Irrational roots and infinitely many primes
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| proof by contradiction/pruːf baɪ ˌkɒntrəˈdɪkʃn/ | phương pháp chứng minh bằng phản chứng |
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.
The method
- 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.
In the √2 proof, why does p²=2q² force p to be even?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
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.
A worked case
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?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
4r²=2q², so q²=2r².
Where it goes wrong
- 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.
Not true: The product of any finite list of primes plus one is prime. The product plus one need not be prime: 2×3×5×7×11×13+1=30031=59×509. It only has a prime factor not in the list.
For the prime list 2,3,5, compute the product-plus-one number.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
2×3×5+1=31.
Proof by contradiction means a proof that assumes the negation of a claim and derives an impossibility.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
True: A proof that assumes the negation of a claim and derives an impossibility.
Applying it
- 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.
Find the remainder when 30031 is divided by 13.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
30031=13×2310+1.
Match each contradiction-proof ingredient to its role.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The contradiction must attack the stated assumption; in the primes proof, N need not itself be prime - 30031=59×509.
In your course
- AQA A-level Mathematics (7357), specification section A.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Proof by contradiction: A proof that assumes the negation of a claim and derives an impossibility.
- To prove √2 irrational, assume √2=p/q for coprime integers p,q with q nonzero.
- Watch for: the even-square step needs a reason: an odd integer 2r+1 has odd square 4r(r+1)+1.