 My Math Forum Divisors of 154 forms a lattice, ordered by divisibility.
 User Name Remember Me? Password

 Abstract Algebra Abstract Algebra Math Forum

February 9th, 2017, 10:25 AM   #1
Senior Member

Joined: Jan 2013
From: Italy

Posts: 154
Thanks: 7

Divisors of 154 forms a lattice, ordered by divisibility.

Hi,

Hi have the following exercise about lattices.

Having the lattice of integer divisors of 154, $L_{154}$, ordered by divisibility:
i) Draw Hasse Diagram of $L_{154}$;
ii) Get, if exists, complements of each element in $L_{154}$;
iii) Establish if the lattice is distributive;
iv) Establish if the lattice is Boolean.

so, this is my development, please can you tell me if is it correct? Many thanks!

divisors of $154$:
First of all, I have computed the divisors of $154$ following this algoritm:
I think it is "scholastic", and I don't know if there is another formal algorithm. Do you know any different one?

However,
I consider the prime factors of $154: 2^1 \cdot 7^1 \cdot 11^1$

I have put the powers of each factors in a table:
$\begin{array}{c|c} 2^0 & 2^1 \\ \hline 7^0 & 7^1 \\ \hline 11^0 & 11^1 \\ \end{array}$

obtaining therefore, this equivalent table:
$\begin{array}{c|c} 1 & 2 \\ \hline 1 & 7 \\ \hline 1 & 11 \\ \end{array}$

now, I multiply each element of the 1st row, with each element of the 2nd row, obtaining:
$1, 7, 2, 14$

now I multiply each element of the above row we've just obtained, with each element of the 3rd row, (I prefer to order the elements obtained based on the decimal digit, so it is simply to write them ordered):
$1, 7, 2 \\ 11, 14 \\ 22 \\ 77 \\ 154$

therefore,
$L_{154} = \left \{ 1,2,7,11,14,22,77,154 \right \}$

i) I have done the Hasse diagram in this way: now, I have a doubt,
following this definition:
Quote:
 An element $1$ in a lattice $L$ is called unity, for $L$ if $a \le 1, \, \forall a \in L$. And an element $0$ in $L$ is called a zero for $L$ if $0 \le a, \, \forall a \in L$. If $1$ and $0$ exists in $L$, then: $a \lor 0 = 0 \lor a = a \quad \mbox{ and } \quad a \land 1 = 1 \land a = a \\ \forall a \in L$. Also any finite lattice has both a unity and zero, and always $0 \ne 1$
so I think it is right to considered as zero, the element $1$ of my set,
as one, the element $154$ of my set. Therefore,
$1 = 0 \\ 154 = 1$

ii) all the complements.
following this definition:
Quote:
 Each element of a lattice of subsets of a set has a complement. In a lattice with $0$ and $1$, an element $a'$ is a complement of an element $a$ if $a \lor a' = 0 \quad \mbox{ and } \quad a \land a' = 1$ A lattice with $0$ and $1$ in which each element has a complement is called a complemented lattice.
I have done this table, where to the left column there are elements $a$, and on top $a'$.
Looking at the Hasse diagram I have done, I put a star where the element has a complement, i.e. when the condition said above is fulfilled:
$a \lor a' = 0 \quad \mbox{ and } \quad a \land a' = 1$

$\displaystyle \begin{array}{c|c} {}_a \ddots {}^{a'} & 1 & 2 & 7 & 11 & 14 & 22 & 77 & 154 \\ \hline 1 & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \star \\ \hline 2 & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \hline 7 & \cdot & \cdot & \cdot & \cdot & \cdot & \star & \cdot & \cdot \\ \hline 11 & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \hline 14 & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \hline 22 & \cdot & \cdot & \star & \cdot & \cdot & \cdot & \cdot & \cdot \\ \hline 77 & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \hline 154 & \star & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \end{array}$

hence,
$1$ is the complement of $154$ and conversely
$154 \land 1 = 1 \land 154 = 0 \quad \mbox{ and } \quad 154 \lor 1 = 1 \lor 154 = 1$
$7$ is the complement of $22$ and conversely
$22 \land 7 = 7 \land 22 = 0 \quad \mbox{ and } \quad 22 \lor 7 = 7 \lor 22 = 1$

