Question

Prove that √3 is irrational. ​

Answer 1

Answer:

Hope this will help you in the clearance of your doubt.

Answer 2

$\huge\underline\bold\purple{QUESTION :}$

Prove that √3 is irrational.

$\huge\underline\bold\purple{ANSWER :}$

Let us assume that √3 is a rational number.

$⟹ \sqrt{3} = \frac{a}{b}$

• [ a, b are integers ]

$⟹ \: { (\sqrt{3}) }^{2} = { (\frac{a}{b}) }^{2}$

• [ squaring on both sides ]

$⟹3 = \frac{ {a}^{2} }{ {b}^{2} }$

$⟹3 {b}^{2} = {a}^{2}$

$⟹ {b}^{2} = \frac{ {a}^{2} }{3}$

∴ 3 divides a² and 3 divides a. Now, “ a = 2c ”

$⟹ {b}^{2} = \frac{ ({3c})^{2} }{ 3}$

$⟹ {b}^{2} = \frac{9 {c}^{2} }{3}$

$⟹ {b}^{2} = 3 {c}^{2}$

$⟹ \frac{ {b}^{2} }{3} = {c}^{2}$

∴ 3 divides b² and 3 divides b.

Thus, 3 is a common factor of a & b.

This contradicts the fact that a and b have no common factor other than 1.

The contradiction arises by assuming that √3 is a rational number.

$\boxed{∴ Hence\;√3\;is\;an\;irrational\;number}$

Step-by-step explanation:

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