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Which order is correct in the proof of irrationality of (\sqrt{2})?

Answer and explanation

Correct answer: Assume rational then square then contradiction

The direct answer is Option C: assume rationality, square, and then obtain a contradiction. The proof uses contradiction. First suppose that √2 is rational, so it can be written as a/b, where a and b are coprime integers and b is not zero. Squaring gives 2 = a²/b², so a² = 2b². Thus a² is even, so a is even; write a = 2r. Substitution then shows b² = 2r², so b is also even. This contradicts that a and b were coprime. Option C is correct. Option A reverses the logical order because a = 2r is obtained after proving a is even. Option B is not a complete proof; a decimal representation does not by itself establish the required contradiction. Option D simply asserts the result without proof. Exam cue: contradiction proofs usually follow assume, transform, contradiction, conclude.

Related tags

Number-SystemsIrrationality-ProofProof-Order

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