alternative proof of condition on a near ring to be a ring

Theorem 1.

Let $(R,+,\\cdot)$ be a near ring with a multiplicative identity $1$ such that the $\\cdot$ also left distributes over $+$ ; that is, $c\\cdot(a+b)=c\\cdot a+c\\cdot b$ . Then $R$ is a ring.

Proof.

All that needs to be verified is commutativity of $+$ .

Let $a,b\\in R$ . Consider the expression $(1+1)(a+b)$ .

We have:

$\\displaystyle(1+1)(a+b)$	$\\displaystyle=(1+1)a+(1+1)b$	$\\displaystyle\\quad\\qquad\\text{by left distributivity}$
	$\\displaystyle=1a+1a+1b+1b$	$\\displaystyle\\quad\\qquad\\text{by right distributivity}$
	$\\displaystyle=a+a+b+b$	$\\displaystyle\\quad\\qquad\\text{since }1\\text{ is a multiplicative identity}$

On the other hand, we have:

	$\\displaystyle(1+1)(a+b)$	$\\displaystyle=1(a+b)+1(a+b)$	$\\displaystyle\\quad\\qquad\\text{by right distributivity}$
		$\\displaystyle=a+b+a+b$	$\\displaystyle\\quad\\qquad\\text{since }1\\text{ is a multiplicative identity}$

Thus, $a+a+b+b=a+b+a+b$ . Hence:

$\\displaystyle a+b$	$\\displaystyle=0+(a+b)+0$	$\\displaystyle\\quad\\qquad\\text{since }0\\text{ is an additive identity ({http://%planetmath.org/AdditiveIdentity})}$
	$\\displaystyle=(-a+a)+(a+b)+(b+-b)$	$\\displaystyle\\quad\\qquad\\text{by definition of additive inverse ({http://%planetmath.org/AdditiveInverse})}$
	$\\displaystyle=-a+(a+a+b+b)+-b$	$\\displaystyle\\quad\\qquad\\text{by associativity of }+$
	$\\displaystyle=-a+(a+b+a+b)+-b$	$\\displaystyle\\quad\\qquad\\text{since }a+a+b+b=a+b+a+b$
	$\\displaystyle=(-a+a)+(b+a)+(b+-b)$	$\\displaystyle\\quad\\qquad\\text{by associativity of }+$
	$\\displaystyle=0+(b+a)+0$	$\\displaystyle\\quad\\qquad\\text{by definition of }$
	$\\displaystyle=b+a$	$\\displaystyle\\quad\\qquad\\text{since }0\\text{ is an }$

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