In: Advanced Math
show that for some ring R, the equality a^2−b^2=(a−b)(a+b) holds ∀a,b∈R if and only if R is commutative.
We want to show that for some ring R, the equality a²−b²=(a−b)(a+b) holds ∀a,b∈R if and only if R is commutative.
Forward:
a²−b²=(a−b)(a+b)∀a,b∈R implies R is commutative
Let x=(a−b). Then x(a+b)=xa+xb
=(a−b)a+(a−b)b
=a²−ba+ab−b²
Then we note that a²−ba+ab−b²=a²−b² iff −ba+ab=0 if and only if ab=ba iff R is commutative.
Backwards:
R is commutative implies a²−b²=(a−b)(a+b)∀a,b∈R.
Let x=(a+b). Then (a−b)x=ax−bx=a(a+b)−b(a+b)=a²+ab−ba−b². R is commutative, so ab−ba=0, so a²+ab−ba−b²=a²−b².
Forward: a²−b²=(a−b)(a+b)∀a,b∈R implies R is commutative
Let x=(a−b). Then
x(a+b)=xa+xb=(a−b)a+(a−b)b=a²−ba+ab−b²
Then we note that a²−ba+ab−b²=a²−b² iff −ba+ab=0 if and only if ab=ba iff R is commutative.
Backwards: R is commutative implies a²−b²=(a−b)(a+b)∀a,b∈R.
Let x=(a+b). Then (a−b)x=ax−bx=a(a+b)−b(a+b)=a²+ab−ba−b². R is commutative, so ab−ba=0, so a²+ab−ba−b²=a²−b².