# Proof of Distributive Property of Vectors

1. Feb 26, 2015

### B18

1. The problem statement, all variables and given/known data
Let u, and v be vectors in Rn, and let c be a scalar.
c(u+v)=cu+cv

3. The attempt at a solution
Proof:
Let u, v ERn, that is u=(ui)ni=1, and v=(vi)ni=1.
Therefore c(ui+vi)ni=1

At this point can I distribute the "c" into the parenthesis? For example:

=(cui+cvi)ni=1
=(cui)ni=1+(cvi)ni=1
=cu+cv.

2. Feb 26, 2015

### Dick

Sure you can. You know you can distribute over real numbers. The components of vectors in $R^n$ are just real numbers.

3. Feb 26, 2015

### B18

Ok I figured. So everything looks in order here? Just was expecting it to be a little more involved!

4. Feb 26, 2015

### Dick

No, it's not more involved. This is an easy one.