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THEOREM:- The Poset  $[D_{n};/] $ is a boolean algebra iff 'n' is a square-free number.

If the Poset $[D_{n};/] $ is a boolean algebra then compliment of $x = \dfrac{n}{x}\: \forall x\in D_{n}$

Please explain this theorem??

and following question 

Q)Which of the following is not a boolean algebra??

$ A) [ D_{110};/ ] $

$ B) [ D_{91};/ ] $

$ C) [ D_{45};/ ]$

$ D) [ D_{64};/ ]$

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2 Answers

9 votes
9 votes

Another way of solving this question can be like this:-

important point:- For a poset [Dn ; /] to be a boolean algebra the number of elements in the lattice should be of the form 2m.


D110 = { 1,2,5,10,11,22,55,110 } , none of the numbers is a perfect square and the number of elements in the lattice is 8=23. So , it is a boolean algebra.

D91 = { 1,7,13,91} none of the numbers is a perfect square so D91 is square free.And the number of elements in the lattice is 4 which is nothing but 22. So, it is  a boolean algebra.

D45 = {1,3,5,9,15,45} 9 is a perfect square.Also the number of elements is not a power of 2.So, it is not a boolean algebra.

D64 = {1,2,4,8,16,32,64} It has got 3 perfect squares.Also the number of elements is 7 which is not of the form 2m. So it is not a boolean algebra.

I hope you got it...

7 votes
7 votes

Let n=p1p2…pk, where the pi are distinct primes.

Then Dn is a Boolean algebra.

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