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In This Part: Base Two
The number systems we’ve been looking at so far in this course use 10 digits (0 through 9), and the value of each position in a number is some power of 10 (1; 10; 100; 1,000; etc.). We refer to this number system as base ten. But numbers can also be written in other bases. In base two, for example, we have two digits (0 and 1), and the value of each position in a number is some power of 2.
To interpret numbers in base ten, we must look at each digit and determine the value of that digit according to its place in the number. The convention we use is that each place value, moving from right to left, represents an increasing power of 10:
In order to understand the value of a number, we need to consider both its face values and its place values. In 2,342, there are two 2s (the face values), but each of these 2s has a different place value (1,000 and 1). Thus, the value of any number is found by multiplying each face value by its place value and then adding the results.
For example, the value of 234,567 in base ten is
(2 • 10^{5}) + (3 • 10^{4}) + (4 • 10^{3}) + (5 • 10^{2}) + (6 • 10^{1}) + (7 • 10^{0}),
or
(2 • 100,000) + (3 • 10,000) + (4 • 1,000) + (5 • 100) + (6 • 10) + (7 • 1).
Similarly, we can consider numbers that are less than 1. The following two numbers may look similar if we look only at their face values. However, they have different place values, which is evident from the following:
0.02 = (0 • 10^{0}) + (0 • 10^{1}) + (2 • 10^{2}) = 2 • 10^{2}
0.002 = (0 • 10^{0}) + (0 • 10^{1}) + (0 • 10^{2}) + (2 • 10^{3}) = 2 • 10^{3}
Thus, 0.02 and 0.002 are two distinct numbers in the base ten system.
Interpreting numbers in base two works the same way, except that the place value of each digit is some power of 2 instead of 10. We determine the value of digits in a base two number in a similar way. Remember, though, we can only use two digits in this system as face values, the digits 0 and 1, and the place values are powers of 2.
Here are base two place values, written as base ten numbers:
… 
2^{5} 
2^{4} 
2^{3} 
2^{2} 
2^{1} 
2^{0} 
2^{1} 
2^{2} 
… 
… 
32 
16 
8 
4 
2 
1 
1/2 
1/4 
… 
Note that both 10^{0} and 2^{0} equal 1. This and other rules about exponents will be explored further in the next part of this session.
We can also look at a base two number and find its value in base ten. For example, the base ten value of the base two number 101110 is:
(1 • 2^{5}) + (0 • 2^{4}) + (1 • 2^{3}) + (1 • 2^{2}) + (1 • 2^{1}) + (0 • 2^{0}),
or
(1 • 32) + (0 • 16) + (1 • 8) + (1 • 4) + (1 • 2) + (0 • 1) = 46.
Problem A1
Try counting in base two. Explain the patterns you see, and compare them to our base ten system.
Look at what happens to particular place values as you count.
In This Part: Converting Between Bases
When converting numbers from one base to another, it is important to remember that in a positional system, we group quantities into the largest place value possible. For example, in the number 234 in base ten, we have 2 hundreds rather than 20 tens.
An analogy you can use is packaging. Imagine that you are trying to package a quantity of items in the most efficient way, using the least number of boxes (groups). Thus, in base ten you would fill all the boxes that can hold 100 before you start filling the boxes that hold only 10; you’d fill the size 10 boxes before the boxes that hold only one and so on.
The same idea applies to base two. Your packages, however, are of different sizes (32, 16, 8, …). So, when converting between bases, we are essentially repackaging from using one set of boxes (…1,000, 100, 10, 1, … for base ten) to another (…32, 16, 8, 2, 1, …for base two). Note that this process works both ways and for all bases.
So, to convert or repackage a base ten integer such as 52 to base two, first you determine which boxes (groups) would be the most efficient, and then you use the least number of boxes.
Let’s work it out step by step:
52_{ten} = ______{two}
Step 1: Record the base ten powers of 2 from right to left, starting with 2^{0}, or 1. Continue until you reach the place where the next power would be greater than the base ten number you are trying to convert. Using the packaging analogy, these will be your new boxes for repackaging.
Here are the base ten powers of 2:
2^{5} 
2^{4} 
2^{3} 
2^{2} 
2^{1} 
2^{0} 
32 
16 
8 
4 
2 
1 
In our case, we can stop at 2^{5} since 2^{6} = 64, and 64 is greater than 52. This box would be too large for 52 items.
Step 2: Next, record a 1 in the place of the greatest power less than your number (i.e., the biggest box you can use), and subtract that base ten value from your number.
For example, we can fill one size 32 box from the 52 items, so write a 1 in that place:
32 
16 
8 
4 
2 
1 
1 





