Verify by method of contradiction p ∶ \(\sqrt{11}\) is irrational.

1.	Verify by method of contradiction p ∶ \(\sqrt{11}\) is irrational.
Answer» Let the given statement be false i.e., ~ p ∶ \(\sqrt{11}\) Is rational. The \(\sqrt{11}\) = \(\frac{p}{q}\) where p and q are coprime and q ≠ 0. ⇒ 11 = \(\frac{p^2}{q^2}\) ⇒ p² = 11q² ⇒ 11 divides p ...(1) ∴ r ∈ z such that− p = 11r ⇒ p² = 121r ⇒ 11q² = 121r² ⇒ q² = 11r² ⇒ 11 divides q … (2) From (1) & (2), we arrive at a contradiction, since p and q are coprime. ⇒ \(\sqrt{11}\) is irrational.

Answer»

Let the given statement be false i.e., ~ p ∶ \(\sqrt{11}\)

Is rational. The \(\sqrt{11}\) = \(\frac{p}{q}\) where p and q are coprime and q ≠ 0.

⇒ 11 = \(\frac{p^2}{q^2}\)

⇒ p² = 11q²

⇒ 11 divides p ...(1)

∴ r ∈ z such that−

p = 11r

⇒ p² = 121r

⇒ 11q² = 121r²

⇒ q² = 11r²

⇒ 11 divides q … (2)

From (1) & (2), we arrive at a contradiction, since p and q are coprime.

⇒ \(\sqrt{11}\) is irrational.

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