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**NoSash****Member**- Registered: 2006-02-06
- Posts: 7

How can I prove the associative law (multiplication) using vectors:

m(na) = (mn)a where m and n are real numbers, and a is a vector?

There are three cases to prove:

m>0, n>0

m<0, n<0

m>0, n<0 and m<0, n>0

I have no idea where to start.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I'm going to change your scalars to k and l, since n is normally used to signify the length of a vector.

Since you don't know the dimension the vector is in, you have to use infinite sequence notation:

In R^n:

a = <a0, a1, .... an>

Then:

k(la) = k(<la0, la1, .... lan>) = <kla0, kla1, ... klan>

But

l(ka) = l(<ka0, ka1, ... kan>) = <kla0, kla1, ... klan>

And are thus, equal.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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