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Boolean Algebra
67 questions By Tony R. Kuphaldt
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Question 28 of 67
A student makes a mistake somewhere in the process of simplifying the following Boolean expression:
$$AB + A(B + C)$$
$$AB + AB + C$$
$$AB + C$$Determine where the mistake was made, and what the proper sequence of steps should be to simplify the original expression.
Reveal answerAn error was made in the second step (distribution). The correct sequence of steps is as follows:
$$AB + A(B + C)$$
$$AB + AB + AC$$
$$AB + AC$$
$$A(B + C)$$
Notes:An interesting way to sharpen students’ understanding of algebraic techniques is to have them view someone else’s incorrect work and find the error(s). Ultimately, algebraic reduction is really just an exercise in pattern recognition. Anything you can do to help your students recognize the correct patterns will help them become better at using algebra.
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Question 29 of 67
Factoring is a powerful simplification technique in Boolean algebra, just as it is in real-number algebra. Show how you can use factoring to help simplify the following Boolean expressions:
$$C + CD$$
$$A\overline{B}C + A\overline{B}\ \overline{C}$$
$$XY\overline{Z} + XYZ + XYW$$
$$\overline{D}EF + AB + \overline{D}E + 0 + ABC$$
Reveal answerYou will be expected to show your work (including all factoring) in your answers!
$$C + CD = C$$
$$A\overline{B}C + A\overline{B}\ \overline{C} = A\overline{B}$$
$$XY\overline{Z} + XYZ + XYW = XY$$
$$\overline{D}EF + AB + \overline{D}E + 0 + ABC = AB+\overline{D}E$$
Notes:For some reason, many of my students (who enter my course weak in algebra skills) generally seem to have a lot of trouble with factoring, be it Boolean algebra or regular algebra. This is unfortunate, as factoring is a powerful analytical tool. The “trick,” if there is any such thing, is recognizing common variables in different product terms, and identifying which of them should be factored out to reduce the expression most efficiently.
Like all challenging things, factoring takes time and practice to learn. There are no shortcuts, really.
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Question 30 of 67
Shown here are nine rules of Boolean algebra (these are not the only rules, of course).
- $$A + \overline{A} = 1$$
- $$A + A = A$$
- $$A + 1 = 1$$
- $$AA = A$$
- $$A(1) = A$$
- $$A \overline{A} = 0$$
- $$A(B + C) = AB + AC$$
- $$A + AB = A$$
- $$A \overline{A}B = A+B$$
Determine which rule is being used in each step of the following Boolean simplification:
$$\overline{C}F + F(A+\overline{B})+C$$
$$\overline{C}F + AF + \overline{B}F + C$$
$$C + F + AF + \overline{B}F$$
$$C + F(1+A+\overline{B})$$
$$C + F(1)$$
$$C + F$$
Reveal answer$$\overline{C}F + F(A+\overline{B})+C$$
$$Rule: \ \ \ \ \ \ \ \ \ A(B + C) = AB + AC$$
$$\overline{C}F + AF + \overline{B}F + C$$
$$Rule: \ \ \ \ \ \ \ \ \ A + \overline{A}B = A +B$$
$$C + F + AF + \overline{B}F$$
$$Factoring$$
$$C + F(1+A+\overline{B})$$
$$Rule: \ \ \ \ \ \ \ \ \ A +1 = 1$$
$$C + F(1)$$
$$Rule: \ \ \ \ \ \ \ \ \ A (1) = A$$
$$C + F$$
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)