1. Final Proof

For $\displaystyle z, w \in \mathbb{C}$, prove that $\displaystyle z + w = w + z$ and $\displaystyle zw = wz$.

My proof:

Let $\displaystyle z = a + ib$ and $\displaystyle w = c + if$

Then, $\displaystyle (a + ib) + (c + if) = (c + if) + (a + ib)$

$\displaystyle (a + ib)(c + if) = ac + aif + ibc -bf$ and similarly for $\displaystyle (c + if)(a + ib)$

Does this prove it?

2. Originally Posted by Ideasman
For $\displaystyle z, w \in \mathbb{C}$, prove that $\displaystyle z + w = w + z$ and $\displaystyle zw = wz$.

My proof:

Let $\displaystyle z = a + ib$ and $\displaystyle w = c + if$

Then, $\displaystyle (a + ib) + (c + if) = (c + if) + (a + ib)$

$\displaystyle (a + ib)(c + if) = ac + aif + ibc -bf$ and similarly for $\displaystyle (c + if)(a + ib)$

Does this prove it?
hehe, weird questions. it's hard to prove things that seem so obvious.

for the addition i guest you could go the extra step and say by the commutative property of addition, you can rearrange the terms, and make it look exactly like the left hand side

for the multiplication, i'd to as you do. consider the left hand side and expand. then consider the right hand side, and expand. show that you end up with the same expression after expanding