computer science introduction to the number base unit adapted from slides by john owen computer...

Computer Science

Introduction to the Number Base Unit

Adapted from Slides by John Owen

Computer Science Instructor, Rockport-Fulton High School,

Rockport, Texas

John Owen, Rockport Fulton HS 2

Objective

This unit of study is designed to introduce the beginner computer science students to the concept of the computer number bases (2, 8, and 16) and their computation.


Part1

Different Number Bases, specifically about those used by the computer

includes: Base Two – binary Base Eight – octal Base Sixteen – hexadecimal


Base Ten

“because it has ten counting digits, 0,1,2,3,4,5,6,7,8, and 9”

To count in base ten, you go from 0 to 9, then do combinations of two digits starting with 10 all the way to 99


Base Two

To count in base two, which only has 0 and 1 as counting digits, you count 0,1, then switch to two digit combinations, 10,11, then to three digit combos, 100, 101,110,111, then four digit, 1000, _____,_______, …, 1111


Base Three

To count in base three, which has 0, 1, and 2 as counting digits, you count 0,1,2, then switch to two digit combinations, 10,11, 12, 20, 21, 22, then to three digit combos, 100, 101,102, 110,111, 112, etc…


Base Eight

base eight (often called octal)… The base eight counting

sequence 0,1,2,3,4,5,6,7,10,11,12,13,…77

100,101,102,103,104,105,106,107110,111, etc.


Base Sixteen

Base Sixteen, also known as hexadecimal, was especially created by computer scientists to help simplify low-level programming, like machine language and assembly language.


Base Sixteen

To get sixteen counting digits, you use 0-9, but still need six more…so it was decided to use A,B,C,D,E, and F.

0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F Then the two-digit combos:10,11,12,…

19,1A,1B,1C,1D,1E,1F,20,21,22,…2D,2E,2F,30,31,…FF


Base conversion

To convert from base ten to another base, such as base two, eight, or sixteen, is an important skill for computer scientists and programmers.


Base Ten to Base Two

Let’s take the value 27 and convert it into base 2.

Here’s the process: Divide 27 by 2 The answer is 13, remainder 1 Divide 13 by 2 Answer is 6, remainder 1



Continue until the answer is 1. 6 divided by 2 = 3, remainder 0 3 divided by 2 = 1, remainder 1

Now take the last answer, 1, and all of the remainders in reverse order, and put them together…11011

27 base 10 = 11011 base two



Here’s an easy way to do it on paper

27 divided by 2 = 13, R 1



13 / 2 = 6, R 1



6 / 2 = 3, R 0



3 / 2 = 1, R 1



Stop, and write the answer


Exercises

Now try a few yourself (see last slide for answers):

1. 1610 = _________2

2. 4710 = _________2

3. 14510 = _________2

4. 3110 = _________2

5. 3210 = _________2


Base Ten to Base Eight

Let’s again take the value 27 and convert it into base 8.

Same process: Divide 27 by 8 The answer is 3, remainder 3 Stop! You can’t divide anymore

because the answer is less than 8



The last answer was 3, and the only remainder was 3, so the base eight value is 33, base 8.



Use the same method on paper

27 divided by 8 = 3, R 3 27, base 10 = 33, base 8


Exercises

Now try the same values for base eight.

6. 1610 = _________8

7. 4710 = _________8

8. 14510 = _________8

9. 3110 = _________8

10. 3210 = _________8


Base Ten to Base Sixteen

Finally we’ll convert 27 into base 16. Divide 27 by 16 The answer is 1, remainder 11 Stop! You can’t divide anymore

because the answer is less than 16



The last answer was 1, and the only remainder was 11, which in base 16 is the letter B, so the base sixteen value is 1B, base 16.



Again, the same method on paper

27 divided by 16 = 1, R 11 or B 27, base 10 = 1B, base 16


Exercises

And now try base sixteen!

