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.
Best Answer
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 $e$.)