Let $p\in\mathbb N$ be a prime number and define for $x,y\in\mathbb N$, $x\sim y\iff (p^n|x \iff p^n|y \text{ for all }n\in\mathbb N).$ The vertical bar $|$ denotes the divisor-operation. Is $\sim$ an equivalent relation on $\mathbb N$? I've already shown reflexivity and symmetry, but I'm stuck on the transitive property.
Is "$x\sim y$ iff $x$ and $y$ have the same powers of $p$ dividing them" an equivalence relation?
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elementary-set-theory
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0Ah, $p$ is fixed. I missed that. – 2012-06-02
2 Answers
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This is a special case of the equivalence relation induced by a function $f:X\to Y$, defined by saying that $a\sim b$ if and only if $f(a)=f(b)$. In your case, the function would send $x\in\mathbb{N}$ to the largest $n$ such that $p^n|x$, or the valuation of $x$ at $p$. Two numbers $x$ and $y$ satisfy the property $ \forall(n\in\mathbb{N}),\ p^n|x\leftrightarrow p^n|y $ if and only if they have the same valuation at $p$.
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1Said another way, "$x\sim y$ iff $x$ and $y$ have the same whatever" is an equivalence relation pretty much no matter what you put in for "whatever". – 2012-06-03
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If $x\sim y$ and $y\sim z$, then for any $n\in\mathbb{N}$, $p^n\mid x\iff p^n\mid y\qquad\text{and}\qquad p^n\mid y\iff p^n\mid z.$ Therefore $p^n\mid x\iff p^n\mid z$, so that $x\sim z$.