Prove that the following are irrationals : (i) 1 2 (ii) 75 (iii) 62
Step-by-Step Solution
Key Concept: Use proof by contradiction: assume the number is rational (expressible as \frac{p}{q} in lowest terms) and show that this leads to a contradiction using the Fundamental Theorem of Arithmetic (prime factorisation).
(i) \(\sqrt{2}\)
Given: Assume \(\sqrt{2}=\frac{p}{q}\) where \(p,q\) are integers with \(\gcd(p,q)=1\).
Step 1: Square both sides → \(2=\frac{p^{2}}{q^{2}}\) ⇒ \(p^{2}=2q^{2}\).
Step 2: Since \(p^{2}\) is even, \(p\) must be even. Write \(p=2k\).
Step 3: Substitute: \((2k)^{2}=2q^{2}\) ⇒ \(4k^{2}=2q^{2}\) ⇒ \(q^{2}=2k^{2}\).
Step 4: Hence \(q^{2}\) is even, so \(q\) is even.
Step 5: Both \(p\) and \(q\) are even, contradicting the assumption that they are coprime.
Conclusion: The assumption is false; therefore \(\sqrt{2}\) is irrational.
(ii) \(\sqrt{75}\)
Given: \(75=3\times5^{2}\) so \(\sqrt{75}=5\sqrt{3}\).
Step 1: Suppose \(\sqrt{75}\) is rational. Then \(\sqrt{3}=\frac{\sqrt{75}}{5}\) would also be rational.
Step 2: But \(\sqrt{3}\) is known to be irrational (proved similarly to part (i)).
Step 3: This contradiction shows the original assumption is false.
Conclusion: Hence \(\sqrt{75}\) is irrational.
(iii) \(\sqrt{62}\)
Given: Assume \(\sqrt{62}=\frac{p}{q}\) with \(\gcd(p,q)=1\).
Step 1: Square both sides → \(62=\frac{p^{2}}{q^{2}}\) ⇒ \(p^{2}=62q^{2}=2\cdot31\,q^{2}\).
Step 2: In the prime factorisation of the right‑hand side, the primes 2 and 31 appear to the first power (odd exponent).
Step 3: The square of an integer (\(p^{2}\)) must have each prime factor with an even exponent (Fundamental Theorem of Arithmetic).
Step 4: Hence the equality cannot hold; the assumption that \(\sqrt{62}\) is rational leads to a contradiction.
Conclusion: Therefore \(\sqrt{62}\) is irrational.
Overall, each number leads to a contradiction when assumed rational, proving they are all irrational.
Correct Answer: All three numbers \(\sqrt{2}, \sqrt{75}, \sqrt{62}\) are irrational.