Square of Cube Number is Cube
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Theorem
Let $a \in \N$ be a natural number.
Let $a$ be a cube number.
Then $a^2$ is also a cube number.
In the words of Euclid:
- If a cube number by multiplying itself make some number the product will be cube.
(The Elements: Book $\text{IX}$: Proposition $3$)
Proof 1
By the definition of cube number:
- $\exists k \in \N: k^3 = a$
Thus:
\(\ds a^2\) | \(=\) | \(\ds \paren {k^3}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds k^6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k^2}^3\) |
Thus:
- $\exists r = k^2 \in \N: a = r^3$
Hence the result by definition of cube number.
$\blacksquare$
Proof 2
From Cube Number multiplied by Cube Number is Cube, if $a$ and $b$ are cube numbers then $a b$ is a cube number.
The result follows by setting $b = a$.
$\blacksquare$
Historical Note
This proof is Proposition $3$ of Book $\text{IX}$ of Euclid's The Elements.
The proof as given here is not that given by Euclid, as the latter is unwieldy and of limited use.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{IX}$. Propositions