In Exercises 1 through 28, determine whether each of the given sets is a real linear space, if addition and multiplication by real scalars are defined in the usual way. For those that are not, tell which axioms fail to hold. The functions in Exercises 1 through 17 are real-valued. In Exercises 3, 4, and 5, each function has domain containing 0 and 1. In Exercises 7 through 12, each domain contains all real numbers.
1. All rational functions.
By definition, the set $V$ is a linear space if it satisfies the following ten axioms.
$\textit{Closure axioms}$
$\text{Axiom 1.}\quad$ $\text{Closure Under Addition. }$ For every pair of elements $x$ and $y$ in $V,$ there corresponds a unique element in $V$ called the sum of $x$ and $y,$ denoted by $x + y.$
$\text{Axiom 2.}\quad$ $\text{Closure Under Multiplication. }$ For every $x$ in $V$ and every real number $a$ there corresponds an element in $V$ called the product of $a$ and $x,$ denoted by $ax.$
$\textit{Axioms for addition}$
$\text{Axiom 3.}\quad$ $\text{Commutative Law. }$ For all $x$ and $y$ in $V,$ we have $x + y = y + x.$
$\text{Axiom 4.}\quad$ $\text{Associative Law. }$ For all $x,$ $y,$ and $z$ in $V,$ we have $(x + y) + z = x + (y + z).$
$\text{Axiom 5.}\quad$ $\text{Existence of Zero Element. }$ There is an element in $V,$ denoted by $O,$ such that
$$
x + O = x
\quad
\text{for all $x$ in $V$.}
$$
$\text{Axiom 6.}\quad$ $\text{Existence of Negatives. }$ For every $x$ in $V,$ the element $(-1)x$ has the property
$$
x + (-1)x = O
$$
$\textit{Axioms for multiplication by numbers}$
$\text{Axiom 7.}\quad$ $\text{Associative Law. }$ For every $x$ in $V$ and all real numbers $a$ and $b,$ we have
$$
a(bx) = (ab)x.
$$
$\text{Axiom 8.}\quad$ $\text{Distributive Law for Addition in $V$. }$ For all $x$ and $y$ in $V$ and all real $a,$ we have
$$
a(x + y) = ax + ay.
$$
$\text{Axiom 9.}\quad$ $\text{Distributive Law for Addition of Numbers. }$ For all $x$ in $V$ and all real $a$ and $b,$ we have
$$
(a + b)x = ax + bx.
$$
$\text{Axiom 10.}\quad$ $\text{Existence of Identity. }$ For every $x$ in $V,$ we have $1x = x.$
Recall that a rational function is the quotient $p/q,$ where $p$ and $q$ are polynomials of degree greater than or equal to zero, defined for all $x$ such that $q(x) \neq 0$. In other words they satisfy the ten axioms in the same way the real numbers do. Thus, the set of all rational functions is a linear space. $\,\blacksquare$
