# Prove the following is irrational

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?

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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.

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.

Char. Limit
Gold Member
But the sum isn't.

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.

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

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

alxm