Examples in lin.alg. where closure/addition axioms don't hold

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SUMMARY

The discussion focuses on identifying subsets in a coordinate space that demonstrate specific closure properties related to addition and multiplication axioms. The participants suggest examples such as the set {-3, -2, 1} for case (c) where neither axiom holds, and the set {-3, -2, -1} for case (b) where closure addition holds but closure multiplication does not. The importance of distinguishing between complex numbers as vector spaces over real numbers versus complex numbers is emphasized, as it affects the closure properties. A systematic approach is recommended for constructing sets that satisfy the closure axioms.

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Homework Statement



Find examples of subsets in a coordinate space where:

(a) closure addition axiom doesn't hold but closure multiplication does hold,

(b) closure addition axiom does hold but clouser multiplication doesn't hold,

(c) where neither hold.

Homework Equations



None in particular, but what he means by closure addition and multiplication axioms is that you can add two elements of the subset together and that will still be an element of the subset. For multiplication, you're supposed to be able to multiply the element with any number in R, and it the product will also belong to the subset.

The Attempt at a Solution


[/B]
I'm thinking complex numbers have something to do with this. (c) I think could be {-3, -2, 1}, (b) is something like {-3, -2, -1}, but it's (a) that I'm worried about. I simply have no idea for (a).

edit, scratch that I'm being silly about (b)... it's very wrong.
 
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You'll need larger sets.
You can use complex numbers, but be careful if you use them as vector space over the real numbers or as vector space over complex numbers (this influences the options you have for multiplication, and also the dimension of the vector space).

A good way to start for (a) and (b): take an arbitrary element (like "1"), figure out what else has to be in your set to satisfy one of the closure axioms. Then see if the other one is satisfied, if yes you'll have to add another element to your set (because removing one does not work) and repeat the step.
 

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