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Let V denot the set of ordered pairs of reals and $c\in \mathbb{R}$.

If $(a_1,a_2), (b_1+b_2)\in V$ $$(a_1,a_2)+ (b_1+b_2)=(a_1+a_2,a_2b_2) \text { and } c(a_1,a_2)=(ca_1,a_2)$$


Neet to show closed add (new def), closed under new def of scalar, has a zero vectro , each element has an inverse

Closed add since reals are closed with add, mult closed mult since a real times a real is still a real

Existince of zero vector of V $$A+\vec {0} = (a_1,a_2)+(0,1)=(a_1+0,a_2*1)=(a_1,a_2)=A$$ Existance of additive inverse $$A+A^{-1}=(a_1,a_2)+(-a_1,a_2^{-1})=(a_1+(-a_1),a_2*a_2^{-1})=(0,1)=\vec{0}$$

we have shown it is closed under new add, new mult., has a zero vector , and additive inverse therefore it is a vector space


Have doubts about if it is actually a vector space because the book states that for any element in V there is no additive inverse so not a vector space.

Also, Does the zero vector with normal add,mult consisting of 2tuples of real numbers have the same zero as this new one?

Appreciate Constructive critique.


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Consider for instance the pair $(0,0)$, does it have an inverse?

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  • $\begingroup$ Ofcourse not!! $(0,0)+(z_1,z_2)=(0+z_1,0*z_2)=(z_1,0)$ there is no way to get $(0,1)$ $\endgroup$ – Tiger Blood Jan 24 '16 at 23:12
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    $\begingroup$ @Decko That's correct, you assumed originally that for each pair $(a_1,a_2)$ there exists another pair $(-a_1,a_2^{-1})$, so that for each real number $a_2$ there exists an inverse $\tfrac1{a_2}$ however this isn't correct for $0$. $\endgroup$ – Workaholic Jan 24 '16 at 23:14

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