# Prove that $5^{1/3}+7^{1/2}$ is irrational

Goal: Prove that $5^{1/3}+7^{1/2}$ is irrational.

Idea: We can prove this is irrational by supposing it is rational and finding a contradiction. So, $5^{1/3}+7^{1/2} = p/q$ where $p$ and $q$ are integers that have no factors in common other than $1$. The issue here is that I can not seem to find a way manipulate this equation to a form where I can find contradiction. I tried taking each side to the power of 6… That was an unmanageable mess. Any ideas?

#### Solutions Collecting From Web of "Prove that $5^{1/3}+7^{1/2}$ is irrational"

Suppose that our sum is equal to the rational number $r$. Then $5^{1/3}=r-\sqrt{7}$. Cubing both sides we obtain
$$5=r^3-3r^2\sqrt{7}+21r-7\sqrt{7},$$
and therefore
$$r^3+21r-5-(3r^2+7)\sqrt{7}=0.$$
Since $\sqrt{7}$ is irrational, we must have $3r^2+7=0$. This is impossible.

Remark: For completeness we sketch a proof of the irrationality of $\sqrt{7}$. It is much like one of the standard proofs of the irrationality of $\sqrt{2}$. Suppose to the contrary that there are integers $p$ and $q$, with $q\ne 0$, such that $p^2/q^2=7$. Without loss of generality we may assume $p$ and $q$ are relatively prime.

Then $p^2=7q^2$. Since $7$ is prime and divides $p^2$, it follows that $7$ must divide $p$. Let $p=7t$. Substituting and cancelling, we get $7t^2=q^2$. Thus $7$ divides $q$, contradicting the fact that $p$ and $q$ are relatively prime.

Let $t=5^{1/3}+7^{1/2}$. Then $t$ is a root of $x^6-21 x^4-10 x^3+147 x^2-210 x-318=0$. (*)

The rational root theorem tell us that either $t$ is irrational or $t$ is an integer dividing $318$.

Since $1 < 5^{1/3} < 2$ and $2 < 7^{1/2} < 3$, we have $3 < t < 5$, and so if $t$ is an integer, it must be $4$. But $4$ does not divide $318$.

Therefore, $t$ is irrational.

(*) This polynomial can be found manually (if you don’t drown in the algebra) or you can ask WA. This is not cheating (well, not crucially) because the nice argument is still to come. It’s an opportunity to learn the rational root theorem, if you haven’t seen it.

Another solution is to look at $K = \mathbb{Q}(\zeta_3, \sqrt[3]{5}, \sqrt[2]{7})$ (where $\zeta_3$ is primitive cube root of unity) which is Galois over $\mathbb{Q}$. It is easy to see that $(K : \mathbb{Q}) = 6 \times 2 = 12$ and the automorphism in $\text{Gal}(K|\mathbb{Q})$ are easily described: they are of the form $\sigma_{i,j,\pm} : \{\zeta_3 \mapsto \zeta_3^i\} \times \{\sqrt[3]{5} \mapsto \sqrt[3]{5} \zeta_3^j\} \times \{\sqrt[2]{7} \mapsto \pm \sqrt[2]{7}\}$ where $i \in \{1, 2\}, j \in \{0, 1, 2\}$. It is easy to check that one of the automorphism $\sigma_{1,2,+}$ does not fix $\alpha = \sqrt[3]{5} + \sqrt[2]{7}$. Hence, $\alpha \not\in \mathbb{Q}$.

Hint $\ \sqrt7\, =\, r – \sqrt[3]5,\,\ r \in\Bbb Q\,\overset{\rm square}\Rightarrow \sqrt[3]5\,$ is a root of a quadratic polynomial $f(x) \in \Bbb Q[x],$ contra $\,x^3-5\,$ is irreducible over $\,\Bbb Q\$ (e.g. by the Rational Root Test).

Remark $\$ If you know a little algebra then you may recognize that the idea implicit in the above is that $\,x^3 – 5\,$ is irreducible hence the minimal polynomial of $\,\sqrt[3]5.\,$ If you go on to study field theory then you can deduce it from a more general result, viz. $\,\Bbb Q(\sqrt[3]5)$ and $\Bbb Q(\sqrt7)$ have coprime degrees $3,2$ over $\,\Bbb Q,\,$ hence their intersection is $\,\Bbb Q.\,$ So $\,\sqrt[3]5 \in\Bbb Q(\sqrt 7)\Rightarrow \sqrt[3]5\in\Bbb Q$.

Edit: Swapped the cube root and the square root in the display. Fixed. This also altered the subsequent argument.

To use your method, you never want to raise to a power anything that has both $\sqrt[3]{5}$ and $\sqrt{7}$ in it. So, you isolate these one at a time.

Suppose $\sqrt[3]{5} + \sqrt{7} \in \Bbb{Q}$ so there are $p,q \in \Bbb{Z}$ with $q \neq 0$ such that $\sqrt[3]{5} + \sqrt{7} = p/q$. By reducing $p/q$ to lowest terms, we may assume that $\gcd(p,q) = 1$. Since solving quadratics is harder than solving linears (for the second isolation), we start by cubing. We compute \begin{align} \sqrt[3]{5} &= p/q – \sqrt{7}, \\ 5 &= (p/q – \sqrt{7})^3 \\ &= p^3/q^3 – p^2\sqrt{7}/q^2 + 7p/q – 7\sqrt{7}, \text{ and}\\ \sqrt{7} &= \frac{5-p^3/q^3 – 7p/q}{-p^2/q^2 – 7} \\ 7 &= \left( \frac{5q^3 – p^3 – 7pq^2}{-p^2q – 7q^3} \right)^2, \text{ so} \\ 7q^2 \left( p^2 + 7q^2 \right)^2 &= \left( p^3 + 7pq^2 – 5q^3 \right)^2 . \end{align}
Now work modulo $4$. The squares are congruent to $0$ or $1$, so the left-hand side is either $0$ or $3 \pmod{4}$ and the right-hand side is either $0$ or $1 \pmod{4}$. So, either $q$ or $p^2 + 7q^2$ is even and $p^3 + 7pq^2 – 5q^3$ is even.

If $q$ is even, $p^3 + 7pq^2 – 5q^3 \cong p^3 \pmod{4}$. Since $p^3 + 7pq^2 – 5q^3$ is even, we must have $p$ is even, contradicting our assumption that $p/q$ was reduced to lowest terms.

Otherwise, we have $p^2 + 7q^2$ is even. Note that $p^3 + 7pq^2 – 5q^3 = p(p^2+7q^2)-5q^3$, so is even only if $q$ is even. But then $p^2 + 7q^2$ is even only if $p$ is and again we contradict reduction to lowest terms.