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2 votes
2 votes
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 ?

7 Answers

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2 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.

2 votes
2 votes

Yes. The set $D_n$ of all positive integer divisors of a fixed integer $n$, ordered by divisibility, is a distributive lattice.

If You seek a Formal Proof, Here it goes :

We already know that "The set $(D_n,/)$ is a  lattice." (also Can be proved easily)

So, In Order to Prove it Distributive, We need to Prove either one of the Two Distributive rules (Because If One holds then Other also Holds)

So, We need to Prove that $a \vee (b \wedge  c) = (a \vee b) \wedge (a \vee c)$

2 votes
2 votes

A distributive lattice, each element can have at most one complement.

1 votes
1 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.

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