Question

prove that√2 is irraational​

Answer 1

Answer:

proof = let us assume that √2 is rational no.

Step-by-step explanation:

since rational no √2= p/q ; hcf(p,q) = 1 ; q not eqal to 0 and p, q belongs to integers

i. e. p/q = √2

s. o. b. s

p2/q2= 2

p2=2q2

here 2 divides p2

2 divides p......... 1

since 2 divides p

let p be 2m where m is some integer

from above 2m = q2

s. o. b. s

q2= 4m2

q2= 2m2

here 2 divides q2

2 divides q

since here 2 divides both q and p

hcf(p,q)=2

but above hcf of p, q is 1

this contradiction arises if our assumption is wrong

therefore our assumption is wrong

and hence √2 is irrational

Answer 2

Let us assume that $\sf\sqrt{2}$ is an rational number.

So, it can be written in the form of $\sf\frac{a}{b}$

Such that, $\sf\sqrt{2}$ = $\sf\large\frac{p}{q}$

$\sf\sqrt{2}b$ = a

Now, Squaring both Sides:-

$\sf \; 2b^2 \; = \; a$

2 divides a²

So, 2 is also Divisible by a.

We can write a = 3c for some integer c.

$\implies\sf \; 2b^2 \; =\; 4c$ that b² = 2c².

This means, 2 divides b² and 2 is also divisible by b.

Therefore, a and b both have 2 as a common factor.

it arises contradiction because of our wrong assumption that $\sf\sqrt{2}$ is rational number.

Hence, $\sf\sqrt{2}$ is irrational number.

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