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Boolean Algebra
67 questions By Tony R. Kuphaldt
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Question 25 of 67
Like real-number algebra, Boolean algebra is subject to certain rules which may be applied in the task of simplifying (reducing) expressions. By being able to algebraically reduce Boolean expressions, it allows us to build equivalent logic circuits using fewer components.
For each of the equivalent circuit pairs shown, write the corresponding Boolean rule next to it:
Note: the three short, parallel lines represent “equivalent to” in mathematics.
Reveal answerIn order, from top to bottom, left to right:
$$A+A = A$$
$$AA = A$$
$$A+0 = A$$
$$A+1 = 1$$
$$\overline{\overline{A}} = A$$
$$A \ x \ 0 = 0$$
$$A \ x \ 1 = A$$
$$A\overline{A}=0$$
$$A+\overline{A}=1$$
$$A+AB = A$$
$$A+\overline{A}B=A+B$$
Notes:Most of these Boolean rules are identical to their respective laws in real number algebra. These should not be difficult concepts for your students to understand. Some of them, however, are unique to Boolean algebra, having no analogue in real-number algebra. These unique rules cause students the most trouble!
An important benefit of working through these examples is to associate gate and relay logic circuits with Boolean expressions, and to see that Boolean algebra is nothing more than a symbolic means of representing electrical discrete-state (on/off) circuits. In relating otherwise abstract mathematical concepts to something tangible, students build a much better comprehension of the concepts.
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Question 26 of 67
Shown here are six rules of Boolean algebra (these are not the only rules, of course).
- $$A+\overline{A} = 1$$
- $$A+A = A$$
- $$A+1 = 1$$
- $$AA = A$$
- $$A+AB = A$$
- $$A+ \overline{A}B = A+B$$
Determine which rule (or rules) are being used in the following Boolean reductions:
$$\overline{DF}+\overline{DF}C=\overline{DF}$$
$$1+G = 1$$
$$B+AB = B$$
$$\overline{FE}+\overline{FE}=\overline{FE}$$
$$XYZ+\overline{XYZ}=1$$
$$GQ+Q = Q$$
$$\overline{H} \ \overline{H}=\overline{H}$$
$$\overline{CD} + \overline{CD}=\overline{CD}$$
$$EF(EF) = EF$$
$$CD+\overline{C}=\overline{C}+D$$
$$LNM+ML = LM$$
$$A\overline{G}F\overline{C}+F\overline{C}\ \overline{G}=F\overline{C}\ \overline{G}$$
$$\overline{M}+1=1$$
$$\overline{BC}+BC=1$$
$$ABC+CAB = BCA$$
$$S+STV\overline{Q}=S$$
$$\overline{DE}(R+1)=\overline{DE}$$
$$\overline{RS}\ \overline{SR}=\overline{RS}$$
$$ABC\overline{D}+D=D+ABC$$
$$AC\overline{B}+CAD\overline{B}=A\overline{B}C$$
$$A+T+\overline{W}+\overline{A}+X=1$$
$$X\overline{YZ}+\overline{X}=\overline{X}+\overline{YZ}$$
$$\overline{GFH} \ \overline{HGF}=\overline{FHG}$$
$$C\overline{AB}+AB=AB+C$$
Reveal answer$$\overline{DF}+\overline{DF}C=\overline{DF} \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+AB=A$$
$$1+G=1 \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+1=1$$
$$B+AB=B \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+AB=A$$
$$\overline{FE}+\overline{FE}=\overline{FE} \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+A=A$$
$$XYZ+\overline{XYZ}=1 \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+\overline{A}=1$$
$$GQ+Q=Q \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+AB=A$$
$$\overline{H} \ \overline{H} =\overline{H} \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: AA=A$$
$$\overline{CD} + \overline{CD}=\overline{CD} \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+A=A$$
$$EF(EF)=EF \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: AA=A$$
$$CD+\overline{C}=\overline{C}+D \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+\overline{A}B=A+B$$
$$LNM+MK=LM \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+AB=A$$
$$A\overline{G}F\overline{C}+F\overline{C}\ \overline{G}=F\overline{C} \ \overline{G} \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+AB=A$$
