Question

prove that √3 is a irrational number ​

Answer 1

Answer:

let us assume root 3 be rational no.

root 3 = a/b ( where a and b are co prime numbers )

a= b×root 3

squaring both the sides

a^2 = 3b^2

a^2/ b^2 = 3

=> a is divisible by b

=> but this contradicts the fact that a and b are co prime

=>This is due to our wrong assumption that root 3 is rational.

This proves that root 3 is an irrational number.

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Answer 2

GIVEN:

• √3

TO FIND:

• Prove that √3 is an irrational number.

SOLUTION:

Let √3 be a rational number, which can be written in the form of p/q, where p and q are integers and q ≠ 0

$\sf{\hookrightarrow \sqrt{3} = \dfrac{p}{q}}$

Squaring on both sides

$\sf{\hookrightarrow (\sqrt{3})^2 = \bigg(\dfrac{p}{q} \bigg)^2 }$

$\sf{\hookrightarrow 3 = \dfrac{p^2}{q^2}}$

$\bf{\hookrightarrow 3q^2 = p^2....1) }$

⁂ 3 divides p²

⁂ 3 divides p

Let p = 3x

Put the value of 'p' in equation 1)

$\sf{\hookrightarrow 3q^2 = (3x)^2 }$

$\sf{\hookrightarrow 3q^2 = 9x^2 }$

$\sf{\hookrightarrow q^2 = 3x^2 }$

⁂ 3 divides q²

⁂ 3 divides q

Thus, 3 divides p and q

✏ It means 3 is a common factor of p and q. This contradicts the assumption as there is no common factor of p and q.

Hence, √3 is an irrational number.

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