Question

Prove that
$prove that \sqrt{3 \: is \: an \: irrational \: \: number}$
3 is an irrational number. ​

Answer 1

Answer:

If possible , let √3 be a rational number and its simplest form be $\frac{a}{b}$

then, a and b are integers having no common factor

other than 1 and b ≠ 0

Now,

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

$⟹3 = \frac{ {a}^{2} }{ {b}^{2} }$

⠀⠀⠀⠀⠀⠀⠀⠀ (On squaring both sides )

$⟹3 {b}^{2} = {a}^{2} \: \: \: \: \: \: \: \: .....(1)$

$⟹3 \: divides \: {a}^{2}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀(as 3 divides 3b²)

$⟹3 \: divides \: a$

Let a=3c for some integer c

Putting a=3c in (1), we get

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

$⟹ {b}^{2} = 3 {c}^{2}$

$⟹3\:divides\:b² \\</p><p> (∵3\:divides\:3c² )$

$⟹3\:divides\:$

Thus 3 is a common factor of a and b

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

The contradiction arises by assuming √3 is a rational.

Hence, √3 is irrational.

Answer 2

Question :

Prove that $\sqrt{3}$ is an irrational number.

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Solution :

ㅤㅤㅤㅤㅤㅤWe need to prove this by contradiction method. So, let us assume that √3 is a rational number.

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$\tt { \sqrt{3} = \dfrac {a}{b} }$

( a, b are co-prime numbers. )

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

Squaring both sides:

$: \implies \tt {a^2 = 3b^2 }$............... (1)

↦ a² is multiple of 3.

↦ a is also multiple of 3.

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$\tt { a = 3c}$...........................(2)

Squaring both sides :

$\tt {a^2 = 9c^2 }$ .....................(3)

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{from (1) and (3) }

$\tt {\cancel {3b^2 } = \cancel {9c^2} }$

$: \implies \tt { b^2 = 3c^2 }$

↦b² is multiple of 3.

↦ b is also multiple of 3.

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$\because$ , a and b are multiples of 3 which means our supposition is wrong.

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•°• , √3 is an irrational number.