Why this reductio ad absurdum works to show $\sqrt7$ is irrational but it does not work for $\sqrt[3]8$?

Question

I'm trying to refresh my mathematics after 10 years and I'm having some problems understanding this. The following is used to show how the square root of 7 is not a rational number:

$\sqrt{7} = \frac{m}{n}$ where $m,n \in \Bbb Z; n \neq 0; m, n$ are coprimes.

$m^2 = 7n^2$

$m^2$ is a multiple of $7$, so $m$ is also multiple of $7$.

$m = 7k; ~ m^2 = 7(7k^2)$

$ 7n^2 = 7(7k^2) $

$ n^2 = 7k^2 $

Thus, both n and m are multiples of 7 which cannot be possible.

But with the same logic I end up with this for $ \sqrt[3]8 $:

$\sqrt[3]{8} = \frac{m}{n}$ where $m,n \in \Bbb Z; n \neq 0$

$m^3 = 8n^3$

$m = 8k;~m^3 = 8(8^2k^3)$

$8(8^2k^3) = 8n^3$

$8^2k^3 = n^3$

$8(8k^3) = n^3$

$m^2 = 8(8^2k^3)$ and $n = 8(8k^3)$

By the same reasoning, both $m$ and $n$ are multiples of 8 and it should be invalid, but as we all know it is $2$, a completely rational number. What am I missing here?

Answer 1

This is because $7$ is a square free number but $8$ is not a cube free number. In general, such proofs to show irrationality only work for $\sqrt[k]n$ if $n$ is a k-free number. If $n$ is not a $k$ free number, then you need to write $\sqrt[k]n = r \sqrt[k]s$ where $s$ is k-free. Let's see why this is so.
Suppose $n$ is not k-free. So, $n = r^ks$ where $s$ is k-free. We will work by our usual proof using contradiction:

Suppose $\sqrt[k]n$ is rational. Then $$\sqrt[k]n = \frac ab$$ for some $a,b \in \Bbb Z, b \neq 0, \gcd(a,b) = 1$. We get: $$a^k = nb^k$$ $n$ divides $a^k$, but may or may not divide $a$. To see this, consider $$\color{blue}{a^k} = \color{blue}{r^k}sb^k$$ It is clear that $r^k$ may or may not divide $a$, although it divides $a^k$, but $s$ certainly divides $a$. $a = r$, $s = 1$ is an easy example.

---

We have established that numbers which are not k-free may lead to fallacies in the proof. We will now show that the proof works for a k-free integer. Suppose $n>1$ is k-free. We need to show that $\sqrt[k]n$ is irrational. Assume the opposite - it's rational. Then, $\sqrt[k]n = \frac ab$ for some $a,b \in \Bbb Z, b \neq 0, \gcd(a,b) = 1$. $$a^k = nb^k$$ $n$ is k-free, $$n = p_1^{q_1}p_2^{q_2} \cdots$$where $p_i$ are prime and $q_i<k$. $$a^k = xp_1^{km_1}p_2^{km_2}\cdots$$Since $q_i\le k-1$, $q_i\le (k-1)m_i$ hence we get $$xp_1^{q_1+l_l}p_2^{q_1+l_2}\cdots = a^k = b^kp_1^{q_1}p_2^{q_2} \cdots$$ Hence $p_i^{l_i}$ divides $b$, $l_i$ is positive, thus $a$ and $b$ are not coprime, a contradiction.

---

Thus, for the proof, we need to write it in the form of $r\sqrt[k]s$. For $\sqrt[3]8$, it immediately fails since $\sqrt[3]8 = 2\sqrt[3]1 = 2$ which is rational. Let's try an example for, say $\sqrt[3]{32} = 2\sqrt[3]4$.

Assume $2\sqrt[3]4$ is rational. Then, $$2\sqrt[3]4 = \frac ab$$ for some $a,b \in \Bbb Z, b \neq 0, \gcd(a,b) = 1$. So, $$\sqrt[3]4 = \frac{a}{2b} = \frac{a'}{b'}$$where $a'/b'$ is the lowest form of of $a/(2b)$. Then $$a'^3 = 4b'^3$$ Since $4$ divides $a'^3$, $2$ divides $a'$. Write $a' = 2x$ to get $$8x^3 = 4b'^3 \iff 2x^3 = b'^3$$ Since $2$ divides $b'^3$, it divides $b'$, hence the fraction $a'/b'$ is not in the lowest form, a contradiction.

Answer 2

The key fact is that for any prime $p$ and integers $a,b$, if $p|a\cdot b$, then $p|a$ or $p|b$. (You can also apply this repeatedly to show that for any number of factors, if $p$ divides the product it divides one of the factors.) Composite numbers do not have this property. In fact, this property of primes is so important that in abstract algebra it is the definition of being prime.

Applying this fact to the problem at hand, in the proof for $\sqrt{7}$, it allows you to deduce $7|n$ from $7|n\cdot n$, because $7$ is prime. However, for $\sqrt[3]{8}$, the proof fails at the step where you try to deduce $8|n$ from $8|n\cdot n\cdot n$. $8$ is not prime, so it is not true that if it divides a product it must divide one of the factors. Indeed, clearly $8|2\cdot 2 \cdot 2$ but $8\not|2$.

Now that's not to say you can't use a similar proof for composite numbers. Just that you have to be more careful. For a prime power like $8$, you show $\sqrt{8}$ is irrational by showing that $8 |n^2$ implies $2|n^2$ which implies $n=2a$ for some $a$. From $8|4a^2$ you can then deduce $2|a$, meaning $n=4b$ for some $b$. Then you have $(4b)^2=8m^2\Longrightarrow 2b^2=m^2$, and proceed as with $\sqrt{2}$. For higher roots, you may need to do this multiple times before you reach an equation of the form $n ^x = p m^x$. This method works in general for $n^x = p^y m^x$ provided that $x\not| y$. If $x|y$, such as with $n^3=8m^3$, the factors of $p$ will entirely cancel out and the process ends. Which is good, because in this case $\sqrt[x]{p^y}$ actually is rational.

Finally, for $n^x= k m^x$, where $k$ is a composite number with more than one prime factor (such as $45$ or $216$), you just check each prime factor separately. The reduction will work on at least one of them unless every exponent in the prime factorization is divisible by $x$. But of course, that's exactly the condition that $k = \ell^x$ for some integer $\ell$. So the proof that $\sqrt[x]{k}$ had better fail in this case.