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Notice that numbers in the value column are all powers of sixteen. If we wrote this number without using the chart, it would be written as:
A x 1 + 9 x 16 + 3 x 256 + 8 x 4096
Another way of writing this would be:
A x 160 + 9 x 161 + 3 x 162 + 8 x 163
Since the number 16 is represented in octal notation by 1016, we could write the expanded notation in another way. It could be written as:
A x 100 + 9 x 101 + 3 x 102 + 8 x 103
or putting in the subscript to identify the number as a base sixteen number, it could be written as:
A x 10160 + 9 x 10161 + 3 x 10162 + 8 x 10163
We call this expanded notation.
Notice the similarities between what has been written above and the manner in which decimal numbers are written in expanded notation. They are the same except for the subscript.
Figure 1 ~ 4.1
Place Values for Hexadecimal Number 839A16
8 3 9 A
|__________10160
| | | _________________10161
| | __________________________10162
|______________________________10163
Example: Convert 51F1C316 to expanded notation.
Solution: First, identify the value of each digit. The first digit on the right, the digit 3, is in the units place or the sixteen to the zero power place; the digit C is in the sixteen to the first power place; the next digit 1 is in the sixteen to the second power place; the digit F is in the sixteen to the third power place; the next digit 1 is in the sixteen to the fourth power place; the final digit 5 is in the sixteen to the fifth power place.
Table 2 ~ 4.1
Digit
Power of Sixteen
=
Value
=
316
=
C016
=
10016
=
F00016
=
1000016
=
50000016
51F1C316
If we wrote the number without using the chart, the number would be
3 x 160 + C x 161 + 1 x 162 + F x 163 + 1 x 164 + 5 x 165
The number is now in expanded notation. Since the number 16 in hexadecimal notation is represented by 1016, we could write the expanded notation in another way. It could be written as:
3 x 100 + C x 101 + 1 x 102 + F x 103 + 1 x 104 + 5 x 105
or putting in the subscript, it could be written as:
3 x 10160 + C x 10161 + 1 x 10162 + F x 10163 + 1 x 10164 + 5 x 10165
Figure 4.1.2
Place Values for Binary Number 51F1C316
5 1 F 1 C 3
|_____10160
| | | | |_____________10161
| | | | ___________________10162
| | |__________________________10163
| |______________________________ 10164
|_____________________________________10165
4.1 ~ Practice Problems
Directions: Turn off the video. Change the following numbers from standard notation to expanded notation. When you have finished working the problems, check your answers in the back of the book or click here to view the Answer Key. Turn on the video to view the solutions to the practice problem.
4.1 ~ Homework Problems
Directions: Turn off the video. Convert the following numbers from standard notation to expanded notation. Keep your work in a notebook to be turned in to your instructor at the conclusion of the course. When you have finished working the problems, check your answers in the back of the book or click here to view the Answer Key. If you have any questions, contact your instructor.
1. 7D14
2. AC824
3. 112E9C
4. C4
5. 703
6. 4C85
7. 1F9D6B05
8. 5000CF3
9. E
10. 6B14F
4.2 - Converting from Expanded Notation to Standard Notation
We can change numbers that have been written in expanded notation to standard notation by reversing the process. Suppose we have the notation
A x 160 + 9 x 161 + 3 x 162 + 8 x 163
Recalling the information in Figure 4.1 and extending the idea, we could make a table for the powers of sixteen in hexadecimal notation.
Table 1 ~4.2 Table of Powers of 16 vs. Hexadecimal Notation
We might think of our number in another way as
A x 10160 + 9 x 10161 + 3 x 10162 + 8 x 10163
Table 2 ~ 4.2
Adding these values together, we get 839A16 in standard notation.
Example: Convert 3 x 10160 + C x 10161 + 1 x 10162 + F x 10163 + 1 x 10164 + 5 x 10165 to standard notation.
Solution: If we follow the rules of order of operations, we must first evaluate the powers of sixteen.
3 x 116 + C x 1016 + 1 x 10016 + F x 100016 + 1 x 1000016 + 5 x 10000016
Next, we perform the multiplication.
316 + C016 +10016 + F00016 + 1000016 + 50000016
Finally, we add the numbers together to obtain the sum:
51F1C316
which is the number in standard form.
4.2~ Practice Problems
Directions: Turn off the video. Change the following numbers from expanded notation to standard notation. When you have finished working the problems, check your answers in the back of the book or click here to view the Answer Key. Turn on the video to view the solutions to the practice problem.
4.2 ~ Homework Problems
Directions: Turn off the video. Convert the following numbers from expanded notation to standard notation. Keep your work in a notebook to be turned in to your instructor at the conclusion of the course. When you have finished working the problems, check your answers in the back of the book or click here to view the Answer Key. If you have any questions, contact your instructor.
