Let A = {2, 3, 4, 5} and B = {3, 6, 7, 10}. A | vedclass

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(D) Two numbers are relatively prime if their greatest common divisor $(GCD)$ is $1$. We check each element $x \in A$ to see if there exists at least

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$\{2, 3, 4, 5\}$

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(D) Two numbers are relatively prime if their greatest common divisor $(GCD)$ is $1$. We check each element $x \in A$ to see if there exists at least one $y \in B$ such that $	ext{gcd}(x, y) = 1$: $1$. For $x = 2$: $	ext{gcd}(2, 3) = 1$,so $(2, 3) \in R$. $2$. For $x = 3$: $	ext{gcd}(3, 7) = 1$,so $(3, 7) \in R$. $3$. For $x = 4$: $	ext{gcd}(4, 3) = 1$,so $(4, 3) \in R$. $4$. For $x = 5$: $	ext{gcd}(5, 3) = 1$,so $(5, 3) \in R$. Since every element in $A$ has at least one image in $B$ under the relation $R$,the domain of $R$ is the set $A = \{2, 3, 4, 5\}$.

Let $A = \{2, 3, 4, 5\}$ and $B = \{3, 6, 7, 10\}$. $A$ relation $R$ is defined from $A$ to $B$ such that $xRy \iff x$ is relatively prime to $y$. What is the domain of $R$?

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