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The operations of addition and scalar multiplication have been defined as: $$(a_1, a_2) + (b_1, b_2) = (a_1 + b_2, a_2b_2)$$ and $$k \cdot (a1, a2) = (k\,a1, a_2^k).$$

This was a problem on one of my tests, and I missed all possible points. When we went over the test, I paid attention to how it was supposed to be done, but did not write it on my test for reference. I tried looking for examples in my textbook, but there aren't any. Any help would be appreciated. My attempt:

zero vector: (a1, a2) + (0,0) = (a1 + 0, a2(0)) = (a1, 0) additive inverse: (a1, 0) + (0,0) = (a1, 0)

This is clearly wrong.

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  • $\begingroup$ Welcome to MSE. please remember to use mathjax. Also it would be better if you write your try on every question :) $\endgroup$ – Arman Malekzadeh May 15 '16 at 15:22
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For the zero vector: you can determine it from any of the two equations (which would make you check that they are compatible, which they are).

If $(a_1,a_2)+(z_1,z_2)=(a_1,a_2)$, this becomes $$(a_1+z_1,a_2z_2)=(a_1,a_2).$$ So you have $a_1+z_1=a_1$, $a_2z_2=a_2$. As you can do this for any choice of $a_1,a_2$, it follows that $(z_1,z_2)=(0,1)$.

Or, easier: $$(z_1,z_2)=0\cdot(1,1)=(0,1^0)-(0,1).$$

For additive inverse: if $(a_1,a_2)+(b_1,b_2)=(0,1)$, we have $$ (a_1+b_1,a_2b_2)=(0,1), $$ that is $a_1+b_1=0$, $a_2b_2=1$. So $b_1=-a_1$, $b_2=1/a_2$. So for $(a_1,a_2)$ with $a_2\ne0$, its additive inverse is $$\left(-a_1,\frac1{a_2}\right).$$

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  • $\begingroup$ During my attempt today, I got it to (a 1 +z 1 ,a 2 z 2 )=(a 1 ,a 2 ). I knew that z 1 = 0, but for reasons unknown couldn't figure out that z 2 was 1. So in regards to the additive inverse, I only need to use the redefined addition and set it equal to the zero vector that was found, and then simplify. $\endgroup$ – Kevin May 15 '16 at 15:36

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