The Irrationality of Problem:
Prove that 
 is an irrational number.
The number, 
, is irrational, ie., it cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us assume that is rational so that we may write
 = a/b1.for a and b = any two integers. We must then show that no two such integers can be found. We begin by squaring both sides of eq. 1:3 = a2/b22.or3b2 = a22a.If b is odd, then b2 is odd; in this case, a2 and a are also odd. Similarly, if b is even, then b2, a2, and a are even. Since any choice of even values of a and b leads to a ratio a/b that can be reduced by canceling a common factor of 2, we must assume that a and b are odd, and that the ratio a/b is already reduced to smallest possible terms. With a and b both odd, we may writea = 2m + 13.andb = 2n +14.where we require m and n to be integers (to ensure integer values of a and b). When these expressions are substituted into eq. 2a, we obtain3(4n2 + 4n + 1) = 4m2 + 4m + 15.Upon performing some algebra, we acquire the further expression6n2 + 6n + 1 = 2(m2 + m)6.The Left Hand Side of eq. 6 is an odd integer. The Right Hand Side, on the other hand, is an even integer. There are no solutions for eq. 6. Therefore, integer values of a and b which satisfy the relationship  = a/b cannot be found. We are forced to conclude that  is irrational.

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Let us assume √3 as rational number for sometime ...

⇒ √ 3 = a /b { Where a & b are some integers , co - prime numbers and b ≠ 0}

Now square on both the sides ,

 (\sqrt{3})   ^{2} =  (\frac{a}{b} )  ^{2}

3 =  \frac{a^{2} }{ b^{2} }

3 b^{2}= a ^{2}

→ Where 3 divided's a² and also 3 divides b too. ( Let 7 be a prime number and   a be a positive number ,so it divides a and a²) ⇒ Equation 1

Now , Let a = 3c
Now , substiute this in eq - 1

3 b^{2} =  (3c)^{2}

3 b^{2} = 9 c^{2}

 b^{2} =   3c^{2}

3c ^{2} =  b^{2}

Where 3 divides b² and also 3 divided b (Let 3 be a prime number and a be a positive   number , so, it divides a and a² )

⇒ So , 3 divided a and b..

⇒ Where a and b are co- prime numbers and 7 is  factor of a and b

→ Hence , it is contradicting .

∴ √3 is an irrational number