Verify by the method of contradiction that √7 is irrational

1.	Verify by the method of contradiction that √7 is irrational
Answer» Let p : √7 is irrational Let us assume p is not true i.e.., √7 is rational . ⇒ √7 = a/b, where a and b are integers having no common factor. ⇒ 7 = a²/b² ⇒ a² =7b² ⇒ 7 divides a² ⇒ 7 divides a ⇒ a = 7c, for some integer c. ⇒ a² = 49c² ⇒ 7b² = 49c² ⇒ b² = 7c² ⇒ 7 divides b² ⇒ 7 divides b Thus, 7 is common factor of both a and b. This contradicts that a and b have no common factor. So, our assumption is wrong. Hence, √7 is irrational is true.

Answer»

Let p : √7 is irrational

Let us assume p is not true i.e.., √7 is rational .

⇒ √7 = a/b, where a and b are integers having no common factor.

⇒ 7 = a²/b²

⇒ a² =7b²

⇒ 7 divides a²

⇒ 7 divides a

⇒ a = 7c, for some integer c.

⇒ a² = 49c²

⇒ 7b² = 49c²

⇒ b² = 7c²

⇒ 7 divides b²

⇒ 7 divides b

Thus, 7 is common factor of both a and b. This contradicts that a and b have no common factor.

So, our assumption is wrong.

Hence, √7 is irrational is true.

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