Question

Theorem 1.5

Class 10

√3 is an irrational no prove​

Answer 1

Solution:

Let us assume that, √3 is a rational number of simplest form $\frac{a}{b}$, having no common factor other than 1.

         ∴√3 = $\frac{a}{b}$

On squaring both sides, we get ;

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

⇒ a² = 3b²

Clearly, a² is divisible by 3.

So, a is also divisible by 3.

Now, let some integer be c.

⇒ a = 3c

Substituting for a, we get ;

⇒ 3b² = 3c

Squaring both sides,

⇒ 3b² = 9c²

⇒ b² = 3c²

This means that, 3 divides b², and so 2 divides b.

Therefore, a and b have at least 3 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 √3 is rational.

So, we conclude that √3 is irrational.

Answer 2

$\huge \bold \pink{heya \: mate}$

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )

=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q

Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number