Let V be the set of all ordered pairs of real numbers. Is V a vector space over the Real in the case below ? The rules for vector addition and scalar multiplication are shown below.
(a,b)+(c,d) = (a,b)
k(a,b)= (ka,kb)
I have checked all the conditions for a group but I don't know if it is commutative or not. I think it isn't as.
(a,b)+(c,d)= (a,b) but (c,d)+(a,b)=(c,d).
I think it is not a vector space due to another reason too.
As (k+k')(a,b) is not equal to k(a,b)+k'(a,b).
Are both reasons true for V not being a vector space ?
Yes, both reasons are correct. Arguably, to be slightly more formal, you should give a precise counterexample rather than a general form of counterexample. For instance, you have that $(a, b) + (c, d) = (a, b)$ while $(c, d) + (a, b) = (c, d)$, but this isn't strictly speaking a contradiction because it might be true that $a = c$ and $b = d$ and therefore $(a, b) = (c, d)$. To be more precise, you should point out that, for example, $(1, 2) + (3, 4) \neq (3, 4) + (1, 2)$, because it is certainly true that $1 \neq 3$.