Question

proof √2 is irrational no.

Answer 1

Let us assume that √2 is rational .Then it will be of the form p/q where p and q are coprime (p and q have no common factor other than 1),q not equal to 0

Now,√2=p/q
On squaring both sides
2= p^2/q^2

2q^2=p^2----(1)
From theorem that states :if p divides a,^2, then p divides a.

So, 2 divides p^2
2 divides p.
Then p can be written as 2c where C is an integer .
On putting p=2c in equation (1)
We get , m 2q^2=(2c)^2
2q^2 = 4c^2
q^2=2c^2

So 2 divides q^2
2 divides q

Thus 2 is a common factor for p and q. But this contradict the fact that p and q are coprime .

Hence √2 is irrational.



Hope it helps!!!

Answer 2

$\huge\star\mathfrak\pink{{HEY MATE...!!}}$

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\pink{R}}}}$

PROOF:

. let us assume that$\sqrt{2}$ is not an irrational.

$\sqrt{2}$ is rational.

$\sqrt{2}$ = $\frac{p}{q}$ where p,q= 0 q is $\cancel{=}$ 0. p,q are co - primes.

squaring on both sides.

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

2= $\frac{p{^2}}{q{^2}}$

${p}^2$= ${2×{q}^2}$

${p}^2$ is divisible by 2.

${p}$ is also divisible by 2.

$\boxed{let p= 2k.....1}$

from equation 1

$({2k}^2)$ = ${2×{q}^2}$

$\cancel{4}$${k}^2$= $\cancel{2}$${q}^2$

${2{k}^2}$ = ${q}^2}$

= ${q}^2}$ =${2{k}^2}$

${q}^2$ is divisible by 2 .

${q}$ is also divisible by 2.

2 is the common factor for both p,q but p,q are co - primes

It is contradiction to our assumption

oure assumption is wrong.

$\sqrt{2}$ is not a rational

$\sqrt{2}$ is an irrational.

$<marquee>#Be Brainly...✔ </marquee>$.