prove that there does not exists a boolean algebra containing only three element

Question

please prove that there does not exists a Boolean algebra containing only three elements .prove it with example so that i can understand easily.i cant understand the question and i could not tried to prove it

Answer 1

If we define a boolean algebra as having at least two elements, then that algebra has a minimal element, i.e., $0$ and a maximal element, i.e., $1$. Each element has a unique complement, i.e., $0^* = 1$ and $1^* = 0$. Let us add a third element $x$, which is distinct from $0$ and $1$. It has to have a complement $x^*$, and $x^*$ is not equal to $0$ or $1$ (since $x$ is not equal to $0$ or $1$). Also, $x^*\neq x$ (if it were $x \lor x^* = 1$ would not hold). So this algebra has $4$ elements, and not $3$. Quite a "long" proof!

Here's a simpler way: any finite Boolean algebra is isomorphic to a field of sets, so has $2^n$ elements (why?). As a consquence of Ston'e representation theorem, any Boolean algebra is isomorphic to a field of sets, proving that no Boolean algebra can have three elements.