Boolean algebra theorem(Lattices)

Question

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};/ ]$

Comments

source of this theorem? i haven't read this

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I see in this theorem in some of the youtube videos.

Answer 1

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

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

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 2^m. So it is not a boolean algebra.

I hope you got it...

Comments

Thank you so much nice explanation

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Much needed information. Thanks mate.

Answer 2

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

Then Dn is a Boolean algebra.

Comments

Can you explain more about square-free number??

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@Lakshman , A number "n"is called square-free if no prime factor is repeated in its prime factorization ..(or) we can say no prime factor should be in a^m form where a belongs to a prime number and m is a positive integer which is >= 2  ..for example , 9 is not a square -free because 9 = 3² ..since prime factor 3 is repeating here 2 times , so it is not a square free number.. 9 is squarefulnumber..similarly 18 is not a square-free number because prime factorization of 18 = 2 * 3² ..Since prime factor 3 is repeating here also ..So it is not a square free number..  similarly 64 = 2⁶   is not a square free number because prime number 2 is repeating here..

For 2^nd part of the theorem, you can refer below image...

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c and d would not be boolean algebra as in the prime factorization numbers are repeated

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@ankitgupta.1729 @Lakshman Patel RJIT

what i know is that in a Boolean algebra, every element will have exactly one complement. But in the hasse diagram of $D_{110}$, there are more than one element not having any complement. Plz clarify this.

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@pranay562 in $D_{110}$, which element does not have complement ?

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@ankitgupta.1729

yeah i found my mistake. got it now. plz check if the following pairs of element & its complement are right $(1,110) (2,55) (11,10) (5,22)$

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@pranay562 Yes, correct :)