Question

prove that √3 is an irrational number.​

Answer 1

Answer:

an irrational number✰

➪Let us assume on the contrary that 3 is a rational number.

➪Then, there exist positive integers a and b such that 3= a/b where, a and b, are co-prime i.e.

now \: \sqrt{3} = \frac{a}{b}now3=ba

\sqrt{3} = \frac{ {a}^{2} }{ {b}^{2} }3=b2a2

⇒3 divides a²

⇒3 divides a ²[∵3 divides 3b ]

⇒3 divides a²[∵3 divides 3b²] ⇒3 divides a...(i)

⇒3 divides a 2 [∵3 divides 3b²] ⇒3 divides a...(i)

⇒a=3c² for some integer c.

⇒a=3c for some integer c

⇒a=3c for some integer c⇒a² =9c²

⇒3b²=9c² [∵a² =3b²]

⇒b ²=3c²

⇒3 divides b²

⇒3 divides b²[∵3 divides 3c²] ⇒3 divides b² [∵3 divides 3c²] ⇒3 divides b...(ii)

From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.

hence, \sqrt{3} \: \: is \: \: an \: \: irrational \: \: numberhence,3isanirrationalnumber

hope it helps you

Answer 2

Answer:

M Y. S N A P C H A T I D

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