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Operations in Other Bases
Do you
really understand place value numeration concepts and arithmetic
operations? If you do, you should be
able to apply similar reasoning when naming numbers and performing arithmetic
operations in a different numeration system, such as base five. Figure 2.12 presents a concept model and
expanded algorithm for the base five operation 1323five
– 432
five = 341
five. The operation
consists of several steps.
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Step 1
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Convert one 5x5 flat to five 1x5 strips
and reposition them in the strip column of the minuend. At this point, the strip column of the
minuend contains seven strips.
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Step 2
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Subtract the number strips in the
subtrahend column from that in the minuend column. At this point, the strip column in the
solution contains four strips.
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Step 3
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Convert one 5x5x5 cube to five 5x5 flats
and reposition them in the flat column of the minuend. At this point, the flat column of the
minuend contains seven flats.
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Step 4
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Subtract the number flats in the
subtrahend column from that in the minuend column. At this point, the flat column in the
solution contains three strips.
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Step 5
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Express the solution as a base five
number, 341
five.
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Figure
2.12: Concept Model & Expanded Algorithm for 1323five
– 432
five = 341
five
A
standard algorithm for the same operation is shown in Figure 2.13.
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10
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10
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1
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3
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2
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3
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-
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4
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3
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2
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3
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4
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1
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Standard Algorithm (Base Five)
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Figure
2.13: Standard Algorithm for the Operation 1323five –
432
five = 341
five
Example 2.5
Create a concept model, expanded algorithm, and standard
algorithm for the operation 1323five + 432 five .
Solution 2.5
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Step 1
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In the ones column, add the three and
the two to obtain five ones. Convert
these five ones into a 1x5 strip and reposition it in the strip
column. At this point, there are six
1x5 strips in the strip column. The
ones column is empty.
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Step 2
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Convert five of the six 1x5 strips in
the strip column into a 5x5 flat and reposition it in the flat column. At this point, there is one strip in the
strip column. In the flat column,
there are eight flats.
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Step 3
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Combine five of the 5x5 flats in the
flat column into a cube. Reposition
the cube in the cube column. At this
point, there are three flats in the flat column and two cubes in the cube
column.
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Step 4
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Express the solution as a base five
number, 2310 five
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