Question

Prove that√2 is irrational

Answer 1

Answer:

let us suppose √2 is irrational number

= √2 = p/q   (p and q are any co-prime integer)

  squaring both side

= 2 = p²/q²

= q²= p²/2 -- 1

p² is divisible by two

p is also divisible by two

= p/2 = r  (r is an integer)

= p = 2 r

putting the value of p in eq. 1

= q² = (2 r)²/2

= q² = 4 r²/2

= q²= 2 r

= q²/2 = r²

q² is divisible by 2

q is also divisible by two

therefore: p and q are not co-prime integer

so, our supposition is wrong

so, √2 is an irrational number

Step-by-step explanation:

Answer 2

Solution:

Let us assume that, √2 is a rational number of simplest form $\frac{a}{b}$, having no common factor other than 1.

√2 = $\frac{a}{b}$

On squaring both sides, we get ;

2 = $\frac{a^{2}}{b^{2}}$

⇒ a² = 2b²

Clearly, a² is divisible by 2.

So, a is also divisible by 2.

Now, let some integer be c.

⇒ a = 2c

Substituting for a, we get ;

⇒ 2b² = 2c

Squaring both sides,

⇒ 2b² = 4c²

⇒ b² = 2c²

This means that, 2 divides b², and so 2 divides b.

Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b have no common factor other than 1.

This contradiction has arises because of our assumption that √2 is rational.

So, we conclude that √2 is irrational.