Let $V = R^+$, the set of positive real numbers. Define the "sum" of two elements $x$ and $y$ in $V$ to be their product $x \cdot y$ (in the usual sense), and define "multiplication" of an element $x$ in $V$ by a scalar $c$ to be $x^c$. Prove that $V$ is a real linear space with 1 as the zero element.
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 $1x = x.$
$Proof.\quad$ Let $x$ and $y$ be two positive real numbers, then their sum $x + y = xy$ is yet another positive real number, thus satisfying closure under addition. If $c$ is some real scalar, then we have $cx = x^c.$ But, we can rewrite $x^c = e^{c\log x}$ since $c\log x$ is real-valued for any positive real $x.$ But by definition, if $c\log x$ is real-valued, then $e^{c\log x}$ is a positive real number, thus satisfying closure under multiplication.
Since every element in this set is a positive real number, we can set $y = (-1)x = x^{-1},$ which is another positive real number. With this, we can confirm the existence of negatives since $x + y = xy = xx^{-1} = x/x = 1$ (with $1$ being the zero element in this set). Moreover, since $1$ is the zero element, for all $x$ in the set, we have $x + 1 = 1x = x.$ If we treat the zero element as a scalar, this also confirms existence of identity since $1x = x^1 = x.$ And since multiplication of real numbers is both commutative and associative, we can see that for any elements $x,$ $y,$ and $z$ in $R^+,$ we have $$x + y = xy = yx = y + x$$ and $$x + (y + z) = x(yz) = (xy)z = (x + y) + z.$$
To show that $R^+$ fulfills the remaining axioms on multiplication by numbers, suppose $x$ and $y$ are elements of $R^+$ with $a$ and $b$ being real scalars, then we have $bx = x^b$ and $$a(bx) = (x^b)^{a} = x^{ba} = x^{ab} = (ab)x.$$
We also have $$a(x + y) = (xy)^a = x^ay^a = ax + ay$$ and $$(a + b)x = x^{(a + b)} = x^ax^b = ax + bx.$$
As such, we have shown that the set of all positive real numbers is a real linear space. This completes the proof.
