# Thread:

1. ## integer modulo 4 vectors

Hi;

is there anything wrong in saying that if u is a vector over $Z_4$ then for all $a,b \in Z_4\; au+bv=u+u+...+u+v+v+...+v=(a+b)v$
so that the distributive property holds for vectors over Z_4.

Also if

$c_1v_1+c_2v_2+...+c_kv_k=a_1v_1+a_2v_2+...+a_kv_k$

then by the inverse property of groups and the above distributive property

$(c_1-a_1)v_1+(c_2-a_2)v_2+...+(c_k-a_k)v_k=0$

where 0 is the zero vector.

Is there anything wrong with my post at all no matter how trivial? thanks

2. Originally Posted by Krahl
Hi;

is there anything wrong in saying that if u is a vector over $Z_4$ then for all $a,b \in Z_4\; au+bv=u+u+...+u+v+v+...+v=(a+b)v$
so that the distributive property holds for vectors over Z_4.

Also if

$c_1v_1+c_2v_2+...+c_kv_k=a_1v_1+a_2v_2+...+a_kv_k$

then by the inverse property of groups and the above distributive property

$(c_1-a_1)v_1+(c_2-a_2)v_2+...+(c_k-a_k)v_k=0$

where 0 is the zero vector.

Is there anything wrong with my post at all no matter how trivial? thanks
What exactly are you asking? If it's a vector field?

Well...maybe you mean field? No, it has non-trivial zero divisors (2)

3. no, i know $Z_4^n$ isn't a field, i was just wondering whether any of my ascertions were true. it doesn't form a vector space but it does form a structure (with different definitions of linear independence, $v_1,v_2,...,v_k$ are linearly independent if $c_1v_1+...+c_kv_k=0 \Longrightarrow c_iv_i=0 \forall i$), and i wanted to know whether i could say the distributive property holds in the above sense. Thanks.
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