Proof of the irrationality of cube root 16

Question

I am trying to follow the proof for the irrationality of $\sqrt[3]{6}$ to form a similar proof for $\sqrt[3]{16}$ (proof by contradiction). Going from $16=(a/b)^3$ (with $a,b \in \mathbb{Z}$ and $b\neq 0$ and GCD$(a,b)=1$) I can show that $2|a^3$ and then consequently $2|a \Rightarrow a=2c$.

But I'm unable to show $2|b$ following the step for $16b^3 = (2c)^3$ which leads to $2b^3 = c^3$ as this only shows that $c$ is even. I'm probably either making a mistake or overlooking a key fact. Any suggestions?

Answer 1

The proof you want is a varaint of the Pythagorean proof of the irrationality of $\sqrt 2$

Proof by conradiction: Suppose $\frac pq$ is a fraction in lowest terms such that

$\frac pq = 16\\ p^3 = 16 q^3$

If this is true the $p$ must be even, which case we can say

$p = 2r\\ 8 r^3 = 16 q^3\\ r^3 = 2 q^3$

$r$ is also even

$r = 2s\\ 8s^3 = 2 q^3\\ 4 s^3 = q^3$

Which would imply that $q$ is even, but that would contradict our initial assumption that $\frac pq$ is in lowest terms.

Answer 2

Let $16^{1/3}=\frac ab$ where $gcd(a,b)=1$. Cube both sides and arrive at $16b^3=a^3, 2(2b)^3=a^3$ notice that on the right hand side in the prime factorisation the power of two is a multiple of 3 (e.g. $(2^pf)^3=2^{3p}f^3$ (where $f$ is relatively prime to 2)), whereas on the left hand side the power of 2 in the prime factorisation is of the form $3k+1$ which is a violation of the fundamental theorem of arithmetic and is hence a contradiction,

Answer 3

If $x=\sqrt[3]{16}$, we have $x^3-16=0$ . It is easy to check no rational roots exist by the rational root theorem.

Answer 4

"But I'm unable to show 2|b following the step for $16b^33=(2c)^3$ which leads to $2b^3=c^3$ as this only shows that c is even. "

So $c$ is even. So $c = 2d$ so $2b^3 = (2d)^3 = 8d^3$. So $b^3 = 4d^3$. So $4|b^3$ so $2|b^3$ so $2|b$.

Answer 5

$x=16^{1/3}$ obviously isn't integral, since $2^3=8$, $3^3=27$, and the cube function is strictly increasing.

Let $k$ be the integer such that $k<x<k+1$

If $x$ were rational, then there would exist integers whose product with $x^2$ is also an integer. Let's denote by $n$ the smallest positive one.

Then, $n(x-k)$ is a positive integer smaller than $n$ which also verifies this property, which is a contradiction.

Therefore, $16^{1/3}$ is irrational.