Question

Prove that (6)^1/3 is not rational

Answer 1

Let us suppose if possible 3√6 is a rational number and it is in simplest form of P/q as we did in last example √7 is not a rational number.



Then, p and q are integers having no common factor other than 1, and q ≠ 0



Now, p/q = 3√6 => (p/q)3 = 6                                        (i)

 

13 < 6 < 23 => 13 < (p/q) < 23                                      by using(i)


                  => 1< p/q < 2


If q = 1, then p is an integer such that 1<p<2


But, there is no integer between 1 and 2.


Hence, q ≠ 1 and so q >1


(p/q)3 = 6 => p3/q3 = 6


                 =>P3/q = p3/q = 6q2 [multiplying both sides by q2]



Clearly, 6q2 is an integer (since q is an integer)


Since p and q have no common factor, p3 and q have no common factor.

Also, q > 1.


So, p3/q is not an integer, while 6q2 is an integer.


Consequently, p3/q ≠ 6q2.


Thus, we arrive at a contradiction.


Our supposition is wrong.


Hence, 3√6 is not a rational number or 3√6 ≠ Rational number