11. 1610 = _________16

12. 4710 = _________16

13. 14510 = _________16

14. 3110 = _________16

15. 3210 = _________16


Here are the answers to the exercises, in jumbled order

10 1F 20 20 2F 37 40 57 91 221 1000011111 101111 100000

10010001


Part 2: Other conversions

Now you will learn other conversions among these four number systems, specifically: Binary to Decimal Octal to Decimal Hexadecimal to Decimal


Other conversions

As well as Binary to Octal Octal to Binary Binary to Hexadecimal Hexadecimal to Binary

Octal to Hexadecimal Hexadecimal to Octal


Binary to Decimal

Each binary digit in a binary number has a place value.

In the number 111, base 2, the digit farthest to the right is in the “ones” place, like the base ten system, and is worth 1.

Technically this is the 20 place.


Binary to Decimal

The 2nd digit from the right, 111, is in the “twos” place, which could be called the “base” place, and is worth 2.

Technically this is the 21 place. In base ten, this would be the

“tens” place and would be worth 10.


Binary to Decimal

The 3rd digit from the right, 111, is in the “fours” place, or the “base squared” place, and is worth 4.

Technically this is the 22 place. In base ten, this would be the

“hundreds” place and would be worth 100.


Binary to Decimal

The total value of this binary number, 111, is 4+2+1, or seven.

In base ten, 111 would be worth 100 + 10 + 1, or one-hundred eleven.


Binary to Decimal

Can you figure the decimal values for these binary values? 11 101 110 1111 11011


Binary to Decimal

answers: 11 is 3 in base ten 101 is 5 110 is 6 1111 is 15 11011 is 27


Octal to Decimal

Octal digits have place values based on the value 8.

In the number 111, base 8, the digit farthest to the right is in the “ones” place and is worth 1.



Octal to Decimal

The 2nd digit from the right, 111, is in the “eights” place, the “base” place, and is worth 8.



Octal to Decimal

The 3rd digit from the right, 111, is in the “sixty-fours” place, the “base squared” place, and is worth 64.



Octal to Decimal

The total value of this octal number, 111, is 64+8+1, or seventy-three.


Octal to Decimal

Can you figure the value for these octal values? 21 156 270 1164 2105


Octal to Decimal

Here are the answers: 21 is 17 in base 10 156 is 110 270 is 184 1164 is 628 2105 is 1093


Hexadecimal to Decimal

Hexadecimal digits have place values base on the value 16.

In the number 111, base 16, the digit farthest to the right is in the “ones” place and is worth 1.




The 2nd digit from the right, 111, is in the “sixteens” place, the “base” place, and is worth 16.

Technically this is the 161

place.



The 3rd digit from the right, 111, is in the “two hundred fifty-six” place, the “base squared” place, and is worth 256.

Technically this is the 162

place.



The total value of this hexadecimal number, 111, is 256+16+1, or two hundred seventy-three.



Can you figure the value for these hexadecimal values? 2A 15F A7C 11BE A10D



Here are the answers: 2A is 42 in base 10 15F is 351 A7C is 2684 11BE is 4542 A10D is 41229


Binary to Octal

The conversion between binary and octal is quite simple.

Since 2 to the power of 3 equals 8, it takes 3 base 2 digits to combine to make a base 8 digit.


Binary to Octal 000 base 2 equals 0 base 8 0012 = 18

0102 = 28

0112 = 38

1002 = 48

1012 = 58

1102 = 68

1112 = 78


Binary to Octal

What if you have more than three binary digits, like 110011?

Just separate the digits into groups of three from the right, then convert each group into the corresponding base 8 digit.

110 011 base 2 = 63 base 8


Binary to Octal

Try these: 111100 100101 111001 1100101

Hint: when the leftmost group has fewer than three digits, fill with zeroes from the left:

1100101 = 1 100 101 = 001 100 101

110011101


Binary to Octal

The answers are: 1111002 = 748

1001012 = 458

1110012 = 718

11001012 = 1458

1100111012 = 6358


Binary to Hexadecimal

The conversion between binary and hexadecimal is equally simple.

