Question

How To Prove √3 Is Irrational

Answer 1

$\huge\boxed{\fcolorbox{orange}{orange}{Answer}}$

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

Let us assume that √3 is a rational number.

Rational numbers are 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}

√3 q = p

Now, by squaring both the sides, we get,

($(\sqrt{3}q)^{2} = p^{2} \\3q^{2} = p^{2}$

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

$p^{2} = (3m)^{2}\\p^{2} = 9 m^{2}$

putting the value of p² in equation ( i )

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

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

$\boxed {\boxed {\underline { \bold{Therefore \sqrt{3} \ is \ an \ irrational \ number.}}}}$