1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ How do you define modulus for real numbers? (e.g. what is $8.252\pmod{2}$?) $\endgroup$ – apnorton Jan 10 '14 at 3:41
  • $\begingroup$ Probably 0.252? $\endgroup$ – Nishant Jan 10 '14 at 4:14
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

It means that the Euclidean algorithm stops after two steps.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.