Why is it wrong to assume h and k are divisible by 3 from the beginning in the proof of √3?
Answer and explanation
Correct answer: Because h and k should initially be coprime and in lowest form
A contradiction proof must begin with the strongest legitimate assumption, not with the contradiction that it intends to derive. We assume √3 = h/k, where h and k are integers, k ≠ 0, and gcd(h,k) = 1. Squaring gives h² = 3k²; divisibility arguments then show that 3 divides h and, after substitution, also divides k. This final result contradicts gcd(h,k) = 1. If both numbers were assumed divisible by 3 at the beginning, the contradiction would be presupposed and the proof would become circular. Hence option C is correct. The other options either impose false conditions or use an irrelevant decimal representation.
Frequently asked questions
What is the correct answer to this question?
Because h and k should initially be coprime and in lowest form
Why is this the correct answer?
A contradiction proof must begin with the strongest legitimate assumption, not with the contradiction that it intends to derive. We assume √3 = h/k, where h and k are integers, k ≠ 0, and gcd(h,k) = 1. Squaring gives h² = 3k²; divisibility arguments then show that 3 divides h and, after substitution, also divides k. This final result contradicts gcd(h,k) = 1. If both numbers were assumed divisible by 3 at the beginning, the contradiction would be presupposed and the proof would become circular. Hence option C is correct. The other options either impose false conditions or use an irrelevant decimal representation.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Number Systems. Topic: Proof of irrationality of square root 2 and square root 3.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