Subtract to see what’s left:
52 – 32 = 20
Step 3: Now look at the difference. What is the next biggest power of 2 (the nextsize box) we can use?
In this case, 20 is greater than the nextsmaller power of 2 (i.e., 16) so we can fill one box of 16 items as well. Write a 1 in that place:
32 
16 
8 
4 
2 
1 
1 
1 




Subtract to see what’s left:
20 – 16 = 4
Step 4: Continue filling the remaining boxes until the remainder is 0, recording a 1 or a 0 as required.
For example, you couldn’t fill a box of size 8 with four items, so write a 0 in that place. But you could fill the nextsmaller box with four items. Write a 1 in that place:
32 
16 
8 
4 
2 
1 
1 
1 
0 
1 


Subtract to see what’s left:
4 – 4 = 0
Since we have reached 0, we know we are done. As a result, there is no 2 and no 1, and we can write a 0 in each of those places. We have just efficiently repackaged the number 52, using the fewest new boxes (powers of two):
32 
16 
8 
4 
2 
1 
1 
1 
0 
1 
0 
0 
Thus, 52 in base ten is equivalent to 110100 in base two: 52_{ten} = 110100_{two}.
Problem A2
Translate the following numbers from one base to the other:
a.  38_{ten}= __________{two} 
b.  63_{ten}= ___________{two} 
Problem A3
Translate the following numbers from one base to the other:
a.  1101_{two} = __________{ten} 
b.  11111_{two} = ___________{ten} 
Video Segment
In this video segment, Doug and Tom convert numbers from base two to base ten and vice versa. Doug notices that the number 63 in base ten is one less than the next power of two, and he explains how his finding is reflected in the answer. Watch this segment after you’ve completed Problems A2 and A3.
Can you apply the packaging analogy to explain Doug’s finding?
You can find this segment on the session video approximately 9 minutes and 54 seconds after the Annenberg Media logo.
In This Part: Base Two Numbers in Computing
Base two numbers are very useful in computers and other appliances with circuitry because electricity uses a twovalue system. An electric current is either on (1) or off (0). Thus, all commands to a computer are relayed via circuits that either conduct (1) or do not conduct (0) an electric current. These two states of electric current correspond to two digits in the base two system, 0 and 1. In order to carry out complicated instructions, the circuits must obey the laws of logic.
There are two basic circuits, an “andcircuit” and an “orcircuit.” In the “andcircuit,” both p and q switches must be on to light the bulb. Electricity will flow only if both p and q are closed:
Circuit for p and q (series circuit)
In the “orcircuit,” either p or q must be on to light the bulb. Electricity will flow if either p or q are closed:
Circuit for p or q (parallel circuit)
A branch of mathematics called Boolean algebra deals with the logic that must be applied to create complicated circuits. Calculators and computers use microchips that are made with specific circuits that mimic the rules of Boolean algebra.
Video Segment
In this segment, Deborah Douglas explains the history of base two numbers in early computer technology. The two digits of the binary system, 1 and 0, correspond to the presence or absence of electric current. This was a basic principle behind computer memory.
You can find this segment on the session video approximately 20 minutes and 42 seconds after the Annenberg Media logo.
Problem A1
Counting in base two would look like this:
1  10  11  100  101  110  111  1000  1001 
A pattern you can observe is that each time you’ve reached the highest possible digit in a particular place value (which in this system is 1), you move to the next place value. This is the same pattern that occurs in base ten (as well as any other base), the only difference being that the highest possible digit in the base ten system is 9.
Problem A2
a. The answer is 100110. The highest power of 2 we can make with 38 is 32, so our columns should read 32, 16, 8, 4, 2, and 1. We can use the 32 (38 – 32 = 6), so we record a 1 in that column. We can’t use a 16 or an 8, so we record zeros in those columns. The nexthighest power we can use is 4 (6 – 4 = 2), so we record a 1 in that column as well as the next one. Therefore, 38_{ten} = 100110_{two}.
b. The answer is 111111. The highest power we can make with 63 is 32, so our columns should read 32, 16, 8, 4, 2, and 1, as before. We can use the 32 (63 – 32 = 31), so we record a 1 in that column. We can use the nexthighest power, 16 (31 – 16 = 15), then the nexthighest (15 – 8 = 7), then the nexthighest (7 – 4 = 3), then the nexthighest (3 – 2 = 1), and finally the last power (1 – 1 = 0). All six columns are filled with 1s, so 63_{ten} = 111111_{two}.
Problem A3
a. This value is equal to (1 • 2^{3}) + (1 • 2^{2}) + (0 • 2^{1}) + (1 • 2^{0}) = 13.
b. This value is equal to (1 • 2^{4}) + (1 • 2^{3}) + (1 • 2^{2}) + (1 • 2^{1}) + (1 • 2^{0}) = 31.