Question #267640

Prove for any ring R and a,b∈ R , (a+b)²= a²+2ab+b²

1
Expert's answer
2021-11-18T11:01:12-0500

Any commutative ring RR and a,bRa,b\in R , (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2

Prove:

(a+b)2=(a+b)(a+b)=a(a+b)+b(a+b)==aa+ab+ba+bb==a2+ab+ba+b2(a+b)2=(a+b)(a+b)a2+2ab+b2=a2+ab+ba+b2ab=ba(a+b)^2=(a+b)(a+b)=a(a+b)+b(a+b)=\\ =a\cdot a+a\cdot b+b\cdot a+b\cdot b=\\ =a^2+a\cdot b+b\cdot a+b^2\\ (a+b)^2=(a+b)(a+b)\\ a^2+2ab+b^2=a^2+ab+ba+b^2\\ ab=ba

That a ring RR is commutative


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