Help understanding Vector Space Axioms

Question

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?

Answer 1

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 $0$.

Answer 2

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.

Answer 3

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.