Question

Prove that $\sf \sqrt{3}$ is an irrational number.​

Answer 1

Answer:

Let us assume to the contrary that $\sf \sqrt{3}$ is a rational number.

Therefore, it can be expressed in the form of $\sf \dfrac{p}{q}$. where, p and q are co - primes and q ≠ 0.

$:\implies\sf \sqrt{3} = \dfrac{p}{q}$

✇ $\sf \underline{Now,\; Squaring\;both\;sides\;-}$

$:\implies\sf ( \sqrt{3})^2 = \bigg( \dfrac{p}{q} \bigg)^2$

$:\implies\sf 3 = \dfrac{p^2}{q^2}$

$:\implies\sf \red{3q^2 = p^2\;\;\;\;\;[1]}$

Therefore, we can say that,

$\dashrightarrow$ 3 divides q²

$\dashrightarrow$ Also, 3 also divides q.

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★ $\sf \pink{Lets\; p = 3r}$

where r is some integer.

$:\implies\sf \red{p^2 = 9r^2\;\;\;\;\;[2]}$

✇ $\sf \underline{Now,\;From\; eq(1) \;and \;(2) -}$

$:\implies\sf 3q^2 = 9r^2$

$:\implies\sf q^2 = 3r^2$

Therefore, we can say that,

$\dashrightarrow$ q² is a multiple of 3.

$\dashrightarrow$ Also, q is a multiple of 3.

Now, We can say that, p,a have a common factor of 3.

Therefore, Our supposition is wrong p and q are not co - prime.

Hence, It becomes a contradiction.

p/q is not a rational number.

$\sf {\underline{\purple{\dag\;Hence,\; \sqrt{3}\;is\;an\; irrational\;no.}}}$

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Answer 2

See the above answer for explanation! ✌✌