Question

Prove that √3 is an irrational number​

Answer 1

Let us assume on the contrary that √3 is a rational number.

Then,

There exist positive integer a and b such that

${\small{\boxed{\green{ \sqrt{3} = \frac{a}{b} }}}}$

Where, a and b are co prime i.e.

There HCF is 1

Now,

${\sf{\implies{ \sqrt{3} = \frac{a}{b} }}}$

${\sf{\implies{3 = \frac{a {}^{2} }{b {}^{2} } }}}$

${\sf{\implies{3b {}^{2} = a {}^{2} }}}$

${\sf{\implies{3 \: divides \: a {}^{2} \: \: \: \: \: \: }}}$

${\sf{\implies{3 \: divides \: a \: \: \: \: \: \: ...(1) }}}$

${\sf{\implies{a = 3c \: for \: some \: integer \: c \: }}}$

${\sf{\implies{a {}^{2} = 9c {}^{2} }}}$

${\sf{\implies{3b { }^{2 } = 9c {}^{2} }}}$

${\sf{\implies{b {}^{2} = 3c {}^{2} }}}$

${\sf{\implies{3 \: divides \:b {}^{2} }}}$

${\sf{\implies{3 \: divides \: b \: \: \: \: \: \: ...(2)}}}$

From (1) and (2) we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co prime. This means our assumption is not correct.

Hence, √3 is an irrational number.

$\huge\sf\underline\pink{Hence \: Proved ! }$