Boolean algebra

Overview

Recall questions on the identities and the commutative, associative, distributive and absorption laws, De Morgan's laws, and simplifying an expression step by step.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   What is Boolean algebra used for in computing?

   Model answer

   It is the algebra of the two values 0 and 1, with the operations AND, OR and NOT. It is used to write down what a logic circuit does as an expression, and then to manipulate that expression into an equivalent but simpler one. A simpler expression means a circuit with fewer gates, which is cheaper to make, uses less power, takes up less space on a chip and is faster because a signal passes through fewer gates.

2. 2

   State the notation used for AND, OR and NOT in Boolean algebra.

   Model answer

   AND is written as a dot, A.B, or by simply writing the letters together, AB, and behaves like multiplication. OR is written as a plus sign, A + B, and behaves like addition. NOT is written as a bar over the variable, or as a prime, or in words as NOT A. Brackets work exactly as they do in ordinary algebra, and AND binds more tightly than OR, so A.B + C means (A.B) + C.

3. 3

   State the Boolean identities involving 0 and 1.

   Model answer

   For OR: A + 0 = A, and A + 1 = 1. For AND: A.0 = 0, and A.1 = A. In words, ORing with 0 and ANDing with 1 leave a value unchanged, while ORing with 1 forces the result to 1 and ANDing with 0 forces it to 0.

4. 4

   State the idempotence, complement and double negation laws.

   Model answer

   Idempotence: A + A = A and A.A = A, so repeating a term gains nothing. Complement: A + NOT A = 1, because one of them must be true, and A.NOT A = 0, because they cannot both be true. Double negation: NOT (NOT A) = A.

5. 5

   State the commutative, associative and distributive laws of Boolean algebra.

   Model answer

   Commutative: A + B = B + A and A.B = B.A, so the order of the two operands does not matter. Associative: (A + B) + C = A + (B + C) and (A.B).C = A.(B.C), so the grouping of three terms joined by the same operator does not matter. Distributive: A.(B + C) = A.B + A.C, exactly as in ordinary algebra, and also A + B.C = (A + B).(A + C), which has no ordinary-algebra equivalent and is easy to forget.

6. 6

   State the absorption laws and explain why they hold.

   Model answer

   A + A.B = A and A.(A + B) = A. For the first: if A is 1 the whole expression is 1 whatever B is, and if A is 0 then A.B is also 0 so the expression is 0. Either way the result is just A, and B is irrelevant. The second is the dual and follows the same argument.

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