Question

☺️☺️50 points_________

♣️ Q.- prove that √2 is an irrational no.

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Answer 1

Hey mate,.

Let us assume that √2 is not an irrational number.

√2=p/q ( where (p,q)=1,p,q€Zand qnot equal to zero)

squaring on both sides

2=p²/q²----->1

2divides p²=>2divides p ----->2

Let p=2m

p²=4m²

putting the value of p² in 1

we get,

2q²=4m²

q²=2m²

here,

2. divides q²=>2 divides q--->3

Thus,

Both q and p have same common factor.

So, This contradicts to our assumption.

Hence ,√2 is irrational number.

Hope it will help you.

✨sai.

Answer 2

$\huge \mathbb{Answer}$

Given :√2 is irrational number.

Let √2 = a / b [where a,b are integers b ≠ 0 we also suppose that a / b is written in the simplest form]

Now √2 = a / b

⇒ 2 = a^2 / b^2

⇒2b^2 = a^2

∴ 2b^2 is divisible by 2

⇒ a^2 is divisible by 2

⇒ a is divisible by 2

∴ let a = 2c a^2 = 4c^2

⇒ 2b^2 = 4c^2

⇒ b^2 = 2c^2
∴ 2c^2 is divisible by 2
∴ b2 is divisible by 2
∴ b is divisible by 2
∴a are b are divisible by 2 .

This contradicts our supposition that a/b is written in the simplest form Hence our supposition is wrong

∴ √2 is irrational number.