Question

√3 is a irrational number are not

Answer 1

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

Answer 2

√3 is irrational number

__________ [TO PROVE]

Let us assume that, √3 is a rational number

√3 = $\dfrac{a}{b}$

Here, a and b are co-prime numbers.

• Squaring on both sides, we get

=> (√3)² = $( { \dfrac{a}{b}) }^{2}$

=> 3 = $\dfrac{ {a}^{2} }{ {b}^{2} }$

=> 3b² = a² ________( eq 1)

Clearly;

a² is divisible by 3.

So, a is also divisible by 3.

Now, let some integer be c.

=> a = 3c

=> a² = 9c²

=> 3b² = 9c² [From (eq 1)]

=> b² = 3c² _______(eq 2)

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

3 divides b² and 3 divide a² also.

So, our assumption is wrong.

√3 is irrational number.

________________________________