Question

show that
$\sqrt{3}$
is irrational
​

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

hope it helps

Answer 2

Let √3 be a rational number.

So, we can write √3 in the form a/b , where a and b are co-prime numbers.

$\sqrt{3} = \frac{a}{b}$

Squaring both the sides

(√3)^2 = (a/b)^2

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

3b^2 = a^2

b^2 = a^2/3 -----(1)

So, a is divisible by 3 (from equation 1)

Let us consider a =3c

Now substituting this value in equation 1 .

${b}^{2} = \frac{{(3c)}^{2}}{3}$

${b}^{2} = \frac{ {9c}^{2} }{3}$

b^2 = 3c^2

b^2 / 3 = c^2 ----(2)

So, b is also divisible by 3 . But this contradicts our assumption that a and b are co-prime numbers. This contradiction arised from our supposition that √3 is a rational number.

So, √3 is an irrational number.