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Problem:

Let $R_1$ and $R_2$ be the "divides" and "is the multiple of " relations on the set of all positive integers respectively. That is, $R_1 = \{(a,b) | a \text{ divides }b\}$ and $R_2 = \{(a,b) | a \text{ is a multiple of }b\}$. Find
a. $R_1 \cup R_2$
b. $R_1 \cap R_2$

My work:
For a, I understand the concept of a union. I am trying to apply this example, say you have a relation $R_1$ with ordered pairs $\{(1,1), (2,2), (3,3)\}$ and a relation $R_2$ with ordered pairs $\{(1,1),(1,2),(1,3),(1,4)\}$. The union $R_1 \cup R_2$ would consist of the ordered pairs $\{(1,1),(1,2),(1,3),(1,4),(2,2), (3,3)\}$.

Applying that idea here, I got that the union $R_1 \cup R_2$ relation would be $\{(a,b) | a \text{ divides } b \text{ or } a \text{ is a multiple of }b\}$ or $\{(a,b) | a = cb \text{ or } b = ak \text{ for some integers }k \text{ and }c\}$.

Is there someway to simpify this into one expression?

For b , going off the last example, the intersection $R_1 \cap R_2$ would consist of ordered pairs $\{(1,1)\}$. Now applying that idea here, $R_1 \cap R_2$ relation would be $\{(a,b) | a \text{ divides } b \text{ or } a \text{ is a multiple of }b\}$ or $\{(a,b) | a = cb \text{ or } b = ak \text{ for some integers }k \text{ and }c\}$.

Investigating this further, I found that $a = cb$ and $b = ak$ is only true when $k=c= 1$ so overall the relation would consist of $\{(a,b) | a = b\}$. Is that correct?

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a) You could say that $R_1\cap R_2=\{(a,b): a \text{ divides } b \text{ or } b \text{ divides } a\}$ as well.

b) Yup. $a=cb$ and $b=ka$ implies $a=c(ka)$, so $ck=1$. Since $c,k$ are integers, $c=k=1$.

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  • $\begingroup$ Oh ok thanks. So no way to reduce a divides b or b divides a? $\endgroup$ – committedandroider Mar 8 '15 at 5:36
  • $\begingroup$ Anything else would obfuscate things I think. Technically, $\max(\frac{a}{b},\frac{b}{a})\in\mathbb Z$ does the trick as well, but I can't come up with anything more clever. $\endgroup$ – StevenClontz Mar 8 '15 at 5:48
  • $\begingroup$ On that note, I guess $\min(a,b)$ divides $\max(a,b)$ might be what you're looking for. $\endgroup$ – StevenClontz Mar 8 '15 at 5:49
  • $\begingroup$ Is another way of looking at a be evaluating $\frac ab$ = m and $\frac ba$ = k and saying that the only way $\frac ab$ and $\frac ba$ evaluates to integers is if a = b? $\endgroup$ – committedandroider Mar 9 '15 at 2:09

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