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?
Best Answer
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, $c=k=1$.