Rationality of Cube Roots

Proof by contradiction is a cornerstone of HSC Proofs, and irrationality of roots is one of its most classic applications. This problem not only tests your ability to harness simple divisibility by 2 in part (i), but also invites you to extend the idea in part (ii) by wielding the Fundamental Theorem of Arithmetic — a technique that separates routine contradiction arguments from truly elegant ones.

Problem Statement

• Prove that $\sqrt[3]{10}$ is irrational.
• And hence, or otherwise, prove it is not possible for both $\sqrt[3]{2n}$ and $\sqrt[3]{5n}$ to be rational for any positive integer $n$.

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Hints

• (i) Use a proof by contradiction. Assume $\sqrt[3]{10} = \frac{p}{q}$ where $p$ and $q$ are coprime integers. Cube both sides and analyze the parity (divisibility by 2) of both sides to find a contradiction.
• (ii) If both are rational, set them equal to rational fractions, cube them, and isolate $n$. Equate the two expressions for $n$. Apply the Fundamental Theorem of Arithmetic (specifically looking at the multiplicity of the prime factor 2 or 5) to prove the resulting equation is impossible.

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Solutions

(i) Prove that $\sqrt[3]{10}$ is irrational.

To prove irrationality, we start by assuming the opposite — that the cube root can be written as a fraction in lowest terms — and show this leads to an inescapable clash with simple parity.

Proof by Contradiction

Assume for the sake of contradiction that $\sqrt[3]{10}$ is rational. Then it can be expressed in simplest fractional form as $\frac{p}{q}$, where $p, q \in \mathbb{Z}^+$, $q \neq 0$, and $\gcd(p, q) = 1$.

Cube both sides to remove the root, giving an integer equation:

$$\begin{aligned} 10 &= \frac{p^3}{q^3} \\ p^3 &= 10q^3 \\ p^3 &= 2(5q^3) \end{aligned}$$

Since $5q^3$ is an integer, $p^3$ is a multiple of 2 (it is even). The cube of an odd number is odd, so $p$ must also be even.
Let $p = 2k$ for some integer $k$. Substituting this back:

$$\begin{aligned} (2k)^3 &= 10q^3 \\ 8k^3 &= 10q^3 \\ 4k^3 &= 5q^3 \end{aligned}$$

Now we inspect the new equation. The left side, $4k^3 = 2(2k^3)$, is even. Therefore, the right side, $5q^3$, must also be even. Since 5 is odd, $q^3$ must be even, which implies $q$ is even.

If both $p$ and $q$ are even, they share a common factor of 2. This contradicts our initial assumption that $\gcd(p, q) = 1$. Therefore, $\sqrt[3]{10}$ is irrational.

$\\square$

(ii) Prove it is not possible for both $\sqrt[3]{2n}$ and $\sqrt[3]{5n}$ to be rational.

We again use contradiction, but now the logical engine shifts from basic parity to the uniqueness of prime factorisations. By assuming both cube roots are rational, we can derive an equation that forces two representations of the same power of 2 to disagree.

Proof by Contradiction

Assume for contradiction that for some positive integer $n$, both expressions are rational. Let $\sqrt[3]{2n} = \frac{a}{b}$ and $\sqrt[3]{5n} = \frac{c}{d}$, where $a, b, c, d \in \mathbb{Z}^+$ and the fractions are in simplest form. Cubing both equations eliminates the radicals and isolates $n$:

$$\begin{aligned} 2n = \frac{a^3}{b^3} &\implies n = \frac{a^3}{2b^3} \\ 5n = \frac{c^3}{d^3} &\implies n = \frac{c^3}{5d^3} \end{aligned}$$

Since both expressions represent the same $n$, we equate them:

$$\begin{aligned} \frac{a^3}{2b^3} &= \frac{c^3}{5d^3} \\ 5a^3d^3 &= 2b^3c^3 \end{aligned}$$

Let $X = ad$ and $Y = bc$, where $X, Y \in \mathbb{Z}^+$. Thus, $5X^3 = 2Y^3$. This is a Diophantine equation, and we can bring the full force of unique prime factorisation to bear.

By the Fundamental Theorem of Arithmetic, we analyze the prime factor 2 on both sides:

• In $X^3$, the prime 2 appears a multiple of 3 times (say, $3k$ times). Since 5 is odd, the left side $5X^3$ has exactly $3k$ prime factors of 2.
• In $Y^3$, the prime 2 appears a multiple of 3 times (say, $3m$ times). The coefficient 2 adds one more, meaning the right side $2Y^3$ has exactly $3m + 1$ prime factors of 2.

Equating the multiplicities gives $3k = 3m + 1 \implies 3(k - m) = 1$. Since $k$ and $m$ are integers, their difference cannot be $\frac{1}{3}$. This contradiction proves that both expressions cannot be rational simultaneously.

$\\square$

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Takeaways

• Parity as a Filter: Part (i) demonstrates that analyzing divisibility (specifically by 2) is a highly efficient mechanism for contradiction proofs involving rational roots.
• Fundamental Theorem of Arithmetic: Part (ii) highlights that comparing the exponents of prime factorizations across an equality is a definitive way to prove impossibility in integer equations.

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Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Polys Ext 1, HSC Vectors