# What does this modular relation mean, if anything?

Let $x$ and $y$ be real numbers.

Suppose $\dfrac{x}{y \bmod x}$ is a natural number.

What does that say about the relationship between $x$ and $y$?

If $x$ and $y$ are naturals themselves, then I think it means that $x$ is some multiple of $y$ plus some divisor of $y$, but I'm a little fuzzy on what it would mean if $x$ and $y$ are reals.

Thank you.

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How do you define modulus for real numbers? (e.g. what is $8.252\pmod{2}$?) – apnorton Jan 10 '14 at 3:41
Probably 0.252? – Nishant Jan 10 '14 at 4:14

## 3 Answers

I think it means that $y$ is $x$ divided by an integer plus some integral multiple of $x$.

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It means that the Euclidean algorithm stops after two steps.

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Assuming that $(y \bmod x) = y - n x$ for some $n \in \mathbb{Z}$ such that $0 \leq y - n x < x$, in the case that $x > 0$, at least.

Then $\dfrac{x}{y \bmod x} = m$ for some integer $m$ means

$$\dfrac{x}{y - n x}=m$$ $$\Rightarrow (mn+1)x = m y$$

so we can say, the ratio of $x$ and $y$, where defined, is rational. I'm not sure if anything else less trivial can be said.

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