14 Arithmetic Operations in Bases Other Than Ten

The base-ten arithmetic algorithms discussed in the previous two sections also work in other bases. In this section we apply the algorithms to base ﬁve.

Addition in Base Five In base ﬁve the digits used are 0,1,2,3, and 4. Using blocks one can easily build the following addition table. + 0 1 2 3 4 0 0 1 2 3 4 1 1 2 3 4 10 2 2 3 4 10 11 3 3 4 10 11 12 4 4 10 11 12 13

All numerals in the table are written in base ﬁve with subscripts omitted.

Example 14.1 Compute 12five + 31five using blocks.

Solution. Figure 14.1 shows how to compute 342five + 134five using blocks.

Figure 14.1

Example 14.2 Use a base ﬁve line to illustrate 12five + 20five.

1 Solution. Note that 12five + 20five = 32five.

Example 14.3 Compute the sum 342five + 134five (a) using the lattice algorithm (b) using the expanded algorithm (c) using the standard algorithm.

Solution.

To check that the answer to the above addition is correct, we convert ev- erything to base 10 where we feel comfortable.

2 342five = 3 · 5 + 4 · 5 + 2 = 97 2 134five = 1 · 5 + 3 · 5 + 4 = 44 3 2 1031five = 1 · 5 + 0 · 5 + 3 · 5 + 1 = 141

The result is conﬁrmed since 97 + 44 = 141.

Practice Problems

Problem 14.1 Compute the sum 13five + 22five using (a) base ﬁve blocks (b) expanded algorithm

2 (c) lattice algorithm (d) standard algorithm.

Problem 14.2 Perform the following computations.

Problem 14.3 Complete the following base eight addition table. + 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 Problem 14.4 Compute 132eight + 66eight. Problem 14.5 Computers use base two since it contains two digits, 0 and 1, that correspond to electronic switches in the computer being oﬀ or on. In this base, 101two = 2 1 · 2 + 0 · 2 + 1 = 5ten. (a) Construct addition table for base two. (b) Write 1101two in base ten. (c) Write 123ten in base two. (d) Compute 1011two + 111two. Problem 14.6 For what base b would 32b + 25b = 57b? Problem 14.7 (a) Construct an addition table for base four. (b) Compute 231four + 121four.

3 Problem 14.8 Use blocks to illustrate the sum 41six + 33six. Problem 14.9 Use an expanded algorithm to compute 78nine + 65nine. Problem 14.10 Create a base seven number line and illustrate the sum 13seven + 5seven. Problem 14.11 Construct an addition table in base seven. Problem 14.12 Use the lattice method to compute the following sums. (a) 46seven + 13seven. (b) 13four + 23four.

Subtraction in Base Five The development of subtraction in base ﬁve from concrete to abstract is shown in Figure 14.2.

Figure 14.2 Example 14.4 Use base ﬁve number line to illustrate 32five − 14five. Solution.

4 Practice Problems Problem 14.13 Perform the following subtractions: (a) 1101two − 111two (b) 43five − 23five (c) 21seven − 4seven. Problem 14.14 Fill in the missing numbers.

Problem 14.15 Use blocks for the appropriate base to illustrate the following problems. (a) 555seven − 66seven (b) 3030four − 102four. Problem 14.16 Use both the intermediate algorithm (discussed in Figure 14.2) and the standard algorithm to solve the following differences. (a) 31four − 12four (b) 1102four − 333four. Problem 14.17 Use base five number line to illustrate the difference 12five − 4five. Multiplication in Base Five Next, consider the multiplication algorithm. A base-five multiplication table will be helpful. × 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 11 13 3 0 3 11 14 22 4 0 4 13 22 31

5 Example 14.5 Calculate 4five × 3five using base ﬁve number line. Solution.

So 4five × 3five = 22five Example 14.6 Calculate 43five × 123five using (a) the lattice method for multiplication (b) the expanded algorithm (c) the standard algorithm.

Solution.

Practice Problems

Problem 14.18 Create a base seven number line to illustrate 6seven × 3seven. Problem 14.19 Find the following products using the lattice method, the expanded algorithm, and the standard algorithm. (a) 31four × 2four (b) 43five × 3five

6 Division in Base Five Long division in base ﬁve can be done with a long division analogous to the base ten algorithm. The ideas behind the algorithms for division can be developed by using repeated subtraction. For example, 3241five ÷ 43five is computed by means of repeated-subtraction technique as shown in Figure 14.3(a) and by means of the conventional algorithm shown in Figure 14.3(b). Thus, 3241five ÷ 43five = 34five with remainder 14five.

Figure 14.3 Practice Problems

Problem 14.20 Perform the following divisions: (a) 32five ÷ 4five (b) 143five ÷ 3five (c) 10010two ÷ 11two.

Problem 14.21 For what possible bases are each of the following computations correct?

7 Problem 14.22 (a) Compute 121five ÷ 3five with repeated subtraction algorithm. (b) Compute 121five ÷ 3five with long division algorithm.

Problem 14.23 (a) Compute 324five ÷ 4five with repeated subtraction algorithm. (b) Compute 324five ÷ 4five with long division algorithm.

Problem 14.24 (a) Compute 1324seven ÷ 6seven with repeated subtraction algorithm. (b) Compute 1324seven ÷ 6seven with long division algorithm.

Problem 14.25 Solve the following problems using the missing-factor deﬁnition of division, that is, a ÷ b = c if and only if b · c = a.(Hint: Use a multiplication table for the appropriate base). (a) 21four ÷ 3four (b) 23six ÷ 3six (c) 24eight ÷ 5eight

Problem 14.26 Sketch how to use base seven blocks to illustrate the operation 534seven ÷ 4seven.