Prove the following is irrational

  • Thread starter Pinkk
  • Start date
  • #1
14
0
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:

Answers and Replies

  • #2
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
14
0
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
Char. Limit
Gold Member
1,204
14
But the sum isn't.
 
  • #5
607
0
The real method. Work out the polynomial with integer coefficients satisfied by this. Then there is a finite algorithm for checking rational roots.

More detail. Your number satisfies
[tex]
{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
[/tex]
To find rational roots, consider the factors of [itex]-5[/itex] (and the factors of the leading coefficient, 1) so you just have to try [itex]5, -5, 1, -1[/itex]. Since none of these is a solution of that polynomial, all solutions are irrational.
 
  • #6
841
0
The real method. Work out the polynomial with integer coefficients satisfied by this. Then there is a finite algorithm for checking rational roots.

More detail. Your number satisfies
[tex]
{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
[/tex]
To find rational roots, consider the factors of [itex]-5[/itex] (and the factors of the leading coefficient, 1) so you just have to try [itex]5, -5, 1, -1[/itex]. 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.
 
  • #7
alxm
Science Advisor
1,842
9
The real method. Work out the polynomial with integer coefficients satisfied by this. Then there is a finite algorithm for checking rational roots.
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
607
0
If I may ask a stupid question, what rational roots would you check for to see if

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

Related Threads on Prove the following is irrational

Replies
21
Views
8K
  • Last Post
3
Replies
60
Views
57K
Replies
4
Views
23K
Replies
23
Views
8K
  • Last Post
Replies
15
Views
7K
  • Last Post
Replies
3
Views
2K
Replies
8
Views
6K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
3K
Top