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Comments on How can 3/1 ≡ 1/(1/3), when left side features merely integers, but right side features a repetend?

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How can 3/1 ≡ 1/(1/3), when left side features merely integers, but right side features a repetend?

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On one hand, I know that algebraically, $\dfrac{3}1 ≡ \dfrac{1}{\color{red}{1/3}}$.

On the other hand, they differ in practice, not least because $\color{red}{1/3}$ contains 3 as the repetend. For example, if a scrap of physical material must have a 3:1 ratio and a length of 3 m, then I shall make the width 1 m.

But presuppose a length of 1 m. Then a 3:1 ratio is impossible to accomplish, because it would require a width of $\color{red}{1/3=0.\overline{3}}$. But it's impossible to measure and cut anything physical at a repetend!

Doesn't this physical impossibility due to the reptend belie, or undermine, the theoretical "equality" that $\dfrac{3}1 ≡ \dfrac{1}{\color{red}{1/3}}$? How can this physical impossibility due to the repetend be reconciled with equality?

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The existence of a repetend depends on the base (1 comment)
The existence of a repetend depends on the base
Moshi‭ wrote about 2 years ago

Putting aside the fact that math does not particularly care about physical possibilities, and further the fact that it isn't physically impossible to trisect an object....

The existence of a repetend (aka repeated digits) depends completely on what base we use. Consider for a moment the usage of base three. Then, clearly, $\frac13=0.1$ in base three, by definition. There are no repeated digits here!