Earlier this week we defined Boolean atom: A Boolean atom a is a nonzero element of a Boolean algebra, such that ax = a or ax = 0 for all x.

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1 Friday, April 20 Today we will continue in Course Notes 3.3: Abstract Boolean Algebras. Earlier this week we defined Boolean atom: A Boolean atom a is a nonzero element of a Boolean algebra, such that ax = a or ax = 0 for all x. Here is a summary of things we have proven: If a is a Boolean atom and a = x + y then a = x or a = y. If a, b, are distinct atoms, then ab = 0. If a is an atom, and b, c, are atoms that are different from a, such that a + b = a + c, then b =c. If a Boolean algebra has more than two atoms, then no two atoms are complements of one another. If the sum of all atoms in a finite Boolean algebra is 1, then every nonzero element in the Boolean algebra can be expressed as a sum of atoms.

2 Our first goal today is to prove: In a finite Boolean algebra, the sum of the atoms is 1. Some special terminology is helpful Let x 1,,, x n be any elements of a Boolean algebra. A (general) minterm is any product of the form! y y y where, for each i, either y 1 2 n i = x i or! y = x. i i Examples of general minterms 1. Here are all the minterms for the elements x 1, : x 1 2. Here are all the minterms for the elements x 1,, : x 1

3 Notice what happens when we sum all the minterms (1) above. x 1 = x 1 ( + ) + = x = x 1 =1 ( ) Likewise, a similar result will occur if we sum all the minterms in (2) above. x 1 = x 1 = ( x 1 )+ ( x 1 ) = 1+ 1 //Why?? = + =1

4 Induction suggests the following theorem: Theorem If x 1,,, x n is any collection of n nonzero elements in a Boolean algebra, then the sum of all minterms in these elements is 1. We use induction to prove this theorem. Once this theorem is proven, we use it, and the properties of atoms, to show that the sum of atoms in a Boolean algebra must be 1. Proof of Theorem : (We will skip to the inductive step; the two results from the previous page are basis step proofs, covering the cases where n = 2 and n =3).

5 Now, we apply the result of Theorem Suppose we have the sum of all minterms of all nonzero elements in a finite Boolean algebra. We have just proved that the sum of these minterms is 1. We will now prove the following: Every nonzero minterm is an atom. Let x 1,,, x n be the nonzero elements of a finite Boolean algebra, and let a = y 1 y 2 y n be a nonzero minterm, where, for each i, either y i = x i or! y i = x i. Let x be an arbitrary element of the Boolean algebra. We will prove that a is an atom by showing that ax = a or ax = 0.

6 Combining the previous results, have shown that the sum of all minterms is 1, and the nonzero minterms are atoms, so: The sum of all atoms in a finite Boolean algebra is 1. Now we prove uniqueness: Suppose x is a nonzero element of a Boolean algebra, and x = a 1 + a a m = b 1 + b b n, where all the a i and b i are atoms. We will use the Idempotent law, and the definition of atom, to show the atoms in the sum of the a i are the same collection as the atoms in the sum of the b i, just given different names and possibly different ordering.

7 We have now shown: Every nonzero element in a finite Boolean algebra can be expressed, uniquely, as a sum of atoms. This means that the number of distinct elements in a finite Boolean algebra with n atoms is equal to the number of different ways to combine the n atoms, which is This gives a fundamental characterization: The cardinality of a finite Boolean algebra is, where n is the number of atoms.

8 EXAMPLE Consider the structure D 12 = {1, 2, 3, 4, 6, 12} with operations product, sum and complement defined in any way you desire. It is impossible to design operations on this set that would satisfy the axioms of a Boolean algebra, because

9 One last characterization: The prototype for a set whose cardinality is 2 n is the power set of U = {1, 2, 3,, n}. Note that we use the name 2 n to denote the Boolean algebra on this set. So, in summary: Every finite Boolean algebra with n atoms has 2 n elements and is isomorphic to the Boolean algebra 2 n. For instance, since D 30 has three atoms, D 30 is isomorphic to 2 3. Another example: Recall from last week that B 5 is the Boolean algebra on Boolean 5-tuples. This Boolean algebra has five atoms: (1, 0, 0, 0, 0), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0), (0, 0, 0, 1, 0), and (0, 0, 0, 0, 1). Therefore, this Boolean algebra has 32 elements and is isomorphic to 2 5.