The discussion revolves around the proof of the distributive property of vectors in Rn, specifically the expression c(u+v)=cu+cv, where u and v are vectors and c is a scalar.
Some participants express confidence in the approach taken, suggesting that the distribution is straightforward. However, there is a hint of expectation for a more complex proof, indicating a mix of understanding and curiosity about the depth of the topic.
Participants note that the components of vectors in Rn are real numbers, which may influence their reasoning about the distribution property.
B18 said: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=(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.
B18 said:Ok I figured. So everything looks in order here? Just was expecting it to be a little more involved!