Question

prove that √3 is irrational number┐( ∵ )┌​



step by step ....​

Answer 1

Step-by-step explanation:

Solutions is in the photo

Answer 2

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

Then, there exist positive integers a and b such that

$\sqrt{3} = \frac{a}{b}$

where, a and b, are co-prime i.e. their HCF is 1

Now,

=>

$3 = \frac{ {a}^{2} }{ {b}^{2} }$

${3b}^{2} = {a}^{2}$

3 divides a square [∵3 divides 3b ^2]

⇒3 divides a...(i)

⇒a=3c for some integer c

⇒

${a}^{2} = {9c}^{2}$

${3b}^{2} = {9c}^{2}$

[∵a^2 =3b^2 ]

${b}^{2} = {3c}^{2}$

$3 \: divides \: {b}^{2}$

[∵3 divides 3c^2]

⇒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, √3 is an irrational number.

$\textit{Hope this helps you}$

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