On what basis does the final conclusion in irrationality of (√2) come?
Answer and explanation
Correct answer: Rational assumption makes both (m) and (n) even
The final conclusion comes from contradiction with the original lowest-terms assumption. Suppose √2 = m/n, where m and n are coprime integers. Squaring gives m² = 2n², so m is even; writing m = 2r then leads to n² = 2r², making n even as well. Thus both numerator and denominator have the common factor 2, which contradicts their being coprime. The rational assumption must therefore be false, and √2 is irrational. Option A states the decisive basis. A terminating decimal is not the conclusion of this proof, m = n is never established, and n = 0 is excluded because a fraction cannot have a zero denominator.
Frequently asked questions
What is the correct answer to this question?
Rational assumption makes both (m) and (n) even
Why is this the correct answer?
The final conclusion comes from contradiction with the original lowest-terms assumption. Suppose √2 = m/n, where m and n are coprime integers. Squaring gives m² = 2n², so m is even; writing m = 2r then leads to n² = 2r², making n even as well. Thus both numerator and denominator have the common factor 2, which contradicts their being coprime. The rational assumption must therefore be false, and √2 is irrational. Option A states the decisive basis. A terminating decimal is not the conclusion of this proof, m = n is never established, and n = 0 is excluded because a fraction cannot have a zero denominator.
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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