Question

$\Large \color{lime} \maltese \: \: \pmb{ \mathbb{ \color{maroon}QUE \color{orange}ST \color{magenta}I \ON}} \: \: \maltese$
Prove that $\sf\sqrt{3}$
is an Irrational Number.

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Answer 1

▪ To Prove :-

$\sf\sqrt{3}$ is an Irrational Number.

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▪ Proof :-

Let, us assume that √3 is a rational number.

So,

√3 can be  written in the form of p/q.

⟹ √3 = p/q.

⟹ q√3 = p.

Squaring on both side  we get ,

    (q√3)² = (p)²

⟹ 3q² = p²

⟹ q² = p²/3

It means  p² is divisible by 3

So,  p is also divisible by 3. - - - - - (1).

Let us  assume that,

   

p/3 = r.

Where r is an integer.

⟹ p = 3r.

⟹ 3q² = p².

Putting  value of p  in  eq (1)

⟹3q² = (3r)².

⟹ 3q² = 9r².

⟹ q² = 3r².

⟹ q²/3 = r².

It means  q² is divisible by 3.

So,  q is also divisible by 3. - - - - - (2)

From equation (1) and (2),

It can easily be determined that :

3 is a common  factor of p and q.

But, p and q are not Co-prime.

This contradiction arise due to our wrong as assumption that √3 is rational

∴ √3 is an irrational number.

Hence  Proved!

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Answer 2

$\red{\bigstar}$ T OㅤP R O V E

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• $\sqrt{3}$ is irrational.

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$\blue{\bigstar}$ P R O O F

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• Let us assume that $\sqrt{3}$ is a rational number.

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So, it can be written as ::

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$\sf \quad\dashrightarrow\quad \sqrt{3} = \dfrac{a}{b}\:(Where\:a\:and\:b\:are\:co - prime)$

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Co - prime numbers are numbers having common factor 1.

$\\ \sf \quad\dashrightarrow\quad b\sqrt{3} = a$

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Squaring both sides ::

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$\\ \sf \quad\dashrightarrow\quad (b\sqrt{3})^2 = (a)^2$

$\\ \sf \quad\dashrightarrow\quad 3b^2 = a^2 \qquad(1)$

$\\ \sf \quad\dashrightarrow\quad 3\:\times\:b\:\times\:b = a^2$

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Therefore, a² is divisible by 3. So, a is also divisible by 3. (2)

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Therefore, we can take a = 3n, putting in (1) ::

$\\ \sf \quad\dashrightarrow\quad 3b^2 = (3m)^2$

$\\ \sf \quad\dashrightarrow\quad 3b^2 = 9m^2$

$\\ \sf \quad\dashrightarrow\quad b^2 = {\cancel{\dfrac{9}{3}}}m^2$

$\\ \sf \quad\dashrightarrow\quad b^2 = 3m^2$

$\\ \sf \quad\dashrightarrow\quad b^2 = 3\:\times\:m\:\times\:m$

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Therefore, b² is divisible by 3. So, b is also divisible by 3. (3)

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Therefore, (2) and (3) a and b are divisible by 3. So, they have common factor 3. But this contradict our assumption. Our assumption is that a and b are co - prime.

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Therefore, our assumption is wrong!

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So, √3 is a irrational number.

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$\purple{\bigstar}$ H E N C EㅤP R O V E D

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