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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.

Related tags

Number SystemsSquare Root 2Contradiction ProofProof Of Irrationality Of Square Root 2 And Square Root 3MathematicsClass 9 Mcq

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