$D_{42}=\{1,2,3,6,7,14,21,42\}$. The Hasse diagram is shown in fig. has the greatest element I=42 and least element O=1. Consider $1\in D_{42}$. Its complement be b . Then by definition lub(1,b)=I=42. This is true when b =42. Similarly by definition glb(l,b)=O=1,which is again true when b=42. Thus complement of 1 is 42, that is 1'=42. By symmerty complement of 42 is 1, that is 42'=1. Similarly

$2'=21 \ \ \ \ since \ \ \ \ glb(2,21)=1=O, \ \ \ \ 21'=2 \\ 3'=14 \ \ \ \ since \ \ \ \ glb(3,14)=1=O, \ \ \ \ 14'=3 \\ 7'=6 \ \ \ \ \ since \ \ \ \ \ glb(7,6)=1=O, \ \ \ \ \ 6'=7$