Prove that for $x^3=3$ there isn't rational solution

What I did:

Suppose $x= u /v$ is solution

$$\left(\frac u v\right)^3=3$$

Let's take third root from both sides:

$$\frac u v =\sqrt[3]3$$

and $\sqrt[3]3$ is irrationl ,

my problem is this "and $\sqrt[3]3$ is irrationl " is it well known that this number is irrationl? same as $\pi$?

or maybe there is another why to prove it?

  • 2
    $\begingroup$ Rational root theorem. $\endgroup$
    – Git Gud
    Mar 22, 2016 at 20:57
  • $\begingroup$ en.wikipedia.org/wiki/Rational_root_theorem $\endgroup$ Mar 22, 2016 at 20:58
  • 1
    $\begingroup$ The classic proof of irrationality of $\sqrt{2}$ can be adapted to $\sqrt[3]{3}$. $\endgroup$
    – Crostul
    Mar 22, 2016 at 20:59
  • 3
    $\begingroup$ My guess is that the point of the exercise is to prove that $\sqrt[3]{3}$ is irrational, so using that is probably not acceptable. $\endgroup$ Mar 22, 2016 at 21:00
  • $\begingroup$ @Henrik Yes, you are right $\endgroup$
    – Error 404
    Mar 22, 2016 at 21:01

4 Answers 4


Just start off my assuming $(u,v) = 1$ i.e the fraction is fully reduced. Then it follows that; $$u^3 = 3v^3 \Rightarrow 3 \mid u^3 \Rightarrow 3 \mid u \Rightarrow u = 3M$$

Therefore $(3M)^3 = 27M^3 = 3v^3$ and so $9 \mid v^3$ which implies $3 \mid v^3$ and so $(v,u) \not = 1$ which is a contradiction.

  • $\begingroup$ $u^3=3v^3$, $u=3M$ therefore $(3M)^3=3v^3$ and not $(3M)^3=v^3$ $\endgroup$ Mar 22, 2016 at 21:17

No, you cannot just say that $\sqrt[3]{3}$ is irrational; that is just restating that which you are trying to prove. Adding "it is well known" does not help.

What you need to do is assume $\sqrt[3]{3}$ is rational; then $\sqrt[3]{3} = \frac{p}{q}$ with $p,q \in \Bbb{Z}^+$ adn there exists such a pair $p,q$ such that g.c.d$(p,q) = 1$.

Then $$\frac{p^3}{q^3} = 3 \\ p^3 = 3q^3 \\ p^3 = 3k \implies \exists n: p = 3n \\ 27n^3 = 3q^3 \\ q^3 = 9 n^3 = 3 (3n^3) \implies q = 3m $$ but then both $p$ and $q$ are multiples of $3$, which contradicts the condition that g.c.d$(p,q) = 1$.

So if $\sqrt[3]{3}$ is rational it cannot be expressed as a reduced fraction, therefore it is irrational.


You were on the right track, Let $$\frac{u}{v}=\sqrt[3]3$$ be a rational number, where $v,u$ are coprime and $v\neq 0$ $$\implies\frac{u^3}{v^3}=3$$ $$\implies u^3=3v^3$$ As $v$ and $u$ are coprime, their cubes must also be coprime. But this is a contradiction ($u^3=3v^3$) and hence the assumption that $\sqrt[3]3$ is rational was false.


As an elementary approach, I personally prefer to prove that given any truly rational number $$ \frac ab\notin\mathbb Z $$ we get truly rational powers $$ \left(\frac ab\right)^n\notin\mathbb Z $$ for all $n\in\mathbb N$. Namely, if any prime $p$ divides $b$ but not $a$, the same goes for $b^n$ and $a^n$.

The reasons I like this approach are

  1. simplicity
  2. generality

One can infer from this that an $n$-th root of an integer is either an integer or irrational. So it follows immediately that $\sqrt 2, \sqrt[3]3,\sqrt[5]{17}$, and generally any $n$-th root of an integer that is not a perfect $n$-th power will be irrational.


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