Proof of Distributive Property of Vectors

[B18]

Homework Statement

Let u, and v be vectors in Rⁿ, and let c be a scalar.
c(u+v)=cu+cv

The Attempt at a Solution

Proof:
Let u, v ERⁿ, that is u=(u_i)ⁿ_i=1, and v=(v_i)ⁿ_i=1.
Therefore c(u_i+v_i)ⁿ_i=1

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

=(cu_i+cv_i)ⁿ_i=1
=(cu_i)ⁿ_i=1+(cv_i)ⁿ_i=1
=cu+cv.

 

[Dick]

Homework Statement

Let u, and v be vectors in Rⁿ, and let c be a scalar.
c(u+v)=cu+cv

The Attempt at a Solution

Proof:
Let u, v ERⁿ, that is u=(u_i)ⁿ_i=1, and v=(v_i)ⁿ_i=1.
Therefore c(u_i+v_i)ⁿ_i=1

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

=(cu_i+cv_i)ⁿ_i=1
=(cu_i)ⁿ_i=1+(cv_i)ⁿ_i=1
=cu+cv.

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

 

[B18]

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

 

[Dick]

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

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