Irrational roots and infinitely many primes
導入済み| English | 意味 |
|---|---|
| proof by contradiction/pruːf baɪ ˌkɒntrəˈdɪkʃn/ | 背理法による証明 |
Why cannot √2 be an exact fraction?
- A calculator prints finitely many digits of √2. Those digits cannot decide whether an exact ratio of integers exists.
- This lesson studies proof by contradiction 反证法: A proof that assumes the negation of a claim and derives an impossibility.
Choose the mathematical structure
- To prove √2 irrational, assume √2=p/q for coprime integers p,q with q nonzero. Then p²=2q² forces p even; writing p=2r forces q even too. This contradicts coprimality. To prove infinitely many primes, assume a complete finite prime list and construct a number with no listed prime divisor.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
In the √2 proof, why does p²=2q² force p to be even?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
p²=2q² makes p² even; since odd integers square to odd results, p itself must be even. Writing p=2r then forces q²=2r², so q is even too - contradicting coprimality.
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.
From p²=2q² and p=2r, obtain q²=2r². Both p and q would have factor 2, contradicting lowest terms. For a supposed complete prime list p₁,…,pₖ, set N=p₁⋯pₖ+1. N>1 has a prime divisor, but every listed prime leaves remainder 1 when dividing N. That divisor is a new prime, contradicting completeness. N itself need not be prime: 2×3×5×7×11×13+1=30031=59×509.
Irrational roots and infinitely many primes
To prove √2 irrational, assume √2=p/q for coprime integers p,q with q nonzero
Classify each argument and explain the condition that makes it valid or incomplete.
If p=2r and p²=2q², what coefficient multiplies r² in q²?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
4r²=2q², so q²=2r².
Test a tempting shortcut
- The even-square step needs a reason: an odd integer 2r+1 has odd square 4r(r+1)+1. The prime proof needs a new prime divisor, not a claim that product-plus-one is always prime. A contradiction must attack the stated assumption.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The product of any finite list of primes plus one is always prime. This claim is false. Explain which definition or assumption it violates.
For the prime list 2,3,5, compute the product-plus-one number.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
2×3×5+1=31.
The product of any finite list of primes plus one is always prime.
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 even-square step needs a reason: an odd integer 2r+1 has odd square 4r(r+1)+1. The prime proof needs a new prime divisor, not a claim that product-plus-one is always prime. A contradiction must attack the stated assumption.
Interpret a new situation
- For an unfamiliar proof, state the contrary assumption precisely, preserve its conditions and name the exact contradiction. Decimal approximations or a large finite list do not settle these two claims.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the remainder when 30031 is divided by 13.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
30031=13×2310+1.
Match each contradiction-proof ingredient to its role.
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 contradiction must attack the stated assumption; in the primes proof, N need not itself be prime - 30031=59×509.
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 assumes the negation of a claim and derives an impossibility. Choose the relationship, show the method, check its assumptions and interpret the result.