alternative proof of condition on a near ring to be a ring
Theorem 1.
Let (R,+,⋅) be a near ring with a multiplicative identity 1 such that the ⋅ also left distributes over +; that is, c⋅(a+b)=c⋅a+c⋅b. Then R is a ring.
Proof.
All that needs to be verified is commutativity of +.
Let a,b∈R. Consider the expression (1+1)(a+b).
We have:
(1+1)(a+b) | =(1+1)a+(1+1)b | by left distributivity | ||
=1a+1a+1b+1b | by right distributivity | |||
=a+a+b+b | since 1 is a multiplicative identity |
On the other hand, we have:
(1+1)(a+b) | =1(a+b)+1(a+b) | by right distributivity | ||
=a+b+a+b | since 1 is a multiplicative identity |
Thus, a+a+b+b=a+b+a+b. Hence:
a+b | =0+(a+b)+0 | since 0 is an additive identity (http://planetmath.org/AdditiveIdentity) | ||
=(-a+a)+(a+b)+(b+-b) | by definition of additive inverse (http://planetmath.org/AdditiveInverse) | |||
=-a+(a+a+b+b)+-b | by associativity of + | |||
=-a+(a+b+a+b)+-b | since a+a+b+b=a+b+a+b | |||
=(-a+a)+(b+a)+(b+-b) | by associativity of + | |||
=0+(b+a)+0 | by definition of | |||
=b+a | since 0 is an |
∎
Title | alternative proof of condition on a near ring to be a ring |
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Canonical name | AlternativeProofOfConditionOnANearRingToBeARing |
Date of creation | 2013-03-22 17:20:06 |
Last modified on | 2013-03-22 17:20:06 |
Owner | Wkbj79 (1863) |
Last modified by | Wkbj79 (1863) |
Numerical id | 9 |
Author | Wkbj79 (1863) |
Entry type | Proof |
Classification | msc 20-00 |
Classification | msc 16-00 |
Classification | msc 13-00 |