[Math] Understanding how $\mathcal{P}(A)$ is a Boolean algebra

boolean-algebra

I'm currently doing work on discrete mathematics in my free time and am having some difficulties with understanding Boolean algebra. To be specific, I'm stuck on the following practice question:

Let $A = \{a, b\}$ and list the four elements of the power set $\mathcal{P}(A)$. We consider the operation $+$ to be $\cup$, $\cdot$ to be $\cap$, and complement to be set complement. Consider $1$ to be $A$ and $0$ to be $∅$.

  1. Explain why the description above defines a Boolean algebra.
  2. Find two elements $x$, $y$ in $\mathcal{P}(A)$ such that $xy = 0$, $x \neq 0$ and $y \neq 0$.

Starting with the power set: $$\mathcal{P}(A) = \{∅, \{a\},\{b\},\{a,b\}\}.$$

How would I go about finding the elements of $x$ and $y$ to satisfy part two of the question using algebraic axioms? Also, for explaining how the above defines a Boolean algebra, do you think it would suffice to simply mention how there are two binary operations and a set associated with the Boolean algebra?

Best Answer

Let $x=\{a\}$ and $y=\{b\}$. We have $x\ne\emptyset$ and $y\ne\emptyset$. We also have $xy=x\cap y=\{a\}\cap\{b\}=\emptyset=0$, so that seems to work.

A boolean algebra has variables that can only be true or false (or $1$ or $0$). Normal algebraic operations can continue as usual, unless of course they don't work.

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