1
$\begingroup$

So I was wondering that why a root that is not rational needs its conjugate for the polynomial's root in order to make a polynomial with only integer coefficient? Why can't a root, say, $1-(2)^{1/2}$ , have a root that is $2+2(2)^{1/2}$? Isn't that also possible? Any help will be appreciated!

  • 0
    For square roots, remember the quadratic formula, in particular the $\pm \sqrt{\Delta}\,$ part. For higher order radicals, lookup rational conjugates or Galois conjugates. For your specific example, note that the sum of those two numbers is an irrational, so by Vieta's formulas they can't be both roots of the same quadratic with integer coefficients. – 2017-02-04
  • 0
    But let's say that there is a high order polynomial that has multiple irrational roots. Couldn't the polynomial simultaneously contain both roots $1-2^{1/2}$ and $2+2*2^{1/2}$ and have some other roots that cancel out all all the irrational part when summing? I know that there are second symmetric sum, third symmetric sum etc. that restrict it, but is there a way of rigorous proof, that if $1-2^{1/2}$ is a root of the polynomial with integer coefficients, its conjugate $1+2^{1/2}$, must also be a root????? – 2017-02-04
  • 1
    The fundamental property that you are looking for is that every algebraic irrational (which includes all expressions with radicals of rationals) has a minimal rational polynomial. Then every other rational polynomial that has that algebraic irrational as a root, must also have *all* the rest of the roots of its minimal polynomial as own roots. In the case of $1-\sqrt{2}$ the minimal polynomial is $x^2-2x-1=0$ which has the roots $1 \pm \sqrt{2}$. So any polynomial with rational coefficients which has $1 - \sqrt{2}$ as a root must also have $1 + \sqrt{2}$ as a root. – 2017-02-04
  • 0
    Is there a specific theorem that you are talking about, concerning to this idea of minimal rational polynomial?If so, Is this concept out of reach for a pre-calculus student? – 2017-02-04
  • 0
    I posted a fully worked-out example for the $1-\sqrt{2}$ case. For the more general context, you may want to lookup [minimal polynomials](https://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory)) and [splitting fields](https://en.wikipedia.org/wiki/Splitting_field). – 2017-02-04

2 Answers 2

1

To answer the $1-\sqrt{2}$ question: let $P(x)$ be a polynomial with rational coefficients having $1-\sqrt{2}$ as a root i.e. $P(1-\sqrt{2})=0\,$. Note that $1 \pm \sqrt{2}$ are the roots of $x^2-2x-1=0$. The Euclidian division of $P(x)$ by $x^2-2x-1$ will give a quotient $Q(x)$ and a remainder $R(x)$ with $\deg R \lt 2\,$, both with rational coefficients:

$$P(x)=(x^2-2x-1)Q(x)+r_0+r_1 x \quad\big|\quad r_0,r_1 \in \mathbb{Q}$$

Substitute $x=1-\sqrt{2}\,$:

$$0 = P(1-\sqrt{2})=0\cdot Q(1-\sqrt{2})+r_0+r_1 (1-\sqrt{2}) \;\;\implies\;\; r_0+r_1 - r_1 \sqrt{2} = 0$$

Since $\sqrt{2}$ is (known to be) irrational, the latter equality implies $r_0=r_1=0\,$. It follows that $x^2-2x-1 \mid P(x)\,$, and therefore $P(x)$ has $1+\sqrt{2}$ as a root as well.

0

Note that in the example which you give, roots $1-\sqrt{2},\,2+2\sqrt{2}$ that this would give corresponding factors $[x-(1-\sqrt{2}]$ and $[x-(2+2\sqrt{2})]$ with product

\begin{eqnarray} [x-(1-\sqrt{2}]\cdot[x-(2+2\sqrt{2})]&=&[(x-1)-\sqrt{2}]\cdot[(x-2)-2\sqrt{2}]\\ &=&(x-1)(x-2)-2\sqrt{2}(x-1)-\sqrt{2}(x-2)+4\\ &=&x^2-(3+3\sqrt{2})x+(6+4\sqrt{2}) \end{eqnarray}

  • 0
    what if these two are not the only two roots of the polynomial? What if it is a polynomial with multiple irrational roots that just simply contain these two roots? – 2017-02-04
  • 0
    That's a valid objection to my response. See the comment of @dxiv above following your post concerning the minimal rational polynomial. – 2017-02-04