Question

Prove that 3underroot6 is not a rational number

Answer 1

Step-by-step explanation:

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

Remark: Similarly, we can prove that none of the number 3√2, 3√3, 3√4, 3√5, 3√7 etc. are in rational number.

In fact, if m is a positive integer which is not a perfect cube then 3√m is not a rational number.