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!!

  • $\begingroup$ The zero vector is not (0,0) but it is (-1,-1). $\endgroup$ Jan 18, 2014 at 23:54
  • $\begingroup$ The correct zero vector is $(-1, -1)$. $\endgroup$
    – user61527
    Jan 18, 2014 at 23:54
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    $\begingroup$ Commenters: of course you know the answer. Could you perhaps work a little harder to lead the OP to the answer on his/her own? (I call it "teaching".) $\endgroup$ Jan 18, 2014 at 23:57
  • $\begingroup$ I mean I can define vector 0=(-1,-1) then it will work $\endgroup$
    – user108701
    Jan 18, 2014 at 23:57

1 Answer 1


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.

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    $\begingroup$ By the way, I do plan on coming back to address at least the last paragraph in my answer later. But I want to give the OP some time to think things through. $\endgroup$ Jan 19, 2014 at 0:16
  • $\begingroup$ Thank you very much for the help! I've solved that (y1,y2)=(-1,-1), so this is the new 0 over the vector space? It makes x+0=x ! $\endgroup$
    – user108701
    Jan 19, 2014 at 0:25

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