Question

prove that √3,√5,√7 is an irrational no.​

Answer 1

it is an irrational numbers because when we divide them they never terminate or repeat

Answer 2

$\bold{TO\:PROVE:}$ √3 is an irrational number.

This type of of proofs are done taking as a case of $\bold{CONTRADICTION}$

$\bold{PROOF:}$

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 )

Now,

if 3 is the factor of p²

then,

3 is also a factor of p . .... ( ii )

=> Let p = 3m { where m is any integer }

now, squaring on both sides,

we get,

=> p² = (3m)²

=> p² = 9m²

putting the value of p² in equation ( i ),

we get,

=> 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,

√$\bold{3}$ is an $\bold{irrational}$ number.

Similarly,

√5 and √7 can also be proved as irrational numbers.