1. Can you divide by vectors? If so, how? For instance, in F=ma, F and a are vectors, so to solve for mass, you need to divide a vector by a vector, how!?!?  2.

3. Originally Posted by Muon321 Can you divide by vectors? If so, how? For instance, in F=ma, F and a are vectors, so to solve for mass, you need to divide a vector by a vector, how!?!?
No, you cannot.  4. Originally Posted by xyzt  Originally Posted by Muon321 Can you divide by vectors? If so, how? For instance, in F=ma, F and a are vectors, so to solve for mass, you need to divide a vector by a vector, how!?!?
No, you cannot.
Yes, you can. jocular  5. Originally Posted by jocular Yes, you can. jocular
xyzt is right, division by vectors is not defined ( obviously ! ); you can only divide the magnitudes of vectors, i.e. you can relate their "lengths" to each other.  6. Division isn't an actually existing operator for vectors. But you can multiply with the inverse vector. However, that would be a transposed vector. And hence not really a vector as the original vector was.  7. Actually Kerling, this is wrong.

A field is defined in abstract algebra as a set set equipped with 2 closed binary operations ( and ) each with their own inverses and their own identities.

A vector space is defined as an abelian group together with an associated scalar field.

The first condition says that for every element in this group there exists another element such that where one takes to be the identity, defined by It is usual for an abelian group to write the group operation as (this is what "abelian" means

It follows that, if arithmetic addition is our group operation, then and the identity must be the zero by . Notice that "subtraction" is not defined.

The space of all such abstract group elements becomes a vector space when, for any field and any that Notice the quotient is well defined in , BUT

This does not mean that is, for this would imply that and thus our vector space would be a field, which we do not want  8. Originally Posted by Kerling Division isn't an actually existing operator for vectors. But you can multiply with the inverse vector. However, that would be a transposed vector. And hence not really a vector as the original vector was.
This is funny. I said something similar to this in another forum and got flamed for it. Of course part of the flame was because I made the goof of solving it as a = F/m when I really meant to obtain m = F/a. Since you weren't insulted for saying this I guess I should stay here.

The vector he is referring to doesn't have an inverse but it does have a transpose. There is only a limited class of matrices whose inverse is their transpose. An n-tuple is not one of them.

However if you did multiply each side by it's transpose then you'd get You could then take the positive square root to get F = ma and then solve to obtain m = F/a.  Posting Permissions
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