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:

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

Title alternative proof of condition on a near ring to be a ring AlternativeProofOfConditionOnANearRingToBeARing 2013-03-22 17:20:06 2013-03-22 17:20:06 Wkbj79 (1863) Wkbj79 (1863) 9 Wkbj79 (1863) Proof msc 20-00 msc 16-00 msc 13-00