Question

algebraic identities geometrical proof a2 - b2 =[a+b] [a-b]

Answer 1

We want to show that for some ring RR, the equality a2−b2=(a−b)(a+b)a2−b2=(a−b)(a+b) holds ∀a,b∈R∀a,b∈R if and only if RR is commutative.

Here's my proof --- I'm not sure if the first part stands up to examination. I'd be grateful if someone could take a look.

Forward: a2−b2=(a−b)(a+b)∀a,b∈Ra2−b2=(a−b)(a+b)∀a,b∈R implies RR is commutative

Let x=(a−b)x=(a−b). Then

x(a+b)=xa+xb=(a−b)a+(a−b)b=a2−ba+ab−b2x(a+b)=xa+xb=(a−b)a+(a−b)b=a2−ba+ab−b2

Then we note that a2−ba+ab−b2=a2−b2a2−ba+ab−b2=a2−b2 iff −ba+ab=0−ba+ab=0 if and only if ab=baab=ba iff RR is commutative.

Backwards: RR is commutative implies a2−b2=(a−b)(a+b)∀a,b∈Ra2−b2=(a−b)(a+b)∀a,b∈R.

Let x=(a+b)x=(a+b). Then (a−b)x=ax−bx=a(a+b)−b(a+b)=a2+ab−ba−b2(a−b)x=ax−bx=a(a+b)−b(a+b)=a2+ab−ba−b2. RR is commutative, so ab−ba=0ab−ba=0, so a2+ab−ba−b2=a2−b2a2+ab−ba−b2=a2−b2.