Correction: Ali, M., et al. Study on the Development of Neutrosophic Triplet Ring and Neutrosophic Triplet Field. Mathematics 2018, 6, 46

We have found the following errors in the article which was recently published in Mathematics [1]:

1. In Example 1, 3 gives rise to the neutrosophic triplet (3, 3, 3). However, 3 has two neutrals: neut(3) = {3, 5}, but 3 does not give rise to a neutrosophic triplet for neut(3)= 5, since anti(3) does not exist in ${\mathbb{Z}}_6$ with respect to neut(3) = 5.

2. In Example 2, ${\mathbb{Z}}_{10}$ is not a neutrosophic triplet group. 7 is the classical unitary element of the set ${\mathbb{Z}}_{10}$. Therefore ${\mathbb{Z}}_{10}$ is a neutrosophic extended triplet group.

3. In classical ring theory, for any ring $\left(R,+,.\right)$, 0 is the additive identity element. However, in a neutrosophic triplet ring $\left(N,\ast ,\#\right)$, 0 is an ordinary element and the element 0 is not used in definition. Also N may not have such an element. So, in Definition 8 and subsequent parts of the paper, when using the element 0, the element 0 should be defined.

4. In classical ring theory, for any ring $\left(R,+,.\right)$, n∙a is defined by $a+\dots +a$ and $a^n$ is defined by $a\dots a$ (n times). In neutrosophic triplet ring (NTR), we do not know the definition of $a^n$. So before Definition 11, the element $a^n$ should be defined.

5. For the proof of Theorem 3, Theorem 1 was used. So Theorem 3 must satisfy the hypothesis of Theorem 1. Also according to definition of $a^n$, Theorem 3 should be rewritten.

6. Proposition 1 and its proof are not true. The sentences “if a is not a zero divisor, so a is cancellable” and “if a is cancellable, a is not a zero divisor” are not true. These statements cannot be obtained from the given definitions and theorems.

7. The set P(X) in Example 3 is not neutrosophic triplet field. P(X) has identity elements X and $\varnothing$ for the operations $\cup \mathrm{and}\cap$, respectively. Therefore P(X) is a neutrosophic extended triplet group.

8. The counterexamples given for Theorem 5 do not satisfy the distributive law since $1\#(1\ast 2)\ne (1\#1)\ast (1\#2)$.

9. In the proof of Theorem 6, the set N is not NTF since $5\#(5\ast 5)\ne (5\#5)\ast (5\#5)$.

10. The proof of Theorem 7(2) is not true. If $c\in U$, then $f^{-1}(c)$ is a set. If f is not a function, $f^{-1}(c)$ can be equal to an empty set. Then $f^{-1}(c)\ast f^{-1}(d)$ is not in $f^{-1}(U)$. We can prove it by the following:

Let $a,b\in f^{-1}(U)$. Then $f(a),f(b)\in U$ and $f(a)\oplus f(b)=f(a\ast b)\in U$. Hence we get $a\ast b\in f^{-1}(U)$. The proof of $a\#b\in f^{-1}(U)$ is similar. Also, since $f(a)\in U$ and $neut*\left(f\left(a\right)\right)=f\left(neut*\left(a\right)\right)\in U$, we have $neut*(a)\in f^{-1}(U)$. The proof of $neu{t}^{\#}(a)\in f^{-1}(U)$ is similar.

11. The proof of Theorem 7(3) is not true. If $i\in I\mathrm{and}r\in NT{R}_2$, then $f^{-1}(i)\mathrm{and}{f}^{-1}(r)$ is a set. If f is not a function, $f^{-1}(i)\mathrm{and}{f}^{-1}(r)$ can be equal to an empty set. Then $f^{-1}(i)\ast f^{-1}(r)$ is not in $f^{-1}(I)$. We can prove it by the following:

Let $a\in f^{-1}(U)$ and $r\in NT{R}_1$. Then $f(a)\in Iandf(r)\in NT{R}_2$ and $f(a)\oplus f(r)=f(a\ast r)\in I$. Hence we get $a\ast r\in f^{-1}(I)$. The remaining part of the proof is similar.

12. The proof of Theorem 7(4) should be proven as the following:

Let $j\in f\left(J\right)$ and $r\in NT{R}_2$. Since f is onto, then $\exists h\in J$ exists such that $f\left(h\right)=j$ and $\exists s\in NT{R}_1$ such that $f\left(s\right)=r$. Then $h\ast s\in J$ and we get $f\left(h\ast s\right)=f\left(h\right)\oplus f\left(s\right)=j\oplus r\in f(J)$.