Prove in 2 ways that the cube root of 18 is irrational

Question

I believe I proved it one way but I am not sure how to prove it a different way. Here is my first proof:

Assume that the cube root of 18 is rational; than it can be written as $\frac pq$ for integers $p$ and $q$ such that $p$ and $q$ share no common divisor.

$(\frac pq)^3$=18

so $(p^3)$/$ (q^3)$=18

so $p^3$=18($q^3$)

Now, $p$ must be a multiple of $3$ else $p^3$ would not be a multiple of $3 or(9)$. Let $r$ be the integer such that $(3r)^3=p^3$

So,$(3r)^3=18(q^3)$

$3*3*3*r^3=18(q^3)$

$3*r^3=2(q^3)$

hence $q$ is a multiple of 3.

If $p$ and $q$ are both multiples of 3 then, $\frac pq$ is not the simplest ways to express the cubed root of $18$ so we have a contradiction.

$therefore$, the cubed root of $18$ must be $irrational$.

If any can show me another way to prove this I would highly appreciate it, also I feel as if the proof I gave can be made more solid if anyone can help there also.

Answer 1

Suppose that $a$ and $b$ are positive integers such that $\sqrt[3]{18}=a/b.$ Then $a^3=18b^3=2\cdot3^2b^3.$ Now let $e_p(n)$ denote the exponent of the prime $p$ in the prime decomposition of $n.$ Then $3e_2(a)=1+3e_2(b)$ which implies that $3\mid1,$ which is clearly false.

Yet another proof is as follows. Let $p$ and $q$ be arbitrary positive integers. Then we have $$\begin{aligned}\left|\dfrac{p}{q}-\sqrt[3]{18}\right|&=\left|\dfrac{p-q\sqrt[3]{18}}{q}\right|\\\\&=\left|\dfrac{p^3-18q^3}{q\left(p^2+pq\sqrt[3]{18}+\sqrt[3]{18^2}q^2\right)}\right|\\\\&\geqslant\dfrac{1}{\left|q\left(p^2+pq\sqrt[3]{18}+\sqrt[3]{18^2}q^2\right)\right|}\\\\&=\dfrac{1}{q\left(p^2+pq\sqrt[3]{18}+\sqrt[3]{18^2}q^2\right)}\\\\&>0,\end{aligned}$$ when the first inequality holds because $p^3$ and $18q^3$ are distinct integers because the exponent of $2$ in the prime decomposition of $p^3$ is a multiple of $3$ and the exponent of $2$ in the prime factorization of $18q^3$ is one more than a multiple of $3.$ Thus, the distance from $\sqrt[3]{18}$ to any positive rational number is positive so necessarily it is an irrational number.

Answer 2

Let $x=\sqrt[3]{18}$, then $x^3=18$ so $x$ is a root of $x^3-18=0$. But that's a polynomial with integer coefficients, so by the rational root theorem any rational root must be an integer divisor of $18$ i.e. among $\{\pm 1, \pm 2, \pm 3, \pm 6, \pm 9\}$. By brute force, none of those verify the equation, so there is no rational root, therefore $x$ must be irrational.

Of course, one could shorten the list before trying all divisors, for example by observing that an integer $x$ satisfying the equation would need to be positive and even, which leaves just $\{2, 6\}$ to try. Or simply note that $2^3 \lt 18 \lt 3^3$, which would require $2 \lt x \lt 3$ i.e. not an integer.