Question

Prove that √3 is an irrational number.

class 10​

Answer 1

Answer:

Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m,n∈N. ... Therefore there exists no rational number r such that r2=3. Hence the root of 3 is an irrational number.

Answer 2

Question:- Prove that √3 is an irrational number.

Answer:- Let is assume that√3 is a rational number.

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

P and Q and hence co-primes, having common factor 1.

Bringing q to the LHS,

$\sqrt{3} q = p$

squaring on both the sides,

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

${ \sqrt{9} }q^{2} = {p}^{2}$

$3 {q}^{2} = {p}^{2} - - - - - (1)$

$3 \: divides \: {p}^{2} = > \: 3 \: divides \: p$

$p = 3m - - - - - (2)$

Substitute equation (2) in equation (1)

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

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

Cancelling the values,

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

Bringing 3 to LHS,

$3 \: divides \: {q}^{2} = > \: 3 \: divides \: q$

$q = 3n - - - - - (3)$

From equations 2 and 3,

We know that P and Q have common factor 3.

Therefore they are not co-primes.

There is a contraction to our assumption.

Hence it's irrational.

Hence, Proved.