# Prove the following is irrational

1. Dec 5, 2009

### Pinkk

Prove that http://img705.imageshack.us/img705/2408/aaa12.png [Broken] is irrational. A user on another forum suggested the following:

http://img130.imageshack.us/img130/1352/abcde.png [Broken]

I follow that up to the last sentence. Can anyone clarify for me how to show this proof?

Last edited by a moderator: May 4, 2017
2. Dec 5, 2009

### nomather1471

if x^6 is not element Q x is not element Q too, you can see it...So it is a contradiction and x is irrational.

3. Dec 5, 2009

### Pinkk

Yes, but how do I prove that x^6 is irrational. I cannot simply say so since it is possible for the sum of two irrationals to be rational.

4. Dec 7, 2009

### Char. Limit

But the sum isn't.

5. Dec 7, 2009

### g_edgar

The real method. Work out the polynomial with integer coefficients satisfied by this. Then there is a finite algorithm for checking rational roots.

$${X}^{12}-{X}^{11}-{X}^{10}+{X}^{9}-3\,{X}^{7}+2\,{X}^{6}+2\,{X}^{5}+{X }^{4}-{X}^{3}+{X}^{2}+2\,X-5 = 0$$
To find rational roots, consider the factors of $-5$ (and the factors of the leading coefficient, 1) so you just have to try $5, -5, 1, -1$. Since none of these is a solution of that polynomial, all solutions are irrational.

6. Dec 7, 2009

### ramsey2879

If I may ask a stupid question, what rational roots would you check for to see if

15X^{12} + ..... + 14 had rational roots.

7. Dec 8, 2009

### alxm

Dang that's so frustrating. I saw this yesterday and was thinking of the rational root theorem, but was too tired to work it out :)

8. Dec 8, 2009

### g_edgar

The factors of 14 are 1,2,7,14 and the factors of 15 are 1,3,5,15, so you have to try just 32 possibilities: plus or minus a fraction, the numerator is one of 1,2,4,7 and the denominator is one of 1,3,5,15.