Question

Prove that √2 is irrational.
Plz answer fast......​

Answer 1

Hey buddy here is your answer

Let us assume that root 2 is a rational number.

Then that implies that

Root 2 = p/q where p and q are coprimes.

Squaring on both sides

2=p^2 /q^2

2 q^2= p^2

This implies that 2 is a factor of p^2

This also implies that 2 is a factor of p

Then there exists 2c=p where c belongs to positive integers

That implies,

(2c)^2=2q^2

4c^2 = 2q^2

2c^2 = q^2

This implies that 2 is a factor of q^2

This also implies that 2 is a factor of a

But this contradicts the fact that p and q are co primes and do not have ea common factor.

This is because of our wrong assumption.

Therefore root 2 is an irrational number.

Hope it helps

Please mark it as the brainliest.

Cheers

-Hermione

Answer 2

$\huge \boxed{ \bf{Answer}}$



$\bf{Proof} \: \div$


Let, √2 be a rational number which can be written in the form of p/q where p and q are co-prime and q ≠ 0 .


$\therefore \: \sqrt{2} = \frac{p}{q}$

$= > \: 2 = \frac{ {p}^{2} }{ {q}^{2} } \: \: \: \: \: \: \: (squaring \: both \: sides)$

$= > \: {p}^{2} = 2 {q}^{2} \: \: \: \: \: \: \: \: \: \: \: - - - > (1)$

$\: \therefore \: 2 \: is \: a \: factor \: of \: {p}^{2}$

$\therefore \: 2 \: is \: a \: factor \: of \: p$


Let, p = 2m, where m is a non-zero integer.


$= > \: {p}^{2} = {(2m)}^{2}$

$= > \: {p}^{2} = 4 {m}^{2}$

$= > 2 {q}^{2} = 4 {m}^{2} \: \: \: \: \: \: \: -- - > (from1)$

$= > \: {q}^{2} = 2 {m}^{2}$


$\therefore \: 2 \: is \: a \: factor \: of \: {q}^{2}$

$\therefore \: 2 \: is \: a \: factor \: of \: q$


So, we can see that both p and q have common factor 2, which contradicts our supposition that p and q are co - prime.


$\therefore \: \sqrt{2} \: is \: an \: \bf{irrational} \: number$