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I know that to be a linear space A1-A10 must be satisfied. I said that this set and operations do not define a linear space over $\mathbb{R}$ because (A2) is not satisfied as $$ (x_1,x_2)\boxplus(y_1,y_2) = (x_1+y_2, x_2+y_1) $$ but $$ (y_1,y_2)\boxplus(x_1,x_2) = (y_1+x_2,y_2+x_1) $$ and $(x_1+y_2, x_2+y_1)$ does not equal $(y_1+x_2,y_2+x_1)$. Is this correct, and would it enough to answer the question?

Edit: with numbers an example would be $x = (1,2)$ and $y = (2,4)$. $$ (1,2)\boxplus(2,4) = (5,4) $$ but $$ (2,4)\boxplus(1,2)= (4,5) $$ and $(5,4)$ is not equal to $(4,5)$. Would this be enough to prove it's not a linear space?

  • $\begingroup$ Looks good to me! $\endgroup$ –  mheldman Aug 26 '18 at 15:58

You might want to be very explicit to make things clear.

For example, we want to make sure $x_1 + y_2 \ne y_1 + x_2$. We can pick $x_1 =y_1=x_2=0$ and $y_2=1$.

$$(0, 0)\boxplus (0,1) = (1,0) $$


$$(0, 1)\boxplus (0,0) = (0,1) $$


I just see your edit, yes, it's ok. just that for some reason your second vector doesn't appear.

  • $\begingroup$ Thanks for your help $\endgroup$ – Randy Rogers Aug 26 '18 at 16:35

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