Question

how to prove that square root of any prime number is irrational​

Answer 1

Answer:

Going with the answers here, you already have a2=p×b2

Since a2 is a perfect square, the right hand side of the equality p×b2 must also be a perfect square.

Since, b2 is already a perfect square, p must be a perfect square. And we know that no prime number is a perfect square.

Answer 2

To Prove:

Square root of any prime number is irrational.

Proof:

Let's take prime number 3 .

Then,

we need to prove $\sqrt{3}$ is an irrational number.

Step-by-step explanation:

This type of Question are solved by Contradiction

So,

Let us assume that √3 is a rational number.

Then,

as we know that,

a rational number should be in the form of $\frac{p}{q}$

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 sides

we get,

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

$3{q}^{2} = {p}^{2}$ ........ ( i )

So,

if 3 is the factor of p²

then,

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

Therefore,

Let,

p = 3m { where m is any integer }

Now squaring both sides,

we get,

$=>{p}^{2} = {(3m)}^{2}$

$=>{p}^{2}=9{m}^{2}$

Therefore,

putting the value of p² in equation ( i )

we get,

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

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

Hence,

√3 is an irrational number

Thus,

Proved