Q1.
Try to prove the irrationality of √3 using the approach of proof by contradiction. Will the same approach work for √5, √7, or √10?
Answer
Claim: √3 is irrational. Follow the eight steps of Section 3.5.1, replacing “even” by “divisible by 3”.
Step 1. Suppose √3 = p/q in lowest terms, q ≠ 0, gcd(p, q) = 1
Step 2. square: 3 = p²/q²
Step 3. 3q² = p²
Step 4. so 3 divides p². Since 3 is prime, 3 divides p
write p = 3k
Step 5. 3q² = (3k)² = 9k²
Step 6. divide by 3: q² = 3k²
Step 7. so 3 divides q², hence 3 divides q
Step 8. p and q share the factor 3 — contradicting gcd(p, q) = 1
Step 2. square: 3 = p²/q²
Step 3. 3q² = p²
Step 4. so 3 divides p². Since 3 is prime, 3 divides p
write p = 3k
Step 5. 3q² = (3k)² = 9k²
Step 6. divide by 3: q² = 3k²
Step 7. so 3 divides q², hence 3 divides q
Step 8. p and q share the factor 3 — contradicting gcd(p, q) = 1
The assumption is the only thing that can be wrong, so √3 is irrational.
Does it work for √5, √7, √10? Yes, all three.
| Number | Key step | Verdict |
|---|---|---|
| √5 | 5 prime: 5 | p² ⇒ 5 | p | irrational |
| √7 | 7 prime: 7 | p² ⇒ 7 | p | irrational |
| √10 | 10 not prime, but 2 | p² ⇒ 2 | p works | irrational |
For √10, run the argument with the prime 2 alone: from 10q² = p² we get p even, p = 2k, so 10q² = 4k², i.e. 5q² = 2k², so q² is even and q is even — contradiction.
Why it happens: the engine of the whole proof is the step “n divides p² implies n divides p”, and that step is valid precisely when n is prime (or, more generally, square-free). It fails for n = 4: 4 divides 6² = 36 but 4 does not divide 6 — which is exactly why √4 = 2 is not irrational. So the method proves √n irrational for every positive integer n that is not a perfect square, and honestly refuses to prove it when n is one.