Relation Between Near Ring and Ring Definition 1: a non-empty set $R$ is ring if:


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*$(R,+)$ abelian group

*$(R,.)$ semigrup

*$a(b+c)=ab+ac$ and $(a+b)c=ac+bc, \forall a,b,c\in R$ 


Definition 2: a non-empty set $NR$ is near ring if:


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*$(NR,+)$ group not necessary abelian

*$(NR,.)$ semigroup

*$(a+b)c=ac+bc, \forall a,b,c\in R$


Definition 3: a non-empty set NF is near field if $(NF,+,.)$ is near ring with unity 1 and every non-zero element has inverse under operation "."
Definition 4: a non-empty set $SR$ is Smarandache ring if:


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*$(SR,+,.)$ ring

*has proper subset $F \subset SR$ is field


Definition 5: a non-empty set $SNR$ is Smarandache near ring if:


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*$(SNR,+,.)$ near ring

*has proper subset $NF \subset SNR$ is near field


I want to know every relation of those, here what i get:


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*every ring is near ring, but not the converse

*every s ring is ring, but ring in general is not s ring

*every s ring is s near ring, but not the converse

*every s near ring is near ring, but near ring in general is not s near ring


I end up with, what is the relation between ring and s near ring? Then, i see ring $\mathbb{Z}$, subgroup of $\mathbb{Z}$ is in form $n\mathbb{Z}$, but the only subgroup which include 1 is only $\mathbb{Z}$ and it's not proper subset of  $\mathbb{Z}$, hence ring $\mathbb{Z}$ is not s near ring. So, not every ring is s near ring.
I think s near ring in general is not ring, but from point 2 and 3 (every s ring is ring and every s ring is s near ring), then i think there must be intersection between ring and s near ring since s ring is contained in both of them.
Here the relation by me:

Is it true? cause i'm doubt about my own thinking and opinion.
As usual, any comment and answer will be very highly apreciated. Thank you.
I've added my answer to the question. But, i still need your comment/answer to make this clearer. Thank you.
 A: In case others need to know, i'll write what i know recently about it.


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*to prove ring in general is not s near ring, by counter example, notice ring $\mathbb{Z}$, it hasn't any proper subset which is ring or near ring, hence $\mathbb{Z}$ failed to be s near ring.

*to prove s near ring in general is not ring, by counter example, notice $(\mathbb{Z_2[x]},+,.)$ under usual addition and define "." by $f(x).g(x)=f(x)$ for all $f(x),g(x) \in \mathbb{Z_2[x]}$. It it straightforward to prove $\mathbb{Z_2[x]}$ is near ring under operation defined. Since $\mathbb{Z_2} \subset \mathbb{Z_2[x]}$ and $\mathbb{Z_2}$ is near field too under operation defined, then $\mathbb{Z_2[x]}$ is s near ring. But $\mathbb{Z_2[x]}$ isn't satisfying the left distributive law, hence $\mathbb{Z_2[x]}$ failed to be ring.

*to prove s ring is both ring and s near ring, it's directly from the definition of s ring. Since s ring must be ring first, then s ring is ring too and yes it is also near ring. Since s ring must have a proper subset which is field, then the proper subset must be near field too, hence s ring is also s near ring. Example, notice ring $\mathbb{Z_{12}}$ under usual addition and multiplication. It has $\{0,4,8\}$ as proper subset and it's not hard to prove $\{0,4,8\}$ is field with $4$ acts as unit. Hence $\mathbb{Z_{12}}$ is s ring. Since every ring is near ring and every field is near field, then $\mathbb{Z_{12}}$ is also s near ring.
from above, imho there is good probability that the relation i asked in question is true. Thank you.
