In the proof of (\sqrt{2}), what is shown false by the rational assumption?
Answer and explanation
Correct answer: (\sqrt{2}) is rational
The proof starts by assuming that \(\sqrt{2}\) is rational. Write it as \(\frac{p}{q}\) in lowest terms, with q nonzero. Squaring gives \(p^2=2q^2\). This makes \(p^2\) even, so p is even. Substituting \(p=2k\) into the equation gives \(4k^2=2q^2\), hence \(q^2=2k^2\), so q is also even.
The conclusion that both p and q are even contradicts the fact that \(\frac{p}{q}\) was chosen in lowest terms. This contradiction does not show that \(\sqrt{2}\) is non-real, non-positive, or unequal to a positive number. In fact, \(\sqrt{2}\) is a positive real number. What fails is only the starting assumption that it is rational. Therefore option C is correct.
Frequently asked questions
What is the correct answer to this question?
(\sqrt{2}) is rational
Why is this the correct answer?
The proof starts by assuming that \(\sqrt{2}\) is rational. Write it as \(\frac{p}{q}\) in lowest terms, with q nonzero. Squaring gives \(p^2=2q^2\). This makes \(p^2\) even, so p is even. Substituting \(p=2k\) into the equation gives \(4k^2=2q^2\), hence \(q^2=2k^2\), so q is also even.
The conclusion that both p and q are even contradicts the fact that \(\frac{p}{q}\) was chosen in lowest terms. This contradiction does not show that \(\sqrt{2}\) is non-real, non-positive, or unequal to a positive number. In fact, \(\sqrt{2}\) is a positive real number. What fails is only the starting assumption that it is rational. Therefore option C is correct.
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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