Can someone please tell me how this looks. Thanks in advance! $$(R,+)$$
Take $R(x)$ and $G(x) \in \mathbb{R}$ then, $R(x) + G(x) \in \mathbb{R}$ and $R$ is closed under addition.
Also, $0 + R(x) = R(x) \in \mathbb{R}$ is the identity, and $R(x) + (-R(x)) = 0 \in \mathbb{R}$ is the additive inverse.
$[R(x) + G(x)] + H(x) = R(x) + [G(x) + H(x)]$ implies $R$ is associative.
$ R(x) + G(x) = G(x) + R(x)$ implies $R$ is commutative.
Hence, $R$ is an abelian group.
$$(R,\times)$$
Take $R(x) \times G(x) = G(x) \times R(x)$ implies $R$ is associative, and $R(x) \times G(x) = R(x) \in \mathbb{R}$. Hence, $R$ contains the identity $1(=G(x)) \in \mathbb{R}$. Also, $R(x) \times G(x) = G(x) \times R(x) \in \mathbb{R}$ implies $R$ is commutative.
Then, $$R(x)[G(x) + H(x)] = R(x) \times G(x) + R(x) \times H(x)$$ and $$[G(x) + H(x)] \times R(x) = G(x) \times R(x) + H(x) \times R(x)$$
Hence, $R$ obeys the distributive laws, and therefore $R$ is a commutative ring with identity $1.$
A ring is a set with two operations that satisfy certain conditions. You've got a set. What are the operations?