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In one text I read that , if n is square free it is DISTRIBUTIVE
in other text I read that if n is square free  it is BOOLEAN ALGEBRA .

Which is most correct ?

Here D36 is not square free then... what conclusion can I make ?
asked in Set Theory & Algebra by Veteran (20.2k points)   | 171 views

We talk about distributivity and boolean algebra only iff it's a lattice(POSET). So, Relation needs to be provided with Set under discussion.

So, the correct question would go like : let $[D_{36},/]$ (or) Set $D_{36}$ on relation divides. (although its know that we talk about division mostly)

3 Answers

0 votes
Dn if n is a square free number,then  it will be a boolean algebra because if it is perfect square ,then its factors will repeat and then it may lead to more than one complement of an element which is actually not a boolean algebra. ...so D36 is not a boolean algebra.
answered by Boss (8.1k points)  
How do you account for distributive ?
0 votes
When we draw the  hasse diagram, 6 doesnt have a complement. Now, for a lattice to be distributive, every element must have unique complement.

Hence, not distributive.
answered by Veteran (14.8k points)  
0 votes

In Dn ,if n is square free number then it will be a boolean algebra along with the numbe of vertex should be 2^n and number of edges should be 2*2^n-2.

Above is most important condition to identify whether a relation is boolean algebra or not.Above rule is not for distributive lattice.

Distributive lattice fallow the distributive properties and sublattice properties.

Example:: D64 not Boolean algebra but D110 is boolean algebra.

answered by Loyal (4.5k points)  


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