Question

Prove that √3 is an irrational number​

Answer 1

assume root 3 as rational number

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )

=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q

Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number

Answer 2

$\huge{\underline{\mathfrak{\textcolor{blue}{Answer :}}}}$ $\LARGE\boxed{\fcolorbox{pink}{pink}{irrational}}$
$\huge{\underline{\mathrm{\textcolor{red}{Step-by-step \: \: explanation :}}}}$

$\LARGE{\underline{\mathtt{\textcolor{violet}{Given :-}}}}$
⚪ root over 3 [ √3 ]

$\LARGE{\underline{\mathtt{\textcolor{green}{To \: \: prove :-}}}}$
⚪ $\sqrt{3}$ is an irrational number.

$\LARGE{\underline{\mathtt{\textcolor{teal}{Concept \: \: used :-}}}}$
⚪ Real numbers
⚪ Irrational numbers

$\LARGE{\underline{\mathtt{\textcolor{blue}{Proof :-}}}}$
✒ If possible, $\sqrt{3}$ be a rational number.

let $\sqrt{3} = \Large{ \frac{a}{b} }$, where a and b are co-primes and b ≠ 0.

Then,
✒ $\sqrt{3} = \Large{ \frac{a}{b} }$

On squaring both the sides :

$\Rightarrow { \left( \sqrt{3} \right) }^{2} = \Large{ { \left( \frac{a}{b} \right) }^{2} }$

$\Rightarrow 3 = \Large{ \frac{ {a}^{2} }{ {b}^{2} } }$

$\Rightarrow 3 {b}^{2} = {a}^{2}$

$\Rightarrow {a}^{2} = 3 {b}^{2}$ ——————— ( 1 )

Therefore, ${a}^{2}$ is divisible by 3.
Therefore, a is also divisible by 3.

let a = 3c, for some integer c. ——————— ( 2 )

On substituting ( 2 ) in ( 1 ) , we get :

${(3c)}^{2} = 3 {b}^{2}$

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

$\Rightarrow {}^{3} \: \cancel{9} {c}^{2} = {}^{1} \: \cancel{5} {b}^{2}$

$\Rightarrow {b}^{2} = 3 {c}^{2}$

Therefore, ${b}^{2}$ is divisible by 3.
Therefore, b is also divisible by 3.

Therefore, a and b have a common factor 3.
This contradicts the fact that a and b are co-primes.
This contradiction arises on assuming $\sqrt{5}$ to be a rational number.
So, our assumption is wrong.
Hence, $\sqrt{3}$ is an $\sf\green{\underline{\blue{irrational \: number}}}$.

$\LARGE{\underline{\mathtt{\textcolor{magenta}{Conclusion :-}}}}$
⚪$\sqrt{3}$ is an $\sf\green{\underline{\blue{irrational \: number}}}$.

${\textcolor{purple}{\fcolorbox{teal}{cyan}{Your \: required \: answer \: is \: above......... keep \: smiling........ ;-D}}}$
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$\LARGE{\fcolorbox{orange}{orange}{brainliest answer}}$

$\mid \underline{\LARGE\bf\green{Brainliest \: Answer}}\mid$