Good day.

Suppose $a$ is rational number, $p$ is positive integer and $a^{1/p}$ is irrational. If we want to eliminate irrationality in the denominator of the fraction $\frac{1}{a^{1/p}}$, then there is obvious solution $\frac{a^{1-1/p}}{a}$.

Some situations like $\frac{1}{a^{1/n} - b^{1/n}}$ are also nice. But what if we have a more general situation - $\frac{1}{a_1^{\frac{1}{p_1}} + ... +a_n^{\frac{1}{p_n}}}$? I think that it's always possible to eliminate irrationality but i don't have an idea yet how to prove it and how to implement it algorithmically.

Any ideas will be highly appreciated. Thanks in advance.

  • $\begingroup$ I think you meant $\dfrac{1}{\sqrt[p_1]{a_1} + ... +\sqrt[p_n]{a_n}}$ instead of $\dfrac{1}{a_1^{p_1} + ... +a_n^{p_n}}$. $\endgroup$ – barak manos Mar 8 '15 at 14:15
  • $\begingroup$ @barakmanos Oh, certainly. Misprint. Thanks $\endgroup$ – Igor Mar 8 '15 at 18:54

This is always possible; Let $P$ be the minimal polynomial over $\mathbb{Q}$ of the element $\alpha=a_1^{\frac{1}{p_1}}+\cdots+a_n^{\frac{1}{p_n}}$ the coefficient of $P$ are all rationals,in particular the product $p$ of its roots $\alpha_1=\alpha,\alpha_2,\cdots,\alpha_n$ is rational so : $$\frac{1}{\alpha}=\frac{\alpha_2\cdots\alpha_n}{p}$$

and this is equivalent to eliminating the irrationality in denominator.


  1. why we multiply by $2-\sqrt3$ to eliminate $2+\sqrt3$? because the minimal polynomial of $2+\sqrt3$ is: $$P(x)=x^2-4x+1=(x-2+\sqrt3)(x-2-\sqrt3) $$
  2. The minimal polynomial of $3^{1/3}+1$ is : $$P(x)=x^3-3x^2+x-4=(x-1-\sqrt[3]3)(x-1-j\sqrt[3]3)(x-1-j^2\sqrt[3]3)$$ with $j$ the $3$th complex root of the unity, this method involves complex number but after simplification thy will disapear: $$\frac{1}{1+\sqrt[3]3}=\frac{(1+j\sqrt[3]3)(1+j^2\sqrt[3]3)}{4}= \frac{1-\sqrt[3]3+\sqrt[3]9}{4}$$
  3. The minimal polynomial of $\sqrt[p]{a_1}$ is $x^p-a$ and the roots are $\alpha_i=w_p^i.\sqrt[p]{a}$ with $w_p$ is the $p$th primitive root of the unity and using the same method $\alpha_2\cdots\alpha_p=\sqrt[p]{a^{p-1}}$ which explains your "obvious" example.
  • $\begingroup$ Oh, very nice. Thanks for your answer. I just didn't get how you find minimal polynomials. Is it slways a trick or there exists some routine to do it? $\endgroup$ – Igor Mar 14 '15 at 1:47
  • $\begingroup$ And how do we know that for such numbers minimal polyomial over Q always exists? I haven't worked much with polynomials and field extensions. $\endgroup$ – Igor Mar 14 '15 at 2:01
  • $\begingroup$ I think i caught the last question. Roots of rational numbers are algebraic over Q and sum of algebraic numbers is again algebraic. Thus the minimal polynomial over Q exists. $\endgroup$ – Igor Mar 14 '15 at 2:20
  • $\begingroup$ Like you said roots of rational numbers are algebraic over $Q$ and the sum of two algebraic numbers is algebraic so the minimal polynomial always exists for such numbers, There is no expression which gives you directly the minimal polynomial, but fortunately there some methods that we can use for the roots of unity we use the cyclotomic polynomials and we can also deduce the polynmial of $a+b,ab$ given P_a and P_b, and there is a general algorithms $\endgroup$ – Elaqqad Mar 14 '15 at 13:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.