Krdytkyu

I was reading an example from a textbook giving a vector space. It goes with all properties to check if it is a vector space, and finally concludes it is. I am worried about existence of zero.

Now, let V be R(+) and define addition and scalar multiplication as follow:
X +Y = XY, for all x and y in R(+), and kx = x^(k). It justifies the existence of zero by saying: From the definition, X + 0 = X•0 implies X + 0 = X which implies X = X•0 which again implies 1 being equal to 0, by canceling the X from both sides. And concludes that the zero element is 1. But can it not that this very statement imply that division by zero is fine? Or maybe since 0 is 1, it doesn't matter. But then X + 1 = X would be equal, and we would again say 1 is 0, so it is fine. So which is it, is 1 zero or zero 1? If 1 is 0, then which number has taken the place of 1?

It is my first encounter with the topic vector space. And also I am finding out I don't know what it means by zero. I just thought it was just defined nothing. Also, any tips how to think about spaces and sets and definition differently.

First, $0$ is not the same as "nothing." It is a specific real number.

In a vector space, you have first a set of elements ("vectors.") In the example this set is $mathbb{R}^+,$ the set of positive real numbers, so that $0$ isn't even an element of the space. Then you have operations called "addition" and "scalar multiplication" that have to satisfy some rules. Here the addition rule is $$X+Y=XY.$$ On the left-hand side the plus sign means addition the vector space; on the right-hand side we have ordinary multiplication of real numbers. One rule we must have is $$X+mathbf{0}=Xtag{1}$$ where I have written $mathbf{0}$ in bold face to indicate that it is the zero element of the vector space, not the real number 0. Since the addition in the vector space is ordinary multiplication, $(1)$ means the same thing as $$Xmathbf{0}=X$$ and $Xne0$ so we must have $mathbf{0}=1.$

As a first encounter with a vector space, this is obviously confusing, as both the field over which you construct the vector space is $mathbb R$, and the elements of the vector space are positive reals $mathbb R_+$, and then you chose/are given an operation for vector space addition that is multiplation in $mathbb R_+$.

But you made the correct conclusion: The neutral element of the vector space addition (the "zero") is the real number $1$. Note that the real number $0$ is not part of the vector space, as $0 not in mathbb R_+$.

When you talk about "division" (by 0) or "the place of 1" you have to remember that vector spaces do not define the concept of multiplication of 2 vector space elements. That means, there is no concept of $2times3$, with both $2$ and $3$ being elements of the vector space. There is a concept of multiplaction of a field element and a vector space element. In this sense $2times3$ is defined, and we have by definition $2times3 = 3^2 = 9$.

Note that in this sense of multiplication of field element and a vector space element the real number $1$ is still the neutral element: $1times x = x^1 = x$.

I can totally understand if this is confusing, as there are now 2 operations named "addition" and 2 operations named "multiplaction" that apply to (some subset of) real numbers and it is hard to distinguish between them if this is all new to you. I'd suggest first looking at other examples of vector spaces where the field and the vector space elements are not (partially) the same.

@saulsplatz wrote a good answer. I will try to add to it.

Think of the "$0$ vector" in a vector space not as the number zero, not as "nothing" but as "the vector that behaves like the number $0$ for addition".

So in ordinary addition, what makes the number $0$ useful and important is the fact that
$$
r + 0 = 0
$$
for every real number $r$,

So in the vector space $mathbb{R}^2$ of pairs of numbers, the $0$ vector is
$(0,0)$ because
$$
(r,s) + (0,0) = (r,s)
$$
for every vector $(r,s)$.

In the strange vector space in your question, where "$+$" is defined to be multiplication, the thing that "behaves like $0$" is $1$, because
$$
r "+" 1 = r times 1 = r
$$
for every positive real number $r$.

(PS If I were teaching linear algebra I would not introduce this strange example early in the curriculum. It's potentially confusing in just the way you were confused.)
$$