Question

prove root3 is irrational​

Answer 1

$\huge \purple {Answer}$

Let √3 is a rational number.

Therefore, It can be expressed in the form of p/q, where p and q are co-primes and q≠ 0.

⇒ √3 = $\frac {p}{q}$

By squaring both sides

⇒ 3 = $\frac {p²}{q²}$

⇒ 3q² = p² -(i)

This means that 3 divides p². This means that 3 divides p because each factor should appear two times for the square to exist.

So, we have p = 3r, where r is some integer.

⇒ p² = 9r² -(ii)

from equation (i) and (ii)

⇒ 3q² = 9r²

⇒ q² = 3r²

Where q² is multiply of 3 and also q is multiple of 3.

Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, $\frac {p}{q}$ is not a rational number. This demonstrates that √3 is an irrational number.

$\boxed {Hence, \: proved}$

Answer 2

$\huge\underline\mathbb{\red S\pink{O}\purple{L} \blue{UT} \orange{I}\green{ON :}}$

Let us assume that √3 is a rational number.

$⟹ \sqrt{3} = \frac{a}{b}$

• [ a & b are co - primes ]

$⟹b \sqrt{3} = a$

• [ Squaring on both sides. ]

$⟹ {(b \sqrt{3} )}^{2} = {(a)}^{2}$

$⟹ {3b}^{2} = {a}^{2}$

$⟹ {b}^{2} = \frac{ {a}^{2} }{3}$

Therefore, 3 divides a² & 3 divides a.

$\boxed{∴ By\:theorem\::\:If\:q\:is\:a\:prime\:number,\:and\:q\:divides\:a²\:and\:q\:divides\:a.}$

Now take the value of a = 3c.

$⟹ {b}^{2} = \frac{ {(3c)}^{2} }{3}$

$⟹ {b}^{2} = \frac{ {9c}^{2} }{3}$

$⟹ {b}^{2} = 3 {c}^{2}$

$⟹ \frac{ {b}^{2} }{3} = {c}^{2}$

Therefore, 3 divides b² & 3 divides b.

$\boxed{∴ By\:theorem\::\:If\:q\:is\:a\:prime\:number,\:and\:q\:divides\:a²\:and\:q\:divides\:a.}$

↪ Hence, 3 divides a & b.

↪ So, 3 is a factor of both a & b.

↪ Therefore, a & b are not co - primes.

↪ But we assume that a & b are co - primes. Therefore, our assumption is wrong.

$\underline{\boxed{\bf{\purple{∴ Hence, \sqrt{3} \: \:is\:an\:irrational\:number.}}}}$

Step-by-step explanation:

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