Consider a relation $R\left ( A,B,C,D,E \right )$ and functional dependencies are 

$F=\left \{ AC\rightarrow B,C\rightarrow D,A\rightarrow E,C\rightarrow B \right \}$

Relation $R$ is decomposed into  $R_{1}\left ( A,B,C \right )$ and $R_{2}\left ( C,D \right )$

Then Is it a lossless decomposition?

I am getting doubt, how it can be not lossless decomposition?

  • $\begingroup$ It looks like you've lost $E$ $\endgroup$ – HEKTO Apr 28 '19 at 23:03
  • $\begingroup$ @HEKTO how? Elaborate plz. $\endgroup$ – Srestha Apr 29 '19 at 3:05
  • $\begingroup$ How far I know, it will be taking common element of $R_{1}$ and R_{2}, i.e. C. Now taking closure of C, $(C)^{+}=CD$ which is in a relation $R_{2}$. So, it is lossless. Is it not correct? $\endgroup$ – Srestha Apr 29 '19 at 3:10
  • $\begingroup$ Neither $R_1$ nor $R_2$ contain $E$ $\endgroup$ – HEKTO Apr 29 '19 at 3:20
  • $\begingroup$ @HEKTO but why we concern only about $E$ , but not concerning about other attribute? I mean $R_{1}$ also not contain D other than B , Similar to $R_{2}$ too. Then why r we only emphasise on E. Moreover, I putted the rule , is that rule not correct? $\endgroup$ – Srestha Apr 29 '19 at 4:27

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