Hi Iam studying Linear Algebra's vector space.
Question : Let $V$ be the vector of all real numbers with operations $u + v = uv - 1 $ and $c * v = v$ where c is a constant and u and v are vectors form a vector space?
I know that this doesn't form a vector space because of associativity but i'm curious about it's additive identity
Now I have found the additive identity as $(1+1/u)$
But This doesn't make the identity unique. Does that mean it's wrong?
On
Most people are pointing out that the operation is not associative, and they're correct, but you asked about the identity:
In the axioms for a vector space the identity axiom is the following:
There exists an element $0 \in V$ such that $v + 0 = v$ for all $v \in V$.
The important point here is that there is one element $0$ that works with every element $v$.
If you have an operation defined by $(u, v) \mapsto uv - 1$ then you've noticed that for any $u$ the element $v = 1 + \frac{1}{u}$ satisfies $(u, 1 + \frac{1}{u}) \mapsto u$. So, for example $(1, 2) \mapsto 1$. But $2$ is not the identity because if I swap $1$ for a different number, say $(3, 2) \mapsto 5$ I don't get $3$.
So the answer to your question is no, you cannot have a different identity element for each vector. The identity must be unique, meaning you must have a single element that functions as the identity for all vectors.
It is not a vector space. Because your addition is not associative!
I'll rewrite $u\oplus v=uv-1, c\odot v=v$ to reduce confusion.
$(u\oplus v)\oplus w=(uv-1)\oplus w=(uv-1)w-1=uvw-w-1$, and $u\oplus (v\oplus w)=u\oplus (vw-1)=u(vw-1)-1=uvw-u-1$. They are not same in general! Especially, $1\oplus(0\oplus-1)\ne(1\oplus 0)\oplus -1$.
Also, when $e$ is identity of (V,$\oplus$), then $0\oplus e=0$ must hold. But $0\oplus e=0e-1\ne 0$. So, there are no $\oplus$-identity.