# Prove/Disprove Vector space s.t $(a_1,a_2)+ (b_1+b_2)=(a_1+a_2,a_2b_2)$ and $c(a_1,a_2)=(ca_1,a_2)$

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.

Consider for instance the pair $(0,0)$, does it have an inverse?
• Ofcourse not!! $(0,0)+(z_1,z_2)=(0+z_1,0*z_2)=(z_1,0)$ there is no way to get $(0,1)$ – Tiger Blood Jan 24 '16 at 23:12
• @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 $0$. – Workaholic Jan 24 '16 at 23:14