Since 2 to the power of 4 equals 16, it takes 4 base 2 digits to combine to make a base 16 digit.


Binary to Hexadecimal 0000 base 2 equals 0 base 8 00012 = 116

00102 = 216

00112 = 316

01002 = 416

01012 = 516

01102 = 616

01112 = 716


Binary to Hexadecimal 10002 = 816

10012 = 916

10102 = A16

10112 = B16

11002 = C16

11012 = D16

11102 = E16

11112 = F16



If you have more than four binary digits, like 11010111, again separate the digits into groups of four from the right, then convert each group into the corresponding base 16 digit.

1101 0111 base 2 = D7 base 16



Try these: 11011100 10110101 10011001 110110101

Hint: when the leftmost group has fewer than four digits, fill with zeroes on the left:

110110101 = 1 1011 0101 = 0001 1011 0101

1101001011101



The answers are: 110111002 = DC16

101101012 = B516

100110012 = 9916

1101101012 = 1B516

1 1010 0101 11012 = 1A5D16


Octal to Binary

Converting from Octal to Binary is just the inverse of Binary to Octal.

For each octal digit, translate it into the equivalent three-digit binary group.

For example, 45 base 8 equals 100101 base 2


Hexadecimal to Binary

Converting from Hexadecimal to Binary is the inverse of Binary to Hexadecimal.

For each “hex” digit, translate it into the equivalent four-digit binary group.

For example, 45 base 16 equals 01000101 base 2


Octal and Hexadecimal to Binary Exercises

Convert each of these to binary: 638

12316

758

A2D16

218

3FF16


Octal and Hexadecimal to Binary Exercises

The answers are: 638 = 1100112

12316 = 1001000112 (drop leading 0s)

758 = 1111012

A2D16 = 1100001011012

218 = 100012

3FF16 = 11111111112


Hexadecimal to Octal

Converting from Hexadecimal to Octal is a two-part process.

First convert from “hex” to binary, then regroup the bits from groups of four into groups of three.

Then convert to an octal number.



For example: 4A316 = 0100 1010 00112 = 010 010 100 0112

= 22438


Octal to Hexadecimal

Converting from Octal to Hexadecimal is a similar two-part process.

First convert from octal to binary, then regroup the bits from groups of three into groups of four.

Then convert to an hex number.



For example: 3718 = 011 111 0012 = 1111 10012

= F98


Octal/Hexadecimal Practice

Convert each of these: 638 = ________16

12316 = ________8

758 = ________16

A2D16 = ________8

218 = ________16

3FF16 = ________8


Octal/Hexadecimal Practice

The answers are 638 = 3316

12316 = 4438

758 = 3D16

A2D16 = 50558

218 = 1116

3FF16 = 17778


Part3: Counting, Place Value

An introduction to the basic idea of counting in different bases and the place value system, associating it with the familiar base 10 system.


Review Base Ten Addition, #1

In Base 10 addition, you learned a very simple process.

Look at this problem: 12

+37 First add the ones column,

then the tens.



12

+37

49

The answer is 49…simple, right?



Now look at this problem: 13 +37 When you add the ones

column values, the result of 10 EQUALS the base value of 10, so you have to CARRY a 1.



1 13 +37 0 When a carry is made, you essentially

divide by 10 (the base) to determine what value to carry, and mod by 10 to determine what value to leave behind.


Review Base Ten Addition,#2

1 13 +37 0 3 plus 7 is 10 10 divided by 10 is 1 (carry) 10 mod 10 is 0 (leave)



1 13 +37 50

Answer is 50



Here’s a third example: 16 +37 When you add the ones

column values, the result of 13 EXCEEDS the base value of 10, so CARRY a 1.