4.3- Adding Numbers in Hexadecimal Notation
Addition in base sixteen is very similar to addition in base ten. The first step is learning the addition facts for base sixteen. Since there are sixteen digits used in the hexadecimal system, there are several different mathematics facts that we must know. When we add any two digits together and their sum is less than the decimal number nine, the sum is the same in both bases. If the two digits have a sum equal to the decimal number ten, we use the hexadecimal digit A. Similarly, if the sum of the digits is eleven, we use the hexadecimal digit B. We use C for a sum of twelve, D for a sum of thirteen, E for a sum of fourteen, and F for a sum of fifteen. The first sixteen facts in the hexadecimal system are the similar to the decimal system, that is 0 + 0 = 0, 0 + 1 = 1, 0 + 2 = 2, 0 + 3 = 3, 0 + 4 = 4, 0 + 5 = 5, 0 + 6 = 6, 0 + 7 = 7, 0 + 8 = 8, 0 + 9 = 9, 0 + A = A, 0 + B = B, 0 + C = C, 0 + D = D, 0 + E = E, and 0 + F = F. (Of course the reflexive property addition holds so that 1 + 0 = 1, 2 + 0 = 2, and so on.) We know that the equivalent of the decimal number 10 is the hexadecimal value 1016. Consequently, F + 1 = 10 in base sixteen. As with addition base ten, a table can provide easy reference to the addition facts in base sixteen. We can refer to the addition table for base sixteen depicted in Table 4.3.1 as we work our examples.
Table 1 ~ 4.3
Addition Table Base Sixteen (Hexadecimal System)
~To view this table Click here~
Example 1: Add: 839A16 + 13516
Solution: The first step is to "line up" the digits, like we do for decimal numbers. We want to make sure that all of the digits in the same place value are beneath one another.
839A16
+ 13516
We start the addition process from the right, just as if we were adding decimal numbers. We can see from Table 4.3.1 that A plus five equals F in the hexadecimal number system. We can write down the F in the units place; we have no numbers to "carry."
839A16
+ 13516
F
The next step is to add the nine and the three. The sum of nine and three is C (see Table 4.3.1) in the hexadecimal number base. We write down the C in the sixteens place and have no numbers to carry.
839A16
+ 13516
CF
The next step is to look at the sixteen to the second power place. Since we have not carried any numbers, all we need to do is add the three and the one to obtain the answer.
839A16
+ 13516
4CF
The final step is to look at the sixteen to the third power place. Since we have not carried any numbers and there is no value in the sixteen to the third power place in the second addend, all we need to do is bring down the eight to obtain the answer.
839A16
+ 13516
84CF16
This shows that the sum of 839A16 plus 13516 equals 84CF16.
Example 2: Add: 4C816 + 7916
Solution: The first step is to "line up" the digits, like we do for decimal numbers. We want to make sure that all of the digits in the same place value are beneath one another.
4C816
+ 7916
We start the addition process from the right, just as if we were adding decimal numbers. We know that eight plus nine equals 11 in the hexadecimal number system as seen in Table 4.3.1. We can write down the one in the units place; we have a one to "carry."
1 "carried one"
4C816
+ 7916
1
The next step is to add the C and the seven. We know that their sum is 13 from Table 4.3.1. At this point, you may wish to carry the one to the next place.
1 1 "carried ones"
4C816
+ 7916
1
Now we must add the remaining three with the carried one. Since 3 + 1 = 4, we write down the four in the sixteens place.
1 "carried one"
4C816
+ 7916
41
The final step is to look at the sixteen to the second power place. We must add four in the first addend with the carried one to get five. We write down the five to obtain then answer.
1 "carried one"
4C816
+ 7916
54116
This shows that the sum of 44C86 plus 7916 equals 54116.
Example 3: Add: 123416 + 5FF16
Solution: The first step is to "line up" the digits, like we do for decimal numbers. We want to make sure that all of the digits in the same place value are beneath one another.
123416
+ 5FF16
We start the addition process from the right, just as if we were adding decimal numbers. We know that four plus F equals 13 from Table 4.3.1. Since there is room for only one digit in the sum line, we must begin to carry numbers. When we carry, we use the same process as the decimal system, that is, we write down the digit on the right, in the units place, and carry the digit on the left (in this case, in the sixteens place.) We can write down the three in the units place and carry the one.
1 "carried one"
123416
+ 5FF16
3
The next step is to add the three and the F. We know that their sum is 12 from Table 4.3.1. At this point, you may wish to carry the one to the next place.
1 1 "carried ones"
123416
+ 5FF16
3
Now we must add the remaining two with the carried one. Since 2 + 1 = 3, we write down the three in the sixteens place.
1 "carried one"
123416
+ 5FF16
33
The next step is to add the two and the five. We know from Table 4.3.1 that their sum is 7. We add the carried one to get eight and write down the eight in the sixteen to the second power place. There is nothing to carry to the sixteen to the third place.
123416
+ 5FF16
833
The final step is to look at the sixteen to the third power place. Because there is nothing in the sixteen to the third power place in the second addend and because there is no value carried from the previous column, we write down the 1 to obtain the answer.
123416
+ 5FF16
183316
This shows that the sum of 123416 plus 5FF16 equals 183316.
4.3 ~ Practice Problems
Directions: Turn off the video.Add the following numbers that are in hexadecimal notation. When you have finished working the problems, check your answers in the back of the book or click here to view the Answer Key. Turn on the video to view the solutions to the practice problem.
4.3 ~ Homework Problems
Directions: Turn off the video. Add the following numbers that are in hexadecimal notation. Keep your work in a notebook to be turned in to your instructor at the conclusion of the course. When you have finished working the problems, check your answers in the back of the book or click here to view the Answer Key. If you have any questions, contact your instructor.
This Concludes Session Four
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Begin this session by watching the video tape. If you have purchased a textbook, follow along in your book. If you are using the Internet version, follow along on the webpage. If you have any difficulties, contact your instructor. After the explanation of place values, turn off the video player and try to work the practice problems. Once you have worked the problems, check your answers in the answer key listed in the back of your book or click on the Answer Key button to the left.
