I don't quite understand how to get the last line in this problem.
I'm assuming the bd = db kind of multiplication is already assumed to be commutative, or else you'd have yourself a circular argument. Multiply out the last line (foil it), and you'll get the second-to-last line. So, from the second-to-last line to the last line is a matter of factoring. Since foiling it out gets you the desired result, the method of factoring, whatever it was, was valid.
Hello, coopsterdude!
I don't understand the proof . . . It's very sloppy!
We are supposed to prove that multiplication is commutive.
But in the proof, they already assumed that mult'n is commutative
. . . as Ackbeet already pointed out.
The first step should look like this:
. .
They simplified the third term like this:
. .
They simplified the fourth term like this:
. .
In the next step, they assume that: .
The second line should be:
. .
. . . .
Factor: .
Factor: .
Like I said, very sloppy!