Question

1. Prove that √3 is irrational. *
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Answer 1

Solution:

${\sf{\underline{Let\;assume\;that\;\sqrt{3}\;is\;rational\;number,}}}$

$\sf{\implies \sqrt{3}=\dfrac{x}{y}\;\;\;(Where\;x\;and\;y\;is\;co-prime\;number)}$

$\sf{\implies \sqrt{3}y=x}$

$\sf{\underline{Squaring\;both\;the\;sides,}}$

$\sf{\implies (\sqrt{3}y)^{2}=x^{2}}$

$\sf{\implies 3y^{2}=x^{2}}$

$\sf{\underline{If\;3\;is\;factor\;of\;x^{2}.\;So,\;3\;is\;also\;factor\;of\;x.}}$

$\sf{Let,\;x=3m\;\;\;(Where,\;m\;is\;integer)}$

$\sf{\underline{Squaring\;both\;the\;sides,}}$

$\sf{\implies x^{2}=9m^{2}}$

$\sf{\underline{Substitute\;the\;value\;of\;x^{2},}}$

$\sf{\implies 3y^{2}=9m^{2}}$

$\sf{\implies y^{2}=\dfrac{9m^{2}}{3}}$

$\sf{\implies y^{2}=3m^{2}}$

$\sf{\underline{If\;3\;is\;factor\;of\;y^{2}.\;So,\;3\;is\;also\;factor\;of\;y.}}$

$\sf{\underline{Hence,\;3\;is\;factor\;of\;both\;x\;and\;y.}}$

$\sf{\underline{Our\;assumption\;is\;wrong.}}$

$\bf{\underline{Hence,\;\sqrt{3}\;is\;irrational\;number.}}$

Answer 2

To Prove :-

• √3 is irrational.

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Solution :-

If possible, let √3 be rational and let it's simplest form be a/b.

Then, a and b are integers having no common factor other than 1, and b ≠ 0

Now,

√3 = a/b => 3 = a²/b² [ on squaring both sides ]

=> 3b² = a² ___1)

=> 3 divides a² [ since, 3 divides 3b² ]

=> 3 divides a

[ since, 3 is prime and 3 divides a² => 3 divides a ]

Let a = 3c for some integer c.

Putting a = 3c in (1) , we get

3b² = 9c² => b² = 3c²

=> 3 divides b² [ since, 3 divides 3c² ]

=> 3 divides b

[ since, 3 is prime and 3 divides b² => 3 divides b ]

Thus, 3 is a common factor of a and b.

But this contradicts the fact that a and b have no common factor other than 1.

The contradiction arises by assuming "√3 is irrational".

Hence, √3 is irrational.

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