is D36 distributive ?

Question

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 ?

Comments

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)

Answer 1

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.

Answer 2

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 $a \vee (b \wedge  c) = (a \vee b) \wedge (a \vee c)$

Answer 3

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

Comments

Complement $c$ of any element $a$ is defined as following :

$a \vee c = I$ And $a \wedge c = O$ ...Where $I$ is the greatest element and $O$ is the least element.

So, here, $I = 36$ and $O = 1$ and $\vee$ represents $LUB$ and $\wedge$ represents $GLB$ .

So, Complement(s) of $4$ = {9}

Complement(s) of $6$ = None.

And It is indeed a Distributive lattice.

Every $(Dn, /)$ is Always a Distributive lattice.

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Complement of 4 is 9

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Yes. That's what I said.

Answer 4

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.

Comments

How do you account for distributive ?