Question

prove root 2 is irrational

Answer 1

$\huge\bf\mathscr\pink{Your\: Answer}$

step-by-step explanation:

Let us assume that √2 is a rational number.

That is,

we can find integers a and b (≠ 0)

such that

√2 = (a/b)

Now,

 Suppose a and b have a common factor other than 1,

then,

we can divide by the common factor,

and,

assume that a and b are coprime. 

Now,

√2b = a 

Now,

squaring on both sides,

we get,

⇒ 2b^2 = a^2 .......................(i)

Therefore,

a^2 is divisible by 2

 Hence ‘a’ is also divisible by 2.

So,

we can write

a = 2c for some integer c.

 Equation (i) becomes, 

2b^2 =(2c)^2

 ⇒ 2b^2 = 4c^2 

∴ b^2 = 2c^2 

This means that b^2 is divisible by 2,

and so

b is also divisible by 2.

Therefore,

a and b have at least 2 as a common factor. 

But,

this contradicts the fact that a and b are coprime.

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

 So, we conclude that √2 is irrational

Answer 2

$\pink{\huge{\bold{Solution:}}}$

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.