0 votes 0 votes Answer is all Doubt ---how (1,4) is present in O/p Databases zeal databases relational-calculus zeal-workbook + – Prince Sindhiya asked Nov 18, 2018 • edited Mar 9, 2019 by ajaysoni1924 Prince Sindhiya 475 views answer comment Share Follow See 1 comment See all 1 1 comment reply Zack Fair commented Dec 30, 2021 reply Follow Share The answer is A.$(2,3)$ we need to include tuple $(x,y)$ from $R$ ,where we do not have $(x,z)(z,y)$ in $R$ That is, Those tuples in $R$ which can not be transitively derived by tuples of $R$. Checking all $6$ tuples $(1,2),(1,3),(1,4),(2,3),(3,4),(4,5)$ $(1,2)\leftarrow $ there is only one $z=1$ such that $(z,2)=(1,2)$, but there is no $(1,z)=(1,1)$ $(1,3)\leftarrow (1,2)(2,3)$ here $z=2$ $(1,4)\leftarrow (1,3)(3,4)$ here $z=3$ $(2,3)\leftarrow$ there is only one $z=3$ such that $(2,z)=(2,3)$, but there is no $(z,3)=(3,3)$ $(3,4)\leftarrow$ there is only one $z=4$ such that $(3,z)=(3,4)$, but there is no $(z,4)=(4,4)$ $(4,5)\leftarrow$ there is only one $z=5$ such that $(4,z)=(4,5)$, but there is no $(z,5)=(5,5)$ Hence among the options only $(2,3) $satisfies the conditions. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes I applied not in the expression which becomes for all z NOT { R(xz) and R(zy) } according to me only option 3 satisfies as its not present in the relation anyways how to proceed further rish1602 answered Dec 30, 2021 rish1602 comment Share Follow See all 0 reply Please log in or register to add a comment.