If √2 = r/s is in lowest form and finally 2 divides r and 2 divides s, which statement is the most precise contradiction?
Answer and explanation
Correct answer: gcd(r,s) = 1 and gcd(r,s) ≥ 2 cannot both hold
A fraction in lowest form is defined by the condition gcd(r,s) = 1, with s nonzero. If the proof establishes that 2 divides both r and s, then 2 is a common divisor of the pair. Consequently gcd(r,s) is at least 2, not 1. These two conclusions cannot simultaneously be true, so the assumption that √2 equals a fraction r/s in lowest terms is impossible. This is the exact logical contradiction, stated in option C. It is more precise than merely saying the numerator and denominator are both even, because it explicitly compares the defining lowest-form condition with the new divisibility result. Options A, B, and D do not follow from the proof.
Frequently asked questions
What is the correct answer to this question?
gcd(r,s) = 1 and gcd(r,s) ≥ 2 cannot both hold
Why is this the correct answer?
A fraction in lowest form is defined by the condition gcd(r,s) = 1, with s nonzero. If the proof establishes that 2 divides both r and s, then 2 is a common divisor of the pair. Consequently gcd(r,s) is at least 2, not 1. These two conclusions cannot simultaneously be true, so the assumption that √2 equals a fraction r/s in lowest terms is impossible. This is the exact logical contradiction, stated in option C. It is more precise than merely saying the numerator and denominator are both even, because it explicitly compares the defining lowest-form condition with the new divisibility result. Options A, B, and D do not follow from the proof.
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.
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