Question

prove irratiobal no. √3​

Answer 1

On the contrary,Let √3 be a rational of the form p/q.

√3=p/q

where p and q are co primes

Squaring on both sides,

→3=p²/q²

→ p²=3q² or q²=p²/3

Implies,3 divides p²

•According to the Fundamental Theorm of Arithmetic,

p should also divide 3

Now,

let p=3r,for some positive integer 3r

Squaring on both sides,

→p²=9r²

→3q²=9r²

→q²=3r²

→r²=q²/3

Thus,q² is divided by 3

and 3 also divides q

It was assumed that p and q are co primes but they more than one factor

So,our assumption is wrong

Hence,√3 is an irrational

Answer 2

Given:-

$\sqrt{3}$

To prove :-

It is an irrational number.

Proof:-

Let us assume that √3 is a rational number then, it can be expressed in the form of p/q where "p" and "q" are co-primes.

Therefore,

$\frac{p}{q} = \sqrt{3}$

$p = q \sqrt{3}$

squaring on both sides.

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

${p}^{2} = 3 {q}^{2}$----eq.1

$\implies$ here, p² is divisible by 3

$\implies$ also, p is divisible by 3.

Let p = 3r for some integers r.

put p = 3r in eq. 1

${(3r)}^{2} = 3 {q}^{2}$

$9 {r}^{2} = 3 {q}^{2}$

$3 {r}^{2} = {q}^{2}$

$\implies$ q² is divisible by 3

$\implies$ also, q is divisible by 3.

since, p and q are both divisible by 3.

$\</strong><strong>t</strong><strong>h</strong><strong>e</strong><strong>r</strong><strong>e</strong><strong>f</strong><strong>o</strong><strong>r</strong><strong>e</strong><strong>$ 3 is the common factor of both "p" and "q".

But, this contradict the fact that

p and q have no common factor than 1.

This contradiction has arisen due to our wrong assumption.

hence, √3 is an irrational number.

$\large</strong><strong>\</strong><strong>r</strong><strong>e</strong><strong>d</strong><strong>{\underline{\boxed{ </strong><strong>Theorm</strong><strong> </strong><strong>used</strong><strong>}}}$

Let p be a prime number.If p divides q² ,then p divides q where q is a positive integer.