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 | |||
On the other hand, we have:
Thus, . Hence:
∎
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 |