Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove or disprove: For all positive integers $ n$ ,

$\sqrt[3]{n}+\sqrt[3]{n+1}$ are irrational numbers.

share|cite|improve this question
Since $x=n^{1/3}$ and $y=(n+1)^{1/3}$ are algebraic integers then $x+y$ is an algebraic integer. Hence, if $x+y$ is rational then it is an integer. – i. m. soloveichik Sep 29 '12 at 2:43
up vote 7 down vote accepted

Note that $(x+y)^{3} = x^{3} +y^{3} + 3xy(x+y).$ If $n$ or $n+1$ is the cube of an integer, the result is clear by the uniqueness of prime factorization, so we assume that neither is the cube of an integer. Then neither is $n(n+1)$. Set $x = n^{\frac{1}{3}}$ and $y = (n+1)^{\frac{1}{3}}.$ If $x+y$ is rational, then so are $(x+y)^{3}$ and $3xy(x+y)$. Hence $3xy$ is rational. Thus $27n(n+1)$ must be the cube of an integer, a contradiction.

share|cite|improve this answer
Why 27n(n+1) not a cube?Can explain in detail Although n and n+1 not a cube, but their product may be ah, for example, 2 and 4 are not cubic number, but 8 cubic number. – tianzhidaosunyouyu Sep 29 '12 at 8:26
But $n$ and $n+1$ are relatively prime. If $p$ is any prime, then if $n(n+1)$ is a cube, the power of $p$ dividing $n(n+1)$ is an integr divisible by $3$. If $p$ divides $n$, then $p$ does not divide $n+1,$ so the power of $p$ dividing $n$ would have to be an integer multiple of $3$. Hence $n$ must be a cube, as $p$ was an arbitrary prime divisor of $n.$ Similarly for $n+1$. – Geoff Robinson Sep 29 '12 at 10:02

Hint $\ $ If $\rm\:n,m\in\Bbb N$ then cubing $\rm\: \sqrt[3]{n}+\sqrt[3]{m}\in\Bbb Q\:\Rightarrow\: nm\in \Bbb N^3\ (\Rightarrow\ n,m\in \Bbb N^3\ if\ (n,m) = 1)$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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