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 \text{}\mathrm{and}\text{}\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 $i\in I\text{}\mathrm{and}\text{}r\in NT{R}_{2}$, then ${f}^{-1}(i)\text{}\mathrm{and}\text{}{f}^{-1}(r)$ is a set. If f is not a function, ${f}^{-1}(i)\text{}\mathrm{and}\text{}{f}^{-1}(r)$ 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 $a\in {f}^{-1}(U)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r\in NT{R}_{1}$. Then $f(a)\in I\text{}and\text{}f(r)\in NT{R}_{2}$ 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 $j\in f\left(J\right)\text{}$ and $\text{\hspace{0.17em}}r\in NT{R}_{2}$. 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 $f\left(h\ast s\right)=f\left(h\right)\oplus f\left(s\right)=j\oplus r\in f(J)$.