Is This Set a Group? Ring? If we consider the set of linear functions that map reals to reals: $$G = \ \{ \ f(x)=mx+b \mid m,b \in \mathbb{R} \}$$ 
Is $\langle G, + \rangle$ a group under function addition?
Is $\langle G, +, * \rangle$ a ring under function addition and function multiplication?
I am sure that the answer to the first question is yes, but I am unsure if this set comprises a ring? From looking at the ring axioms I would say no, but any hints or solutions would be appreciated.
 A: Your set is not a ring since it is not closed under multiplication.  Note, for example,
$$
(x) * (x) = (x^2) \notin G
$$
If you change $*$ to composition rather than multiplication, then $G$ is closed under $*$, but is not a ring since it fails the distributive property.
A: The multiplicaton operation has to that $G \mapsto G$, so we can't have a group if we define 
$$
(f_1 \circ f_2) (x) = f_1(x) f_2(x) $$
since the right side is not linear.
Instead, let's try
$$f_1 \circ f_2  = f_1( f_2(x) )$$ 
Now multiplication is closed, and there is an identity, namely $f(x) = x$.
Now there are multiple functions, not just the additive identity $f(x) = 0$, that have no multiplicative inverse:
If
$f(x) = k$, then there is no function $g(x)$ such that for all $x$, $g(k)=x$. But that does not disqualify $G$ from being a ring, since there is no need for multiplicative inverses.
But consider associativity.  Does
$$
m_1(m_2+m_3)x + b_2+b_3) +b_1 \stackrel{?}{=}(m_1+m_2)(m_xx+b_3) + (b_1+b_2)
$$
Of course not, just look at the terms that have only $b_i$ in them.  On the left you have $b_1$, on the right you have $(b_1+b_2)$.
So this does not form a ring.
A: Take $f,g\in G$ defined by $f(x)=m_1x+b_1$ and $g(x)=m_2x+b_2$.
$(f\cdot g)(x)=(m_1x+b_1)(m_2x+b_2)=m_1m_2x^2+(m_1+m_2)x+m_1m_2\notin G$
Then $G$ is not closed under multiplication.
