 Hello and welcome to the session. In this session we discussed the following question which says show that cube root 6 is irrational. Let's move on to the solution. We need to show that cube root 6 is irrational. So for this let us assume that cube root 6 is rational that we have assumed to the contrary. Now that cube root 6 is rational, so we can find integers a and b where the integers a and b have no common factors other than 1. And also integer b is not equal to 0. And we can write cube root 6 equal to a upon b where a and b are integers such that a and b have no common factors other than 1 and b is not equal to 0. Now cubing both sides we get 6 is equal to a upon b whole cube. Now since 1 cube is equal to 1 and 2 cube is equal to 8 therefore we have that a upon b is greater than 1 and less than 2. Now b would be greater than 1 since if we take b equal to 1 then a upon b will be an integer and we know that there is no integer between 1 and 2 therefore b would be greater than 1 and b cannot be less than 1 since we have taken that b is not equal to 0. Now 6 is equal to a upon b whole cube so this means we have a cube upon b cube is equal to 6. Now multiplying both sides by b square we get a cube upon b is equal to 6 b square or we have 6 b square is equal to a cube upon b. Now that we know that b is an integer so this means that 6 b square would also be an integer then b is greater than 1 and as we had assumed that a and b do not have common factors other than 1 so b is greater than 1 and b does not have a common factor with a and so consequently with a cube. So this means that a cube upon b is a fraction. Now we have 6 b square is equal to a cube upon b and 6 b square is an integer and a cube upon b is a fraction thus 6 b square is not equal to a cube upon b but this is a contradiction as we had that 6 b square is equal to a cube upon b therefore our assumption is wrong thus we have that cube root 6 is irrational since we had assumed that cube root 6 is rational so we finally get that cube root 6 is irrational hence proved so this completes the session hope you have understood the solution for this question.