Linear Algebra: How can the zero vector equal 1? This example is given in my text for linear algebra.

Example 4.1.6. Let $V =\{x\in\mathbb R: x > 0\}$ be the set of all positive real numbers. We will define a strange “addition” and “scalar multiplication” on $V$, and to avoid confusion, we use $\oplus$ for our definition of addition, and $\odot$ for our definition of scalar multiplication.

*

*Given $a,b\in V$, let $a\oplus b = a\cdot b$, i.e. $a\oplus b$ is usual product of $a$ and $b$.

*Given $a\in V$ and $c\in\mathbb R$, let $c\odot a = a^c$ (notice that $a^c > 0$ because $a > 0$).

Finally, let $\vec0 = 1$. With these operations, $V$ is a vector space.

Now the aspect of this that I don't understand, is how the zero vector can be equal to 1.
 A: Observe that the exponential function $x\mapsto e^x$ takes addition to multiplication, and it will actually serve as an isomorphism between $\Bbb R$ with the usual one dimensional vector space structure and the newly defined one on $\Bbb R_{>0}$. 
And, the usual zero vector, i.e. $0\in\Bbb R$ is mapped to $e^0=1$, which is the neutral element for the operation $\oplus$ (multiplication).  Writing out, the identity 
$$\tilde0\oplus x=x$$
translates to
$$1\cdot x=x$$
(Note that any positive real number works as base for the exponential in place of $e$.) 
A: Your vector space needs a $0$ vector. 
For the strange addition that is defined as product, the only candidate is $1$.
I worry more about the statement " Given a ∈ V and c ∈ R, let c  a = a c (notice that a c > 0 because a > 0). "
If $c$ is an arbitrary scalar in $R$,  then how come $ac>0$ because $c>0 ?$  
A: Note I go through proving the whole vector space for anyone who was confused about anything else with this specific vector space, but your specific question deals with properties 5 and 6, so you could just skip to those if you wanted. 
If we have a set called $V$ and define two arbitrary operations we call "addition" ($+$) and "scalar multiplication" ($*$) on $V$ then for $V$ to be a vector space on $\Bbb C$ it has to satisfy the following properties:


*

*Additive Closure: for $\mathbf a, \mathbf b \in V$ then $\mathbf a + \mathbf b\in V$

*(Scalar) Multiplicative Closure: if we have some scalar $c \in \Bbb C$ and $\mathbf a \in V$ then $c* \mathbf a$ must also be in $V$

*Commutativity: $\mathbf a, \mathbf b \in V$ then $\mathbf a + \mathbf b = \mathbf b + \mathbf a$

*Associativity in addition: $\mathbf a, \mathbf b, \mathbf c \in V$ then $(\mathbf a + \mathbf b) + \mathbf c= \mathbf a + (\mathbf b +\mathbf c)$

*Additive Identity Element (Zero Vector): there must be some $\mathbf 0$ such that for all $\mathbf a \in V$ then $\mathbf a + \mathbf 0 = \mathbf a$

*Additive Inverse: there must be some $- \mathbf a$ for all $\mathbf a$ such that $\mathbf a + (-\mathbf a) = \mathbf 0$

*(Scalar) Multiplicative Associativity: if $c, k \in \Bbb C$ and $\mathbf a \in V$ then $(ck)\mathbf a = c(k \mathbf a)$

*Distibutive Over Vector Addition: $c \in \Bbb C$ and $\mathbf a, \mathbf b \in V$ then $c(\mathbf a + \mathbf b) = c \mathbf a + c\mathbf b$

*Distributive Over Scalar Addition: $c, k \in \Bbb C$ and $\mathbf a \in V$ then $(c+k) \mathbf a = c \mathbf a + k \mathbf a$
With your given set and operations all of these properties are satisfied:


*

*When defining $+$ as the usual operation of multiplication then for any two vectors you "add", because all vectors are positive reals and multiplication between two real numbers always gives another positive real number, addition will always give you another vector in $V$

*Since all of your vectors are positive real numbers and "multiplication" is raising to a power all of the vectors that your operation returns will also be positive real numbers and therefore are in $V$ (even if the scalar power was negative, you would just get a fraction) 

*The usual operation of multiplication is commutative so therefore "addition" in this vector space is also commutative

*Since the usual operation of multiplication is associative, so is your operation of "addition"

*One of the properties of a zero vector is to be the identity element under addition in a vector space, which just means that if you do $anything + \mathbf 0 = anything$. That's why in your vector space since we use multiplication as "addition" the real number 1 is the zero vector since it will return anything that gets input i.e. because $1*anything=anything$ your operations give $anything + \mathbf = anything$

*An additive inverse would be the reciprocal of whatever your vector is in this vector space. The inverse has to produce the zero vector, and with these operations you can produce the zero vector. $a*1/a= \mathbf 0$ since the zero vector is the number 1. 

*Because multiplication is associative so is the "addition" operation

*Multiplication is distributive with scalars and with vectors which satisfies properties 8 and 9.


So hopefully that clears up any confusion at all with this really interesting (and odd) example of a vector space. The main reason it's taught is just to abstract the idea of a zero vector and really show that it has less to do with the specific number zero and more to do with the properties that zero has (that it is the additive identity, and that the additive inverse produces it). 
