What is the final contradiction in the proof that √2 is irrational?
Answer and explanation
Correct answer: Both p and q become even
The proof uses contradiction. Assume that √2 is rational and write √2 = p/q in lowest terms, where p and q are integers, q is non-zero, and p and q have no common factor. Squaring gives p² = 2q², so p² is even and p must be even. Let p = 2k. Substitution gives 4k² = 2q², or q² = 2k², which shows that q is also even. Thus both p and q are divisible by 2, contradicting the assumption that p/q was in lowest terms. Therefore option A states the final contradiction. The other choices do not follow from the parity argument and do not contradict the original lowest-form assumption.
Frequently asked questions
What is the correct answer to this question?
Both p and q become even
Why is this the correct answer?
The proof uses contradiction. Assume that √2 is rational and write √2 = p/q in lowest terms, where p and q are integers, q is non-zero, and p and q have no common factor. Squaring gives p² = 2q², so p² is even and p must be even. Let p = 2k. Substitution gives 4k² = 2q², or q² = 2k², which shows that q is also even. Thus both p and q are divisible by 2, contradicting the assumption that p/q was in lowest terms. Therefore option A states the final contradiction. The other choices do not follow from the parity argument and do not contradict the original lowest-form assumption.
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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