16 +37 6 plus 7 is 13 13 divided by 10 is 1 (carry) 13 mod 10 is 3 (leave)



1 16 +37 53 Answer is 53



And finally, a fourth example: 76 +35 The ones column result of 11

EXCEEDS the base value of 10, and you CARRY a 1.


Review Base Ten Addition,#4




1 76 +35 1 1+7+3 is 6 plus 5, which equals

11 11 divided by 10 is 1 (carry) 11 mod 10 is 1 (leave)



1 76 +35 111 Answer is 111, base 10


Base Eight Addition, #1

Now here is an example in base eight:

12 +34 When you add the ones column

values, the answer is 6, and the second column answer is 4.


Base Eight Addition. #1

12 +34 48 Answer is 48, base eight You say, “four eight base eight”,

not “forty-eight” The phrase “forty-eight” is meant

for base ten only.



Now look at this problem: 14 +34 When you add the ones

column values, the result of 8 EQUALS the base value of 8, and you have to CARRY a one.



14 +34 Again you divide by 8 (the base)

to determine what value to carry, and mod by 8 to determine what value to leave behind.



1 14 +34 50 Answer is “five zero, base

eight”! Looks strange, but it is correct!



Here’s a third example: 16 +37 When you add the ones

column values, the result of 13 EXCEEDS the base value of 8, and you have to CARRY a one.



1 16 +37 55 Answer is 55, base eight.



And a fourth example: 76 +35 The ones column result of 11

EXCEEDS the base value of 8, …CARRY a one.



1 76 +35 33 1+7+3 is 6 plus 5 is 11 11 divided by 8 is 1 (carry) 11 mod 8 is 3 (leave)



1 76 +35 133Answer is 133, base 8


Base Two Addition, #1

Base Two Addition is quite interesting, but also fairly simple.

Since the only counting digits in base two are the values 0 and 1, there are only a few situations you have to learn.



We’ll start simple: 1 +1 =10 (“one zero, base two”) This looks strange, but the

same process applies.



1 +1 = 10 Since 1 + 1 is 2, this EQUALS

the base value of 2, which means you carry the “div” answer and leave the “mod” answer



1 +1 = 10 2 / 2 = 1 (carry) 2 % 2 = 0 (leave) That’s it!



Here’s another: 10 +11 = 101 Can you figure it out?



10 +11 = 101 In the ones column, 1 + 0 is 1. In the second column, 1+1 is

2, or 10 in base 2



And another: 101101 +110011

= Can you figure it out?



Step by step… 1 101101 +110011 = 0



Step by step… 1 101101 +110011 = 00



Step by step… 1 101101 +110011 = 000



Step by step… 1 101101 +110011 = 0000



Step by step… 1 101101 +110011 = 00000 Since 1+1+1 is 3, carry 1 and

leave 1



Step by step… 1 101101 +110011 =1100000 All done!


Base Sixteen, Example #1

In base sixteen, remember the digits are 0-9, then A-F, representing the values 0-15

Here’s an example: 29 +12



29 +12 = 3B, base 16 2 + 9 is 11, which is B in base

sixteen 2+1 is 3, so the answer is 3B



1 A9 +47 = F0, base 16 9+7 is 16, equal to the base,

so carry 1 and leave 0 1 + A(10) + 4 is 15, which is F



11 D6 +7C = 152, base 16 6+C(12) = 18, carry 1, leave 2 1+D(13)+7 = 21, carry 1, leave

5



11 EF +2D = 11C, base 16 F(15) + D(13) = 28, carry 1, leave

C(12) 1 + E(14) + 2 = 17, carry 1, leave 1


Exercises

Now try these exercises1. 12 + 12 =

2. 78 + 68 =

3. F16 + F16 =

4. 58 + 58 =

5. 916 + B16 =

6. C16 + D16 =


Exercises

7. 38 + 48 =

8. F16 + 216 =

9. 102 + 102 =

10. 12 + 10112 =

11. 102 + 1102 =

12. 2168 + 3648 =

13. 7778 + 38 =


Exercises

14. ACE16 + BAD16 =

15. 23416 + 97516 =

16. 4216 + F16 + 87616 =


ANSWERS (JUMBLED)

7 11 12 10 14 15 19

1E 100 602 BA9 8C71000

1002 1100 167B


Part 4

In this section you’ll learn how to do subtraction and how to solve simple equations involving Base 2, 8, and 16.

