Question

Prove root 3 is an irrational number

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

Given:

$\sqrt {3}$

To find:

$\sqrt {3}$ is an irrational number.

Solution:

$\sqrt {3} = \frac {p}{q}$ (Were p and q is a co-prime)

$\sqrt {3 \times q} = p$

Squaring both the side in above equation

${ \sqrt {3 \times {q}^{2}} = p^{2}$

$3 \times q^{2} = p^{2}$

if 3 is a factor of $p^{2}$

Then, 3 will also be a factor of p

$\Rightarrow Let\quad p = 3 \times m$ {where m is a integer}

Squaring both sides we get

$p^{2} = {3 \times m}^{2}$

$p^{2} = 9 \times m^{2}$

Substitute the value of $p^{2}$ in the equation

$3 \times q^{2} = p^{2}$

$3 \times q^{2} = 9 \times m^{2}$

$q^{2} = 3 \times m^{2}$

If 3 is a factor of $q^{2}$

Then, 3 will also be factor of q

Hence, 3 is a factor of p & q both

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

So, $\sqrt {3}$ is an "irrational number". Hence proved.