Question

prove​​ ​
$\sqrt{}$
prove that root under 3 is an irrational number ​

Answer 1

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$


$\bf{QUESTION}$

$\sf{prove \: \sqrt{3} \: is \: an \: irrational \: number}$

Let √3 be a rational number

Hence


$\bf{ \sqrt{3} = \frac{p}{q}}$

where p and q are integers and q≠0

$\bf{( \sqrt{3})^{2} = {( \frac{p}{q} })^{2}}$

(squaring both side)


$\bf{3 = (\frac{p}{q} )^{2} }$

$\bf{ \implies \: 3q ^{2} = {p}^{2} }$

$\bf{ \implies \: q^{2} = \frac{ {p}^{2} }{3} }$


here

°•° p² is divided by 3


•°• p is divided by 3


Now

take p= 3r


$\bf{ \implies \: {q}^{2} = \frac{ ({3r})^{2} }{3}}$

$\bf{ \implies \: {q}^{2} = \frac{9r^{2} }{3} }$


$\bf{ \implies \: {q}^{2} = 3 {r}^{2} }$

$\bf{ \implies \: {r}^{2} = \frac{ {p}^{2} }{3} }$

Here

°•° q² is divided by 3


•°• q is divided by 3


Hence 3 is the common factor

hence the contradiction we supposed is wrong .

•°• √3 is an irrational number


Hence proved ✔✔



$\huge \pink{ \boxed { \boxed{ \mathbb{THANKS}}}}$

Answer 2

Let us assume that √3 is a 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

ALTERNATIVE METHOD:

it can also be proved by stating that the square root of any Prime number is irrational.

3=1×3

Hence, 3 is a prime number. Therefore, √3 is irrational