TITLE: Prove that the real root of $x^3 + x + 1$ is irrational
QUESTION [5 upvotes]: Using wolframalpha.com we get that the real root of this polynomial is $-0.68233$
The only way that I have found how to prove it is using the Rational Root theorem.
Using that theorem the possible rationals roots are 1 and -1, but none of them work.
Then the real root has to be irrational. 
My questions are:

Is there other way to prove it without using the Rational Root Theorem?
Is my proof written well?

REPLY [9 votes]: If it were rational it would be $m/n$, where $m$ and $n$ are integers.  Then we would have
$$
\left(\frac m n \right)^3 + \frac m n + 1 =0.
$$
From that we get
$$
m^3 + mn^2 + n^3 = 0.
$$
If the fraction $m/n$ is in lowest terms, then $m$ and $n$ have no prime factors in common.  But
$$
n^3 = -m(m^2+n^2) = (-m\cdot\text{something})
$$
so $n^3$ is divisible by the prime factors of $m$.  That can happen only if $m$ has no prime factors, i.e. $m=\pm1$.  In the same way, we have
$$
m^3 = - n^2(m+n) = (-n^2\cdot\text{something}),
$$
and so $n=\pm1$.  But $1$ and $-1$ are not roots of the equation.