# Help understanding Vector Space Axioms

I am having a difficulty trying to understand an axiom regarding vector spaces.

There exists an element $0$ in $V$ such that $x + 0 = x$ for each $x\in \mathbb{R}$

Two examples, that I don't quite understand.

$1)$ Addition defined as: $x \oplus y = max(x,y)$ for all $x, y \in \mathbb{R} .$

We would prove the axiom in the following way: $x+0=x$. For any $x \in \mathbb{R}$, there exists a number $x-1$ such that $x+0=max(x-1, x)$ so vector $0$ does not exist and axiom fails.

My question is, why are we representing a $0$ vector with anything else but $0$? So far, I heard that $0$ vector is not a conventional $0$ number. What is it then?

$2)$ Another question:

If we define multiplication as: $x \oplus y = x \cdot$ y for all $x, y \in \mathbb{R}$, how can we prove the following axiom?

For each $x \in V$, there exists an element $−x$ in V such that $x+(−x) = 0$.

The proof is as follows:

$x + \frac 1x = x \cdot \frac 1x = 1$ which is the $0$ vector in this case.

Why is $1$ the $0$ vector? How does this equation prove the axiom?

It says that there is an element, represented by the symbol "$0$" such that for all $v\in V$, $v+0=v$. The symbol could be anything really. The axiom could just as well be:

There exists an element, represented by the symbol "$\heartsuit$" such that for all $v\in V$, $v+\heartsuit=v$. The symbol "$0$" is just that-a symbol. It doesn't mean that the element represented by the symbol $0$ is actually $0$. The reason why we give it that symbol is because it BEHAVES like $0$.

The additive identity $\mathbf{0} \in V$ must be a single element of $V$ that satisfies $\mathbf{x} + \mathbf{0} = \mathbf{x}$ for all $\mathbf{x} \in V$.

If you define your operation by $\mathbf{x} \oplus \mathbf{y} = \max(\mathbf{x}, \mathbf{y})$, then the additive identity equation would be saying that there's some vector $\mathbf{0}$ such that $$\max(\mathbf{x}, \mathbf{0}) = \mathbf{x} \quad \text{for all } \mathbf{x} \in V.$$ Since $V = \Bbb{R}$, you're looking for a real number that is less than or equal to every real number. This certainly doesn't exist as the reals are not bounded below.

The term zero vector is the one which is causing confusion for you. Instead of saying zero vector call it "identity". So "identity" of the vector space is that element $i \in V$ such that $v + i = i + v = v$ for all $v \in V$.

In the first case where $V=\mathbb{R}$ and the "addition" has been given as $x \oplus y = \max(x,y)$. For "identity" we want to find an element $i \in \mathbb{R}$ such that $$x \oplus i = i \oplus x = x \qquad \text{ for all } x \in \mathbb{R}.$$ This means we want $i$ such that $$\max(x,i)=\max(i,x)=x \text{ for all } x \in \mathbb{R}.$$
This means $i$ is a real number that is less than or equal to every real number. But no such number can exist in $\mathbb{R}$. This means there is no identity element, i.e. no zero vector.

Now try the same approach to your second problem.