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A question on irrationals.

  1. Aug 3, 2006 #1
    prove for any rational root of a polynomial with integer coeffiecnts,[tex]a_{n}x^n+a_{n-1}x^{n-1}+...+a_{1}x+a_0[/tex] an doesnt equal 0.
    if written in lowest terms as p/q. that the numerator p is a factor of a0 and the denominator q is a factor of an.

    well, what i did is as follows:

    now if anp^(n-1)/q^n+...+a1/q=b/c (where b/c is in its lowest terms), i need to prove that c divides p.
    because a0 is an integer c must divide p, because it doesnt divide b.
    but this doesnt imply that p is a factor of a0, but that p/c is factor of a0.
    i dont know how to proceed from here, i know i must show that c=1, but i dnot know how.

    i would appreciate it if you could help me also with an.
  2. jcsd
  3. Aug 3, 2006 #2


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    How is this a question on irrationals?

    Suppose [itex]x_1= \frac{p}{q}[/itex] is a root of
    x2, ..., xn are the others.
    1) [tex]a_{n}x^n+a_{n-1}x^{n-1}+...+a_{1}x+a_0= a_n(x-x_0)(x-x_1)\cdot\cdot\cdot(x-x_n) [/tex]
    2) [tex](x- x_1)= x- \frac{p}{q}= \frac{1}{q}(qx- p)[/tex]
  4. Aug 3, 2006 #3
    i can see that from (an/q)(x-x0)(qx-p)...(x-xn)=anx^n+...+a1x+a0
    we can conclude that q is a factor of an, but how can i prove that p is a factor of a0?

    btw, this question in the book isnt about irrtional but it's connected to a question about irrationals which employs this fact, i forgot that this question doesnt entail anthing about it.
  5. Aug 3, 2006 #4


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    Once you've factored your polynomial, let's say you're looking at the solution which fits the factored term:


    All your terms will look exactly like that.
    Most people prefer to factor out the q, like HallsofIvy did, and leave the an as a separate constant from the rest of the polynomial, however, no one says you have to do it that way. So:


    And of course, it's clear that once the terms are all multiplied back, q will be a factor of an, and p a factor of a0, because multiplying all the constant terms is a0
  6. Aug 3, 2006 #5


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    If p/q is a rational root of the polynomial, written in lowest terms, then GCD(p,q)=1. We can write:

    [tex] a_n \left(\frac{p}{q}\right)^n+...+a_1 \left(\frac{p}{q}\right)+a_0=0[/tex]

    Multiplying across by qn:

    [tex] a_n p^n + a_{n-1} p^{n-1}q+...+a_1 p q^{n-1}+a_0q^n=0[/tex]

    Now look at this equation mod p and mod q, and use the fact that GCD(p,q)=1.
  7. Aug 4, 2006 #6
    0=(anp^n+...a1pq^n-1+a0q^n)mod q
    each term in the polynomial besides anp^n has q as its factor, so bacuse anp^n should be divisble by q, and gcd(p,q)=1, an is divisble by q. and the same goes with a0 but with p.

    thanks, status.

    i have a follow up question, prove with help of the above statement, that [tex]\sqrt 2+ 2^{\frac{1}{3}} and \sqrt 3+2^{\frac{1}{3}}[/tex]
    are irrationals.

    i guess also here i need to assume that they are rational roots of a polynomial and perhaps to prove that p isnt a factor of a0 and q isnt a factor of an, but how do i find a suiteable polynomial with integer coefficients?
    perhaps if x1 is one of the numbers and x2 is the other, then the polynomial is (x-x1)(x-x2)=0, but im not sure how to apply ad absurdum here.

    thanks in advance.
  8. Aug 4, 2006 #7


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    Consider, for example, [itex]x=2^{1/3}+3^{1/2}[/tex]. If you look at x2, x3, etc, you will quickly see that they are all rational linear combinations of [itex]1, 2^{1/3}, 2^{2/3}, 3^{1/2}, 2^{1/3} 3^{1/2}, 2^{2/3} 3^{1/2},[/itex]. Thinking of these as basis vectors in a 6 dimensional vector space over the rational numbers, the seven vectors 1,x,x2,...,x6, must be linearly dependent, and so there are some rational numbers ai with:

    [tex]a_0+a_1 x + ... + a_6 x^6=0[/tex]

    These can be found by solving a system of 6 linear equations.

    This is pretty tedious though. The only other way I can think of to find this polynomial is using Galois theory. Consider the following simple example. Say we want to find a quadratic polynomial with rational coefficients for which r=a+b[itex]\sqrt{2}[/itex] is a root, where a,b are rational. We write this polynomial as:

    [tex]x^2+cx+d = (x-r)(x-s)[/tex]

    where s is the other root. We could use the above method to find c and d. But Galois theory allows us to determine s by the following argument. First we note that if we take any true equation involving numbers of the form e+f[itex]\sqrt{2}[/itex], e and f rational, and we replace [itex]\sqrt{2}[/itex] by [itex]-\sqrt{2}[/itex] every place it occurs, we get another true equation. This means that this operation is something called an "automorphism".

    If we apply this operation to the equation above, the LHS doesn't change, since c,d are rational, and so the RHS must also remain the same. But this is only possible if s=a-b[itex]\sqrt{2}[/itex]. Expanding this out we get our minimal polymial for r.

    In the case you're aksed to do, the group of automorphisms is more complicated than the example I just gave. If you knew how to compute the group, it would give you a significant shortcut to finding the answer, but if you're not familiar with this process, you should probably try the first way I suggested.
  9. Aug 5, 2006 #8
    your second appraoch i cant still use cause i havent learnt galois theory.

    about your first approach, i said that i need to use my first question in the first post, which means to use the fact that a_is are integers, but you use this polynomial with rational coefficients which differs from this.

    i would like to see a way that incorporates the first question.

    btw, in your approach i should prove that a_i arent rational in order to prove that x isnt rational, right?
  10. Aug 5, 2006 #9


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    Well, you can always multiply through by a common denominator to get a polynomial with integer coefficients. By the way, the first method I suggested guarantees an answer, but you're probably better off just playing around with different powers of x, combining them with integer coefficients to try to construct the polynomial. If this doesn't work, use the first method, but please use a computer to do it.
    Last edited: Aug 5, 2006
  11. Aug 5, 2006 #10
    i don't think the aim of the book from which i took the problem is to solve this question with a computer aid.
  12. Aug 5, 2006 #11


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    Then don't use one. But you'll turn a 5 minute problem into an hour of tedious work, where the chances of making a mistake is just about 100%. By the way, if you do want to try to find the polynomial by just messing around, you might want to try combining the polynomials for [itex]\sqrt{3}[/itex] and [itex]2^{1/3}[/itex]. (eg, plug [itex]\sqrt{3}+2^{1/3}[/itex] into [itex]x^2-3[/itex])
    Last edited: Aug 5, 2006
  13. Aug 5, 2006 #12
    i tried messing with polynomials but didnt go very far.
    anyway the polynomial i should find is of integer coefficients, and then i should show that a0 doesnt have p as its factor and an doesnt have q as its factor, right?
  14. Aug 5, 2006 #13


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    Well, once you find the polynomial, you'd have to show it has no rational roots. By using the conditions on p and q, you can reduce the possible rational roots to some finite number, and check all of these. Then if x is a root of the polynomial, you know it isn't rational.
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