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I need help with proving whether or not the relation $R=\left\{(f,g)\mid \exists k\in\Bbb Z,\forall x\in\Bbb Z, \ f(x)g(x)\lt k\right\}$ is an equivalence relation on $\mathbb{Z}\times\mathbb{Z}$

I understand the basic key components needed in order to determine if the relation is an equivalence relation, like reflexivity, symmetry, and transitivity but i can't seem to find a way to proof reflexivity.

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    $\begingroup$ Something isn't right. You say this is a relation on $\mathbb{Z}\times \mathbb{Z}$, but what you define is not a subset of $(\mathbb{Z}\times\mathbb{Z})\times(\mathbb{Z}\times\mathbb{Z})$. A binary relation on a set $A$ is a subset of $A\times A$ satisfying the properties you mention. Can you edit your definition of $R$ to make it more clear? $\endgroup$ – Jared May 23 '13 at 14:33
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. I would like to ask you to take this into consideration for future questions. Thanks in advance. :) $\endgroup$ – Lord_Farin May 23 '13 at 14:34
  • $\begingroup$ Thanks Lord_Farin, I will :) $\endgroup$ – zv.diego May 23 '13 at 14:38
  • $\begingroup$ @Jared I understand but I'm checking at it again and the problem says: Determine if the relation over the set of functions in $\Bbb Z$ squared is an equivalence relation. The definition of $R$ is the same as well. $\endgroup$ – zv.diego May 23 '13 at 14:44
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    $\begingroup$ Consider the function $f:\mathbb{Z}\to\mathbb{Z}$ defined by $f(x)=x$. There is no $k$ bounding $f(x)\cdot f(x)=x^2$ so the relation is not reflexive. $\endgroup$ – Jared May 23 '13 at 14:58
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Consider the set $A$ of all functions $f:\mathbb{Z}\to\mathbb{Z}$. Define an equivalence relation on $A$ by:

$$R=\{(f,g)|\exists k,\forall x,f(x)g(x)<k\}\subset A\times A$$

I claim the relation is symmetric, but not reflexive or transitive. For symmetry, it suffices to note that $f(x)g(x)=g(x)f(x)$.

As a counterexample to reflexivity, consider $f(x)=x$, and note that $(f,f)\notin R$ because $x^2$ is not bounded.

Finally, consider the functions $f(x)=h(x)=x$ and $g(x)=0$. Then we have $(f,g),(g,h)\in R$, but $(f,h)\notin R$ so the relation is not transitive.

It follows that the relation is not an equivalence relation.

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  • $\begingroup$ Couldn't be any clearer, thanks much Jared. $\endgroup$ – zv.diego May 23 '13 at 16:11

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