Proof by contradiction is effective for proving statements such as “√2 is irrational.”
Steps in a Proof by Contradiction
①Assume the negation of the conclusion
②Reason logically to derive a contradiction
③Since this is a contradiction, the assumption is false → the conclusion is true
Example ① (√2 Is Irrational)
Assume that √2 is rational. Let √2 = p/q（p,q are relatively prime positive integers）.
2q² = p² → p² is even → p is even（p=2m）
2q² = 4m² → q² = 2m² → q² is even → q is even
Both p and q are even → GCD(p,q)≥2 → contradiction to their being relatively prime → √2 is irrational ■
Assume that √2 is rational. Let √2 = p/q（p,q are relatively prime positive integers）.
2q² = p² → p² is even → p is even（p=2m）
2q² = 4m² → q² = 2m² → q² is even → q is even
Both p and q are even → GCD(p,q)≥2 → contradiction to their being relatively prime → √2 is irrational ■
Example ② (√3 Is Irrational)
Assume √3 = p/q（in lowest terms）. 3q² = p² → p² ≡ 0(mod 3) → p ≡ 0(mod 3)（p=3m）
3q² = 9m² → q² = 3m² → q ≡ 0(mod 3) → p and q have common factor 3 → contradiction ■
Assume √3 = p/q（in lowest terms）. 3q² = p² → p² ≡ 0(mod 3) → p ≡ 0(mod 3)（p=3m）
3q² = 9m² → q² = 3m² → q ≡ 0(mod 3) → p and q have common factor 3 → contradiction ■
Key Points
- √p（p is prime）is irrational for every prime p
- Key to the proof: “p² is a multiple of p → p is a multiple of p”
- Proof by contradiction → clearly write the flow of assumption, reasoning, and contradiction
Practice Problems
- Use proof by contradiction to show that √5 is irrational.
- Show that √2 + √3 is irrational (you may use the fact that √6 is irrational).
- Show that log₂3 is irrational.
Answers and Explanations
- Assume √5=p/q（in lowest terms）. 5q²=p² → p=5m → 5q²=25m² → q²=5m² → q is a multiple of 5 → contradiction ■
- Assume √2+√3=r（rational）. √3=r−√2 → 3=(r−√2)²=r²−2r√2+2 → √2=(r²−1)/(2r)（rational）→ contradiction to √2 being rational ■
- Assume log₂3=p/q（a positive rational number in lowest terms）. 3=2^(p/q) → 3^q=2^p. The left side is odd, while the right side is even → contradiction ■