In the proof of √2, if m/n is in lowest form, which situation is impossible?
Answer and explanation
Correct answer: m and n are both even
The governing concept is the lowest-form condition in a rational representation. When m/n is in lowest form, m and n are integers, n is nonzero, and they have no common factor greater than 1. If both m and n were even, each would be divisible by 2, so the fraction could be reduced by cancelling 2. That would prove it was not in lowest form. Therefore option D describes the impossible situation. The conditions in options A, B and C are all part of a valid rational representation: the denominator must not be zero, and numerator and denominator are integers. In the √2 proof, both being even is not an initial assumption; it is the contradiction derived from the equation m² = 2n² after assuming √2 rational.
Frequently asked questions
What is the correct answer to this question?
m and n are both even
Why is this the correct answer?
The governing concept is the lowest-form condition in a rational representation. When m/n is in lowest form, m and n are integers, n is nonzero, and they have no common factor greater than 1. If both m and n were even, each would be divisible by 2, so the fraction could be reduced by cancelling 2. That would prove it was not in lowest form. Therefore option D describes the impossible situation. The conditions in options A, B and C are all part of a valid rational representation: the denominator must not be zero, and numerator and denominator are integers. In the √2 proof, both being even is not an initial assumption; it is the contradiction derived from the equation m² = 2n² after assuming √2 rational.
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.
