Question

Prove that √2 is irrational

Answer 1

Sol:
Given √2 is irrational number.
Let √2 = a / b wher a,b are integers b ≠ 0
we also suppose that a / b is written in the simplest form
Now √2 = a / b ⇒ 2 = a2 / b2 ⇒ 2b2 = a2
∴ 2b2 is divisible by 2
⇒ a2 is divisible by 2
⇒ a is divisible by 2
∴ let a = 2c
a2 = 4c2 ⇒ 2b2 = 4c2 ⇒ b2 = 2c2
∴ 2c2 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.


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Answer 2

Answer :

To prove,

√2 is an irrational no.

Proof :

Let √2 be a rational number in the form of p / q where q is not equal to zero at p and q are co-prime integers.

√2 = p/q

Whole sqauring both sides of this equation :-

2 = p^2/q^2

p^2 = 2q^2 (I)

From (I),

2 divided p^2

So, p divides p. (a)

Now , let p= 2k where k is any integer.

Substituting the values , we get :-

(2k)^2 = 2q^2

4k^2 = 2q^2

q^2 = 2k^2 (ii)

From (ii),

2 divides q^2.

Therefore, 2 divides q also. (b)

From statements (a) and (b) , we can say that :-

p and q have a common factor namely 2.

Hence, our assumption that p and q are co-prime is wrong. Hence , √2 is an irrational no.

Hence proved.

This method is called contradiction method.