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16 votes
16 votes

Let $\wedge $, $\vee $ denote the meet and join operations of lattice. A lattice is called distributive if for all $x, y, z,$

$x\wedge \left ( y\vee z \right )= \left ( x\wedge y \right )\vee \left ( x\wedge z \right )$

It is called complete if meet and join exist for every subset. It is called modular if for all $x, y, z$

$z\leq x\Rightarrow x\wedge \left ( y\vee z \right )=\left ( x\wedge y \right )\vee z$

The positive integers under divisibility ordering i.e. $p\leq q$ if $p$ divides $q$ forms a.

  1. Complete lattice.
  2. Modular, but not distributive lattice.
  3. Distributive lattice.
  4. Lattice but not a complete lattice.
  5. Under the give ordering positive integers do not form a lattice.
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4 Answers

6 votes
6 votes

Consider the lattice with $gcd$ as meet and $lcm$ as join.

It is distributive

  • $gcd(a, lcm(b, c)) = lcm(gcd(a, b), gcd(a, c))$
  • $lcm(a, gcd(b, c)) = gcd(lcm(a, b), lcm(a, c))$

It is complete semi-meet lattice. As there is no upper bound it is not complete semi-join lattice.
ANS: C

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2 votes
2 votes
It will be a complete lattice

12---------3---------1

24------------2

all these lattice are complete lattice, but not distributive or modular always
2 votes
2 votes

as per given wikipedia https://en.wikipedia.org/wiki/Modular_lattice

lhs part of implication in question is i think is typo mistake it shold be x<=z to make it modular , i m giving solution as given in question , according to question it will not be modular , but according to wikipedia it will modular becuse every distributive lattice must be modular , and given lattice is distributive but not showing modular which is contradiction , so i think question is typo mistake 

1 votes
1 votes
it will not be distributive lattice . because for being Distributive lattice it must be complemented lattice and it is not complemented lattice .

ex- 12-------6-----3-----1

now its not complemented lattice . but it will be modular . so it will be modular but not distributive lattice .
Answer:

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