Rationality of Cube Roots

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 $n$.

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.

Solutions

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

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 $\gcd(p, q) = 1$.

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:

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.

\hfill $\square$

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

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 yields:

Equating the two expressions for $n$:

Let $X = ad$ and $Y = bc$, where $X, Y \in \mathbb{Z}^+$. Thus, $5X^3 = 2Y^3$.

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.

\hfill $\square$

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.

Further Readings

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

• HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
• HSC Polys Ext 1: https://vumaths.com/booklets/hsc-polys-ext-1/
• HSC Vectors: https://vumaths.com/booklets/hsc-vectors/

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