$$\overline{M}+1=1 \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+1=1$$
$$\overline{BC}+BC=1 \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+\overline{A}=1$$
$$ABC+CAB=BCA \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+A=A$$
$$S+STV\overline{Q}=S \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+AB=A$$
$$\overline{DE}(R+1)=\overline{DE} \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+1=1$$
$$\overline{RS} \ \overline{SR}=\overline{RS} \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: AA=A$$
$$ABC\overline{D}+D=D+ABC \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+\overline{A}B=A+B$$
$$AC\overline{B}+CAD\overline{B}=A\overline{B}C \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+AB=A$$
$$A+T+\overline{W}+\overline{A}+X=1 \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+\overline{A}=1 \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+1=1$$
$$X\overline{YZ}+\overline{X}=\overline{X}+\overline{YZ} \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+\overline{A}B=A+B$$
$$\overline{GFH} \ \overline{HGF} =\overline{FHG} \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: AA=A$$
$$C\overline{AB}+AB=AB+C \ \ \ \ \ \ \ \ \ \ \ \ \ Rule: A+\overline{A}B=A+B$$
Notes:Quite frequently (and quite distressingly), I meet students who seem to have the most difficult time relating algebraic rules in their general form to specific instances of reduction. For example, a student who cannot tell that the rule A AB = A applies to the expression QR R, or worse yet B AB. This skill requires time and hard work to master, because it is fundamentally a matter of abstraction: leaping from literal expressions to similar expressions, applying patterns from general rules to specific instances.
Questions such as this help students develop this abstraction ability. Let students explain how they “made the connection” between Boolean rules and the given reductions. Often, it helps to have a student explain the process to another student, because they are better able than you to put it into terms the struggling students can understand.
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Question 27 of 67
Shown here are eight rules of Boolean algebra (these are not the only rules, of course).
- $$A+\overline{A}=1$$
- $$A+A = A$$
- $$A+1 = 1$$
- $$AA = A$$
- $$A\overline{A}=0$$
- $$A(B++C) = AB+AC$$
- $$A+AB = A$$
- $$A+\overline{A}B=A+B$$
Determine which rule is being used in each step of the following Boolean simplification:
$$AB+B(B+\overline{C})+\overline{B}C$$
$$AB+BB+B\overline{C}+\overline{B}C$$
$$AB+B+B\overline{C}+\overline{B}C$$
$$AB+B+B\overline{C}$$
$$AB+B+C$$
$$B+C$$
Reveal answer$$AB+B(B+\overline{C})+\overline{B}C$$
$$Rule: \ \ \ \ \ \ \ \ \ A(B+C) = AB+AC$$
$$AB+BB+B\overline{C}+\overline{B}C$$
$$Rule: \ \ \ \ \ \ \ \ \ AA = A$$
$$AB+B+B\overline{C}+\overline{B}C$$
$$Rule: \ \ \ \ \ \ \ \ \ A+AB = A$$
$$AB+B+B\overline{C}$$
$$Rule: \ \ \ \ \ \ \ \ \ A+A\overline{B}=A+B$$
$$AB+B+C$$
$$Rule: \ \ \ \ \ \ \ \ \ A+AB = A$$
$$B+C$$
Notes:Quite frequently (and quite distressingly), I meet students who seem to have the most difficult time relating algebraic rules in their general form to specific instances of reduction. For example, a student who cannot tell that the rule A + AB = A applies to the expression QR + R, or worse yet B + AB. This skill requires time and hard work to master, because it is fundamentally a matter of abstraction: leaping from literal expressions to similar expressions, applying patterns from general rules to specific instances.
Questions such as this help students develop this abstraction ability. Let students explain how they “made the connection” between Boolean rules and the given reductions. Often, it helps to have a student explain the process to another student, because they are better able than you to put it into terms the struggling students can understand.
Question 29, answer 3 has a typo - should be A*B + !(D)*E not A*B+!(D*E)