Question

Prove that roote 3 is an irrationnal 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

Answer:

Let us suppose that root 3 is irrational

Whereas a and b are coprime and b is not equal to 0

root 3 = a/b

root 3 b = a

Squaring Both Sides

( root 3 b)² = a²

3b² = a²

a² / 3 = b²

Hence , 3 divides a²

a/3 = c

So, a = 3c

3b² = a²

Putting a = 3c

3b² = (3c)²

3b² = 9b²

b²/3 = c²

So, 3 divides

3 divides both a and b

Hence 3 is a factor of a and b

Therefore, a and b are not coprime

Hence , our supposition is wrong which contradicts a and b are coprime

root 3 is irrational