Proving a given set is not a vector space

linear algebrasolution-verificationvector-spaces

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.

Best Answer

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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