The connection between Boolean Algebra and Number Theory.

Represent a natural number to multi-set

Consider a $m\in \mathbb{N}$$m\in \mathbb N$ $m=p_1^{i_1}{p}_2^{i_2}{p}_3^{i_3}...p_n^{i_n}$$m=p_1^{i_1}p_2^{i_2}p_3^{i_3}...p_n^{i_n}$, and define $\mathcal{D}$$\mathcal D$ be the divisor set of $m$$m$.

$\mathcal{D}:=\{d\in \mathbb{N},d|m\}$$\mathcal{D}:=\{d\in\mathbb N,d|m\}$

Define $S:=\{p_1,...p_1,p_2,...p_2,...,p_n,...,p_n\}$$S:=\{p_1,...p_1,p_2,...p_2,...,p_n,...,p_n\}$, $S$$S$ is a multi-set.

Then we have $(\mathcal{D},|,gcd,\mathrm{lcm})\cong (P(S),\subseteq ,\cap ,\cup )$$(\mathcal D,|,\gcd,\mathrm{lcm})\cong(P(S),\subseteq,\cap,\cup)$

we need observe that for $|S|=N,|P(S)|\ne 2^N$$|S|=N,|P(S)|\ne 2^N$ for multi-set

for example, $S=\{a,a,b\}$$S=\{a,a,b\}$ , we can consider $(i_a,i_b),i_a\in \{0,1,2\},i_b\in \{0,1\}$$(i_a,i_b),i_a\in\{0,1,2\},i_b\in\{0,1\}$

Thus $|P(S)|=2\times 3=6\ne 8$$|P(S)|=2\times 3=6\ne 8$

Another thing we need to observe is $\begin{array}{c}\{a,a,a\}\cap \{a,a\}=\{a,a\}\\ \{a,a,a\}\cup \{a,a\}=\{a,a\}\end{array}$$\{a,a,a\}\cap\{a,a\}=\{a,a\}\\\{a,a,a\}\cup\{a,a\}=\{a,a\}$

because of $\begin{array}{c}A\cap B=A\iff A\subseteq B\iff A\cup B=B\end{array}$$A\cap B=A\Leftrightarrow A\subseteq B\Leftrightarrow A\cup B=B\\$

by the way, I think the ''disjoint union'', I mean, $\{a,b\}\bigsqcup \{b,c\}=\{a,b,b,c\}$$\{a,b\}\sqcup\{b,c\}=\{a,b,b,c\}$

is a representation of $ab\times bc$$ab\times bc$

The isomorphism $f:\mathcal{D}\to P(S)$$f:\mathcal D\to P(S)$ is given by $f(1)=\mathrm{\varnothing },f(a^2)=\{a,a\},f(ab)=\{a,b\}...$$f(1)=\emptyset,f(a^2)=\{a,a\},f(ab)=\{a,b\}...$

The representation of partial order

we have $d_1|d_2\iff f(d_1)\subseteq f(d_2)$$d_1|d_2\Leftrightarrow f(d_1)\subseteq f(d_2)$, $f(gcd(a,b))=f(a)\cap f(b),f(\mathrm{lcm}(a,b))=f(a)\cup f(b)$$f(\gcd(a,b))=f(a)\cap f(b),f(\mathrm{lcm}(a,b))=f(a)\cup f(b)$

The duality is given by ${\displaystyle \frac{m}{a}}$$\displaystyle\frac{m}{a}$, $\displaystyle f\left(\frac{m}{a}\right)=f(a)^c, (a')'=\frac{m}{\displaystyle \frac{m}{a}}=a$

And we have De Morgan Law (amazing)

$(A\cap B{)}^c=A^c\cup B^c\leftrightarrow {\displaystyle \frac{m}{gcd(a,b)}=\mathrm{lcm}{\displaystyle (\frac{m}{a},{\displaystyle \frac{m}{b}})}}$$(A\cap B)^c=A^c\cup B^c\leftrightarrow \displaystyle\frac{m}{\gcd(a,b)} =\mathrm{lcm}\displaystyle\left(\frac{m}{a},\displaystyle\frac{m}{b}\right)$

