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A comment on How Craig Barton wishes he’d taught maths. Parent comment
In fact, closure under scalar multiplication is there. Pick d in Z and d(a,b,c) = (da,db,dc) is fine.
The real problem is, I need an inverse. So if that exists, we have : e(da,db,dc) = (a,b,c) and e must exist in the set for (a,b,c) != (0,0,0).
Now you are trying to find e that behaves like 1/d , but you've left the set - no good.
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In fact, closure under scalar multiplication is there. Pick d in Z and d(a,b,c) = (da,db,dc) is fine.
The real problem is, I need an inverse. So if that exists, we have : e(da,db,dc) = (a,b,c) and e must exist in the set for (a,b,c) != (0,0,0).
Now you are trying to find e that behaves like 1/d , but you've left the set - no good.