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Doubt ---how (1,4) is present in O/p

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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. $(1,2)\leftarrow$ there is only one $z=1$ such that $(z,2)=(1,2)$, but there is no $(1,z)=(1,1)$
2. $(1,3)\leftarrow (1,2)(2,3)$ here $z=2$
3. $(1,4)\leftarrow (1,3)(3,4)$ here $z=3$
4. $(2,3)\leftarrow$ there is only one $z=3$ such that $(2,z)=(2,3)$, but there is no $(z,3)=(3,3)$
5. $(3,4)\leftarrow$ there is only one $z=4$ such that $(3,z)=(3,4)$, but there is no $(z,4)=(4,4)$
6. $(4,5)\leftarrow$ 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.

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
by

1
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