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Let $V$ denote the set of ordered pairs of real numbers. If $(a_1,a_2)$ and $(b_1,b_2)$ are elements of $V$ and $c\in \mathbb{R}$, define $$(a_1,a_2)+(b_1,b_2)=(a_1+b_1,a_2b_2)$$ and $$c(a_1,a_2)=(ca_1,a_2)$$ Is $V$ a vector space over $\mathbb{R}$ with these operations? Justify your answer.

Here is my answer:

If $V$ were a vector space, then since $(a_1,a_2)+(0,1)=(a_1,a_2)$,$\;$$(0,1)$ would be the zero vector. But for the scalar $0$, we have $0(0,2)=(0,2)\neq(0,1)$.

This violates the following theorem about vector spaces: $\forall \;x\in V\;(0x=0)$.

Is my answer correct? I worry because this is an indirect argument. I don't explicitly show that some vector space axiom fails to hold. Can problems like this one be solved indirectly this way? This is my first time studying linear algebra so I'm trying to be extra careful so I don't mess up my foundations.

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    $\begingroup$ IMHO, it's correct. If you need a more direct argument, which pair is the opposite of $(1,0)$? $\endgroup$
    – Pacciu
    May 26, 2020 at 17:27
  • $\begingroup$ @Pacciu $(1,0)$ has no inverse with respect to the identity $(0,1)$. If it did, say $(x_1,x_2)$, then we'd have, $(1,0)+(x_1,x_2)=(1+x_1,0)=(0,1)$, This gives $0=1$ which is absurd. Thanks :) $\endgroup$
    – DS2830
    May 26, 2020 at 17:38

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Yes, this is completely correct. Well done!

If you want a more direct approach: what axiom(s) in the vector space definitions are not satisfied?

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  • $\begingroup$ Existence of inverse fails. As Pacciu pointed out, $(1,0)$ has no inverse. Thank you :) $\endgroup$
    – DS2830
    May 26, 2020 at 17:41

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