Neutrosophic Fields and Rings

Then, the neutrosophic triplets are: (0,4,0) and (4,4,4). Whence, (0,0,0) is not a neutrosophic triplet.

Neutrosophic Fields and Rings

To Mumtaz Ali 1) In classical ring, a ring R(*, #) has the properties:

R(*) = commutative group;

R(#) = well-defined, associative law; and distributivity of # with respect to *. Then a neutrosophic triplet ring NTR(*, #) should be:

NTR(*) = commutative neutrosophic triplet group;

NTR(#) = well-defined, associative law; distributivity of # with respect to *. Then it will result that a neutrosophic triplet commutative group NTG(*), endowed with a second law # that is well-defined and associative (no need for neutrosophic triplets with respect to #) and distributive with respect to *, will become a NTR. 2) Going further, a classical ring with a unitary unit with respect to # is a unitary ring. This extends to a Unitary Neutrosophic Triplet Ring (UNTR), i.e. a neutrosophic ring such that each element "a" has a neut(a) with respect to #. 3) Then a Commutative Unitary Neutrosophic Triplet

Ring is a UNTR such that # is commutative. 4) A Neutrosophic Triplet Field NTF(*, #) is a set such that:

NTF(*) = neutrosophic triplet commutative group;

NTF(#) = neutrosophic triplet group; # is distributive with respect to *. 5) If # is commutative, then NTF(*, #) is a Neutrosophic

Triplet Commutative Field. 6) The relationship: neut(a)#neut(b) = neut(a#b) is true indeed if a and b are cancellable. Proof: Apply "a" to the left, and "b" to the right above: a#neut(a)#neut(b)#b = a#neut(a#b)#b, and using the associativity one has: (a#neut(a))#(neut(b)#b) = (a#b)#neut(a#b) a#b = a#b which is true. 7) Similarly if a and b are cancellable we can prove that: anti(a)#anti(b) = anti(a#b) by applying "a" to the left and "b" to the right, then the associativity: (a#anti(a))#(anti(b)#b) = (a#b)#anti(a#b) or neut(a)#neut(b) = neut(a#b) which was proved to be true.