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Define the set $\{C = (x,y) : x,y \in \Bbb R\}$ with the operations as $$ (x_1, y_1) \oplus (x_2, y_2) = (x_1 + x_2 + 1, y_1 + y_2 + 1) $$ and $\alpha$ $\otimes$ ($x_1$, $y_1$) = ($\alpha$ $x_1$ + $\alpha$ - 1, $\alpha$ $y_1$ + $\alpha$ - 1), for all ($x_1$, $y_1$), ($x_2$, $y_2$) $\epsilon$ C and $\alpha$ $\epsilon$ ${R}$: Is it a vector space? Justify your answer.

I understand how to figure out if something is a vector space, I'm mostly confused by the use of the direct addition and multiplication signs because I was under the impression that the direct addition of ($x_1$, $y_1$) $\oplus$ ($x_2$, $y_2$) is always ($x_1$ + $x_2$, $y_1$ + $y_2$). So how would I define the set in this situation?

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    $\begingroup$ The notation $\oplus$ in the question does not correspond to your notion of direct sum. It is just a notation for anew operation. $\endgroup$ – Kavi Rama Murthy Feb 4 '20 at 5:25
  • $\begingroup$ Just as $\otimes$ does not have its usual meaning as a tensor product in this context, so does $\oplus$ not have its usual meaning as a direct sum. $\endgroup$ – Ben Grossmann Feb 4 '20 at 7:46
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Hint: (For an expedient proof). Note that for $x_1,x_2 \in \Bbb R$, $$ \alpha x_1 + \alpha - 1 = \alpha(x_1 + 1) - 1\\ x_1 + x_2 + 1 = (x_1 + 1) + (x_2 + 1) - 1. $$


Another perspective: if $\phi:\Bbb R \to \Bbb R$ is defined by $\phi(x) = x - 1$, then we have $$ (x_1,x_2) \oplus (y_1,y_2) = (f^{-1}(f(x_1) + f(x_2)), f^{-1}(f(y_1) + f(y_2))),\\ \alpha \otimes (x_1 ,x_2) = (f^{-1}(\alpha f(x_1)), f^{-1}(\alpha(x_2))). $$ In other words, $\oplus$ and $\otimes$ are addition and mulitplication redefined via a transport of structure. See this question and this question for the same idea with a more complicated $f:\Bbb R \to \Bbb R$.

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