The problem is asking us to show that $V$ is a vector space over $\mathbb{R},$ where $V=\{(x_1,x_2)\mid x_1,x_2\in\mathbb{R}\}$
and addition and scalar multiplication in $V$ are defined as:
$$(x_1,x_2)+(y_1,y_2)=(x_1+y_1+1,x_2+y_2+1),$$
$$c(x_1,x_2)=(c+cx_1-1,c+cx_2-1).$$
While I was trying to verify that $x+0=x,$ I found $x+0=(x_1+0+1,x_2+0+1)=(x_1+1,x_2+1),$ which is not equal to $(x_1,x_2)$.
Did I do something wrong with it? Or should I just define $0=(1,1)$?
Hope you understand what am I talking about, thank you very much!!
Best Answer
You have correctly identified that the vector $(0,0)$ -- which is the additive identity for the "usual" addition operation in $\mathbb{R}^2$ -- is not an identity for the new addition operation -- which I will write as $\oplus$ so as to distinguish it from the usual one -- you've been given.
That's disconcerting but not definitive: maybe some other vector functions as an additive identity for this new operation? In other words, you want to see if there is $(y_1,y_2) \in \mathbb{R}^2$ such that for all $(x_1,x_2) \in \mathbb{R}^2$ we have
$(x_1+y_1+1,x_2+y_2+1) = (x_1,x_2) \oplus (y_1,y_2) = (x_1,x_2)$.
It should be easy enough to solve this for $y_1$ and $y_2$, if possible. And if it is possible, you should go on to check whether the other vector space axioms hold. If any of these axioms gives you trouble, please let us know.
Also you are probably at least subconsciously wondering: "If this funny new $\oplus$ operation does turn out to give a vector space structure on $\mathbb{R}^2$, does it have something to do with the original $+$ operation?" Don't neglect to answer that question too: the answer will be enlightening.