Question

Prove that root 3 is an irrational numver

Answer 1

Answer:

Step-by-step explanation:

here is ur answer hope it helps u

by the method of contradiction

let us assume that √3  as irrational

so ,

√3= a/b

b√3 = a

squaring on both sides  , we get

3b² = a² -------------------------------------------------------- ( 1  )

if 3 divides a² then it also divides a

so ,

a = 3c

squaring on both sides , we get

a² = 9c²---------------------------------------------------------(  2  )

substitute equation( 1 )in equation (2 )

we get

3b² = 9 c²

b² = 3 c²

here a b and c have 3 as common factor

so 3 is rational number

but ,

√3 is irrational

this contradiction has arisen due to our incorrect assumption

this contradicts the fact that ,

√3 is a irrational number

hope it helps u

Answer 2

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

Hope it helps u.