3. Limitations of proposed method for ranking of two SVNS
Let \(\text{\ \ A}_{1}=\left\langle 0.8,0.1,0.6\right\rangle\) and\(\text{\ \ A}_{2}=\left\langle 0.8,0.2,0.4\right\rangle\) be two SVNS, then according to the proposed method [10, Section 3, Def. 3.1, pp. 379], discussed in Section 2.2, \(N{(A}_{1})=N{(A}_{2})=0.5\). Therefore, according to the proposed method [10, Section 3, Def. 3.1, pp. 379], \(A_{1}=A_{2}\). While, it is obvious that\(\text{\ \ A}_{1}\neq A_{2}\). Hence, the proposed method [10, Section 3, Def. 3.1, pp. 379] for the ranking of two SVNS is not valid.
Similarly, let\(\text{\ \ A}_{1}=\left\langle 0.1,0.0,0.1\right\rangle\) and\(\text{\ \ A}_{2}=\left\langle 0.3,0.0,0.3\right\rangle\) be two SVNS, then according to the proposed method [10, Section 3, Def. 3.1, pp. 379], discussed in Section 2.2, \(N{(A}_{1})=N{(A}_{2})=0.5\). Therefore, according to the proposed method [10, Section 3, Def. 3.1, pp. 379], \(\text{\ A}_{1}=A_{2}\). While, it is obvious that\(\text{\ \ A}_{1}\neq A_{2}\). Hence, the proposed method [10, Section 3, Def. 3.1, pp. 379] for the ranking of two SVNS is not valid.