iii) if the lattice is distributive the following law must be fulfilled:
$a \land (b \lor c) = (a \land b) \lor (a \land c), \forall a,b,c \, \in L_{154}$
we have to consider a sublattice.
I have considered this theory:
Quote:
 A subset $M$ of a lattice $L$ is called a sublattice of $L$ if $M$ is closed relative to the operations $\lor$ and $\land$ of $L$. It can be proved that any nondistributive lattice contains a sublattice whose diagram is like one of the two:  hence,
a subset $M$ of the lattice $L_{154}$ is called a sublattice of $L_{154}$, if $M$ is closed relative to the operations $\lor$ and $\land$ of $L_{154}$.

for example:
$M = \left \{ 11, 22, 77, 154\right \}$
is a sublattice, and this one, is distributive because,
if try for each element in the following way we obtain always the two equations have the same result
$a=22, b=77, c=1 \\ a \land (b \lor c) = 22 \land (77 \lor 1) = 22 \land 1 = 22 \\ (a \land b) \lor (a \land c) = (22 \land 77) \lor (22 \land 1) = 0 \lor 22 = 22$
and however, instead of attempt for every value, it can be seen looking at the Hasse diagram, that it can't be retrieved a sublattice that has a form of one of the two seen above in the two images,
therefore, this lattice is distributive.

iv) is it a Boolean Lattice?
considering the lattice $L_{154}$, and the operations $\lor$ and $\land$ on it, the following laws must be all fulfilled.
I) commutative laws
$a \lor b = b \lor a, a \land b = b \land a$
II) associative laws:
$a \lor (b \lor c) = (a \lor b) \lor c, a \land (b \land c) = (a \land b) \land c$
III) distribute laws:
$a \land (b \lor c) = (a \land b) \lor (a \land c), a \lor (b \land c) = (a \lor b) \land (a \lor c)$
IV) existance of zero and unity:
$a \lor 0 = a, a \land 1 = a$
V) existance of complements:
$\forall a \in L$ there is an element $a'$ such that
$a \lor a' = 1 \quad \mbox{ and } \quad a \lor a' = 0$

only the V) law is not fulfilled, so $L_{154}$ it is NOT a Boolean Lattice.

What do you think about my development of the exercise?
Is there any error? -please, can you give me any suggestions? Many Thanks!!

Last edited by beesee; February 9th, 2017 at 10:33 AM. February 11th, 2017, 05:00 AM #2 Senior Member   Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 there are some errors I go to correct, the right Hasse diagram of $L_{154}$ is the following: that is isomorphic to the cube structure of a Boolean Algebra. so the table become: $\displaystyle \begin{array}{c|c} \, & 1 & 2 & 7 & 11 & 14 & 22 & 77 & 154 \\ \hline 1 & - & - & - & - & - & - & - & * \\ \hline 2 & - & - & - & - & - & - & * & - \\ \hline 7 & - & - & - & - & - & * & - & - \\ \hline 11 & - & - & - & - & * & - & - & - \\ \hline 14 & - & - & - & * & - & - & - & - \\ \hline 22 & - & - & * & - & - & - & - & - \\ \hline 77 & - & * & - & - & - & - & - & - \\ \hline 154 & * & - & - & - & - & - & - & - \\ \end{array}$ each element has a complement, so the lattice is Complemented, and this lattice is a Boolean Algebra. But, it is certainly complemented, (so the table, I think that can be omitted) since the Hasse diagram of $L_{154}$ is isomorphic to "the cube", that is a Boolean Algebra, and a Boolean algebra must be complemented, so $L_{154}$ must be complemented. I think that with this corrections, that's all! What do you think? Tags 154, divisibility, divisors, forms, lattice, ordered Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post CEL Calculus 0 October 2nd, 2012 07:38 PM xYlem Applied Math 1 January 27th, 2012 04:24 PM Mathh Abstract Algebra 9 December 13th, 2010 06:25 PM julian21 Algebra 3 July 22nd, 2010 03:48 PM johnny Elementary Math 4 March 12th, 2008 08:37 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      