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I am a bit unclear on vector space properties in terms of ordered pairs of numbers.

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 $c \in\mathbb{R}$,

  1. $(a_1, a_2) + (b_1, b_2) = (a_1 + b_1, a_2b_2)$ and $c(a_1,a_2)=(ca_1, a_2)$

From my understanding, this violates the vector space property of commutativity of addition and associativity of addition? Does it also violate the the zero vector property because if $c = 0$, then $ca_1 = 0$, but if $a_2$ is an arbitrary number, it can also be $0$, violating the uniqueness of the zero vector. Correct?

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I've edited your question to what I think you intended. Please check to make sure I didn't change your meaning. –  Zev Chonoles Apr 8 '13 at 1:22
    
Thank you for the reformatting. This is my first time on match.stackexchange. I now learned a few things from this edit. I was also wondering why my zeroes looked so weird. –  cYn Apr 8 '13 at 1:29

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You would also have problems because $$(6, 0) + (2, 3) = (6, 0) + (2, 1) = (12, 0)$$ so vector addition fails to satisfy the cancellation property.

For similar reasons, as you argue with respect to the zero vector not being unique the zero vector for the definition of scalar multiplication, which is well argued, we would have problems due to the non-cancelling addition because $(0,2) + (0, 0) = (0, 0),$ as does $(0, 4) + (0, 0) = (0, 0)$.

Note also that scalar multiplication we would also run into problem if $c = 0$, $c(1, 2) = (1c, 2) = (0, 2) = (7c, 2) = c(7, 2)$.

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The would-be vector addition is a perfectly well-defined commutative associative binary operation; it just fails, as you point out, to satisfy the cancellation property. –  Branimir Ćaćić Apr 8 '13 at 1:29
    
Sorry if this is trivial, but how did you get c(7,2)? –  cYn Apr 8 '13 at 1:34
    
I meant that if $c = 0$, then $c(7, 2) = 0\cdot (7, 2) = (0\cdot 7, 2) = (0, 2)$, to show that multiplying by $c$ in this way against any two of infinitely many different non-zero vectors gives equal (infinitely many) pairs. –  amWhy Apr 8 '13 at 1:41
    
Ahh thank you for that. I understand it better now. –  cYn Apr 8 '13 at 1:47
    
Sleepy Amy. $_{+}^{+}$ –  Babak S. Apr 8 '13 at 6:27

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