Scientific Forums


 

Add reply · Start new topic ·


> anyone know how to prove this question
MonkiezDevil
  Posted: Dec 27 2006, 09:21 AM


Newbie
*

Group: Members
Posts: 14
Joined: 17-November 06

Positive Feedback: 0%
Feedback Score: 0


Theorem : the nth root of an integer
is either an integer or irrational.
Prove the more general result:
Theorem: If x is a root of the polynomial xm + c1 xm-1 + c2 xm-2+ ... + cm = 0,
(with the coefficients c1, c2, ... , cm all integers,
of course), then x is either an integer or irrational.
(Hint: suppose a/b is a root of the polynomial, and a and
b are relatively prime. Plug into the equation above and multiply
both sides by bm. Show that b must equal 1.)

Incidentally, this result naturally leads to the
definition of a quadratic integer .

Please teach me . Thank you very much
Top
fivedoughnut
  Posted: Dec 27 2006, 09:25 AM


Member of the "forum mafia"
*****

Group: Members
Posts: 1667
Joined: 13-November 05

Positive Feedback: 57.14%
Feedback Score: 32


How do you prove a question?.... only an answer surely! blink.gif
Top
AlphaNumeric
Posted: Dec 27 2006, 11:21 AM


Professional mathematician
*****

Group: Power Member
Posts: 10336
Joined: 16-June 06

Positive Feedback: 84.15%
Feedback Score: 420


Let the n'th root of an integer, k, would be r_i, so (r_i)^n = k, for n different r_i.

Therefore you are solving the polynomial x^n = k.

Suppose x = (a/b) where a and b are COPRIME. You have (a/b)^n = k, so a^n = k(b^n).

Consider prime factorisation, there MUST be the same prime factorisaton on both sides of that equation or they aren't equal. Since a and b are coprime, a contains none of the prime factors b does. But that means they can't possibly be equal unless b contains the only factor it could (ie it has no prime factorisation) is 1 (1 is not a prime!)

Therefore b=1 [i]provided[/b] x is rational. Therefore the root must be integer if rational. If it's not an integer (b not 1), then you've just shown that it can't be rational (A -> B has the equivalent logic statement ¬B -> ¬A). Therefore it's irratinal.

Can you extend this method to the second part of the question?


--------------------
The views in the above post are those of its author and not those of the people who educated him through a degree and masters, supervised him or collaborated with him during his PhD, paid him to teach and mark undergraduate mathematics and physics courses or who pay him to do research now.

Any insults, flames or rants are purely the work of the author and not said people or institutions. Cranks are not suffered well.
Top

Topic Options

Add reply · Start new topic ·


 

Terms of use