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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)

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... it's impossible to measure and cut anything physical at a repetend.

This is incorrect. Given a segment with $1$ unit length, one can easily construct, "physically", a segment with a length of $1/3$ units. In fact, given any line segment, there are many ways to trisect it. See for instance this article or this YouTube video. See also this Wikipedia article on $\sqrt{2}$; one can even get a line segment with the length of an irrational number!

Your confusion essentially boils down to why $0.\overline{9}=1$. Notice that on the right you have the integer $1$, while on the left, you have a repeating decimal which continues infinitely.

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Your equations express rational numbers, and are correct.

Just because in some numbering systems some of these rational numbers can't be represented with finite notation doesn't in any way invalidate the equations.

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