Prove that√6 is irrational

Let √6 be a rational number , then √6 = p÷q , where p,q are integers , q not = 0 and p,q have no common factors ( except 1 ) => 6 = p² ÷ q² => p² = 2q² ................(i) As 2 divides 6q² , so 2 divides p² but 2 is a prime number > 2 divides p..... Let p = 2m , where m is an integer ....... Substituting this value of p in (i) , we get ,...... (2m)² = 6q² ....... => 2m² = 3q² ....... As 2 divides 2m² , 2 divides 3q² ....... => 2 divides 3 or 2 divide q² ...... But 2 does not divide 3 , therefore , 2 divides q²........ => 2 divides q ....... Thus , p and q have a common factor 2 . This contradicts that p and q have no common factors ( except 1 )....... Hence , our supposition is wrong . Therefore , √6 is an irrational number....

Square root hai cube root hai

Assume cube root 6 is rational. Then let cube root 6 = a/b ( a & b are co-prime and b not = 0)
 Cubing both sides : 6=a^3/b^3
 a^3 = 6b^3
 a^3 = 2(3b^3) 
Therefore, 2 divides a^3 or a^2 * a . By Euclid's Lemma if a prime number
 divides the product of two integers then it must divide one of the two integers
 Since all the terms here are the same we conclude that 2 divides a
.
 Now there exists an integer k such that a=2k
 Substituting 2k in the above equation
 8k^3 = 6b^3
 b^3 = 2{(2k^3) / 3)}
 Therefore, 2 divides b^3. Using the same logic as above. 2 divides b.
 Hence 2 is common factor of both a & b. But this is a contradiction of the fact that a & b are co-prime. Therefore, the initial assumption is wrong.
 
cube root 6 is irrational.