$(A\cup B{)}^c=A^c\cap B^c\leftrightarrow {\displaystyle \frac{m}{\mathrm{lcm}(a,b)}=gcd(\frac{m}{a},\frac{m}{b})}$$(A\cup B)^c=A^c\cap B^c\leftrightarrow\displaystyle\frac{m}{\mathrm{lcm}(a,b)}=\gcd\left(\frac{m}{a},\frac{m}{b}\right)$

Absorb Law

$A\cap (A\cup B)=A,A\cup (A\cap B)=A\leftrightarrow gcd(a,\mathrm{lcm}(\mathrm{a},\mathrm{b}))=a，\mathrm{lcm}(\mathrm{a},gcd(\mathrm{a},\mathrm{b}))=a$$A\cap (A\cup B)=A, A\cup(A\cap B)=A\leftrightarrow\gcd(a,\mathrm{lcm(a,b)})=a，\mathrm{lcm(a,\gcd(a,b)})=a$

Associative Law

$(A\cap B)\cap C=A\cap (B\cap C),(A\cup B)\cup C=A\cup (B\cup C)$$(A\cap B)\cap C=A\cap(B\cap C),(A\cup B)\cup C=A\cup (B\cup C)$

$\begin{array}{c}gcd(gcd(a,b),c)=gcd(a,gcd(b,c))=gcd(a,b,c),\mathrm{lcm}(\mathrm{lcm}(\mathrm{a},\mathrm{b}),\mathrm{c})=\mathrm{lcm}(a,\mathrm{lcm}(b,c))=\mathrm{lcm}(a,b,c)\end{array}$$\gcd(\gcd(a,b),c)=\gcd(a,\gcd(b,c))=\gcd(a,b,c),\mathrm{lcm}(\mathrm{lcm(a,b),c)} =\mathrm{lcm}(a,\mathrm{lcm}(b,c))=\mathrm{lcm}(a,b,c)\\$

Distributive Law

$A\cap (B\cup C)=(A\cap B)\cup (A\cap C),A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$$A\cap(B\cup C)=(A\cap B)\cup( A\cap C),A\cup(B\cap C)=(A\cup B)\cap( A\cup C)$

$\begin{array}{c}gcd(a,\mathrm{lcm}(b,c))=\mathrm{lcm}(gcd(a,b),gcd(a,c)),\mathrm{lcm}(a,gcd(b,c))=gcd(\mathrm{lcm}(a,b),\mathrm{lcm}(a,c))\end{array}$$\gcd(a,\mathrm{lcm}(b,c))=\mathrm{lcm}(\gcd(a,b),\gcd(a,c)),\mathrm{lcm}(a,\gcd(b,c))=\gcd(\mathrm{lcm}(a,b),\mathrm{lcm}(a,c))\\$

One interesting application for the distributive law is as follow

Consider this question, if $gcd(a,b)=1$$\gcd(a,b)=1$, how could I prove that $\gcd(a+b,ab)=1?$

$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{f}.$$\rm Proof.$

Observe that $gcd(a,b)=1\iff \mathrm{lcm}(a,b)=ab$$\gcd(a,b)=1\Leftrightarrow \mathrm{lcm}(a,b)=ab$

So $\begin{array}{c}gcd(a+b,ab)=gcd(a+b,\mathrm{lcm}(\mathrm{a},\mathrm{b}))=\mathrm{lcm}(gcd(a+b,a),gcd(a+b,b))\end{array}$$\gcd(a+b,ab)=\gcd(a+b,\mathrm{lcm(a,b))}=\mathrm{lcm}(\gcd (a+b,a),\gcd(a+b,b))\\$( by the distributive law)

And we know that $gcd(a+b,a)=gcd(a+b,b)=gcd(a,b)=1$$\gcd(a+b,a)=\gcd(a+b,b)=\gcd(a,b)=1$

Because ${\displaystyle \frac{a+b}{d},\frac{a}{d}\in \mathbb{N}\iff \frac{a}{d},\frac{b}{d}\in \mathbb{N}}$$\displaystyle\frac{a+b}{d},\frac{a}{d}\in\mathbb N\Leftrightarrow \frac{a}{d},\frac{b}{d}\in\mathbb N$

So we have $\begin{array}{c}\mathrm{lcm}(gcd(a+b,a),gcd(a+b,b))=\mathrm{lcm}(1,1)=1\end{array}$$\mathrm{lcm}(\gcd (a+b,a),\gcd(a+b,b))=\mathrm{lcm}(1,1)=1\\$