# is the following a Boolean Algebra?

Boolean Algebra:

$$D_{30}=\{n:n\mid30\}= \{1,2,3,5,6,10,15,30\}$$

I don't know how to test that this is a boolean algbra (a BA is a distributive lattice with $T,F$ in which every element has a complement)

Why are the "atoms": $2,3,5$? what is the definiton of atoms?

How is is an isomorphism (one-to-one and when subjective)?

i have this question material on my test and i really need help in clarification. Thank You

If $B$ is a Boolean algebra with $\bot$ as its bottom element (your $F$), an element $a\in B$ is an atom of $B$ if $a\ne\bot$ and $\{b\in B:\bot<b<a\}=\varnothing$. In other words the atoms of $B$ are the minimal non-$\bot$ elements.

You can’t show that $D_{30}$ by itself is a Boolean algebra, but you can show that it becomes one when you give it the right operations and order. The join of two elements of $D_{30}$ is their least common multiple, and the meet is their greatest common divisor: $m\land n=\gcd\{m,n\}$ and $m\lor n=\operatorname{lcm}\{m,n\}$. This means that the lattice order on $D_{30}$ is the divisibility relation: if $\preceq$ denotes the lattice order ($m\preceq n$ iff $m\land n=m$), then $m\preceq n$ iff $m\mid n$. It’s not hard to see that this means that $\bot=1$ and $\top=30$. The minimal elements of $D_{30}\setminus\{1\}$ are then $2,3$, and $5$: every other member of $D_{30}\setminus\{1\}$ has a non-trivial divisor and therefore fails to be minimal.

To show that $D_{30}$ with these operations really is a Boolean algebra, you simply have to verify that the operations have the required properties: each distributes over the other, and the lattice is complemented. The only part of this that perhaps isn’t completely routine is figuring out what the complement of $n\in D_{30}$ is. You need an element $m$ such that $n\land m=\bot$ and $n\lor m=\top$, i.e., such that $\gcd\{n,m\}=1$ and $\operatorname{lcm}\{n,m\}=30$; what’s a simple description of that $m$ in terms of $n$?

If $B$ is a Boolean algebra, $b\in B$ is an atom if whenever $b=x+y$ we have that $x=b$ or $y=b$. If you think about power sets with the usual operations, then singletons are the atoms.

Alternatively, if you consider this as an order, $b\in B$ is an atom if and only if $\{x\in B\mid x\leq b\}$ is a linearly ordered set. To see the equivalence recall that $a\leq b\iff a+b=b$.

Now we ask ourselves how $D_{30}$ can be made into a Boolean algebra. Here is a [big] hint, which you might figure out on your own when you know the definition of atoms:

Use $\gcd,\operatorname{lcm}$ to define the addition and multiplication; and note that $a\leq b\iff a|b$ is the order you seek.

To prove this is a Boolean algebra, you need to go through the definition of a Boolean algebra and prove that this interpretation of addition, etc. satisfies the axioms of a Boolean algebra.