odd number after 2n−1?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Successive odd numbers differ by 2.
Check what the reasoning assumes
- This is induction 数学归纳法: a valid starting case plus a step from any case to the next. Both parts matter. A step alone does not establish that the chain begins.
- Ask whether the algebra, definition of the next term and domain are correct. A calculation error or a changed starting sequence can invalidate the argument.
A valid induction step is sufficient even when no starting case is established.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The general transition needs an established case from which the chain begins.
Why is the sum 3+5+7=15 not a counterexample?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
This list differs from the one specified in the claim.
Compare proof with observation fairly
- The reason for confidence is the general chain, rather than the number of tested examples. A diagram of a few squares can explain the idea, but the argument must show why the construction extends.
- Ask how mathematical justification differs from checking repeated observations. Do not infer that proofs settle claims outside their assumptions or that empirical testing has no value in its own field.
Match each part of the proof to its role.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
A proof must connect its start, transition and domain.
Which reasons support the universal claim?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The general reasoning, together with a valid start, establishes the domain.
For n=1, the sum is 1. Assume the sum through 2n−1 is n². Adding 2n+1 gives n²+2n+1, which equals (n+1)². The base case and general transition establish the claim for all positive integers. The list 3,5,7 has sum 15, not 9, because it does not start at 1. That different list is outside the original claim, not a counterexample to it.
Do not use a successful calculator check for n=100 as a substitute for the general transition. It checks one more case.
State the domain, justify a starting case and establish a valid general step; explain exactly what the resulting proof covers.
Write from three sums
- The first odd number is 1. The first two sum to 4. The first three sum to 9.
- Say why these three checks do not prove the claim for every positive integer.
The first one, two and three positive odd numbers sum to 1, 4 and 9. Why does that not prove that the first n always sum to n squared?
A conjecture can fit the tested cases and still be unproved. Confidence comes from a general argument.
What do the three sums show?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
1, 4 and 9 fit the first three cases. A proof has to cover every positive integer, not only the cases already added.