Again, it is essentially the same concept as Base 10, just in a different base!


Review Base Ten Subtraction

In Base 10 subtraction, you use a very simple process.

Look at this problem: 48 -37 = 11


Review Base Ten Subtraction

48 -37 = 11 Each column is subtracted to

get an answer of 11…


Subtraction, Base 10

Now look at this problem: 63 -37

In this problem, you need to borrow.



513 63 -37

Borrowing means taking a value from the next column and adding it to the column you need.



513 63 -37

In this case, borrow from the 6, which becomes five, and add 10 to the 3, making 13.



513 63 -37

When you borrow 1 from one column, it becomes the value of the base in the next column, or 10 in this case.



513 63 -37 26

Then you subtract the two columns with a result of 26.


Subtraction, Base 8

Now let’s try base eight: 63 -37

Again, in this problem, you need to borrow.


Subtraction, Base 8

511 63 -37

Borrow from the 6, which becomes five, and add 8 to the 3, making 11!


Subtraction, Base 8

511 63 -37

When you borrow 1 from a column, it becomes the value of the base in the next column, or 8 in this case.


Subtraction, Base 8

511 63 -37 24

Then you subtract the two columns with a result of 24, base 8.



Now base 16: 519 63 -37

Again, we borrow from the 6, which becomes five, and add 16 to the 3, making 19!



519 63 -37

When you borrow 1 from a column, it becomes the value of the base in the next column, or 16 in this case.



519 63 -37 2C In the ones column, 19 minus 7

is 12, which is C in base sixteen, with 2 in the second column.



Here’s another example in base 16

D6 -3B

How is this one solved? Try it.



C22 D6 -3B

We must borrow from D, which becomes C, then add 16 to 6, which makes 22.



C22 D6 -3B 9B

22 minus B (11) is B. C minus 3 is 9. Answer is 9B


Subtraction, Base 2

Now base 2: 11 - 1 10 This one is easy…answer is 10


Subtraction, Base 2

Another in base 2: 02 110 - 1 Here we need to borrow from

the twos place…


Subtraction, Base 2

02 110 - 1 101Subtract to get the answer.


Subtraction, Base 2

Still another in base 2: 02 110 - 11 1Now borrow again…


Subtraction, Base 2

2 100 - 11 01Final answer is 01, base 2


Simple Equations

Here an equation to solve (base 10):

x + 6 = 14


Simple Equations

Solution…subtract 6 from both sides

x + 6 = 14 -6 -6x = 8


Simple Equations

Now do it in base 8:

x + 6 = 14


Simple Equations

Solution…subtract 6 from both sides

x + 6 = 14 -6 -6x = ?


Simple Equations

Answer is 6, base 8 12x + 6 = 14 -6 -6x = 6


Simple Equations

Here’s an equation in base sixteen (remember, A and F are NOT variables, but base sixteen values):

x + 2A = F3


Simple Equations

Solution?

x + 2A = F3


Simple Equations

Subtract 2A from both sides: E19x + 2A = F3 - 2A -2Ax = C9


Exercises

Now try these exercises1. 12 - 12 =

2. 78 - 68 =

3. F16 - A16 =

4. 158 - 68 =

5. 4916 - 2B16 =

6. CC16 - AD16 =


Exercises

7. 738 - 348 =

8. 3E16 – 2F16 =

9. 1012 - 102 =

10. 11012 - 112 =

11. 10102 - 1112 =

12. 7168 - 3648 =

13. 7768 + 3378 =


Exercises

Now let’s mix it up a bit!14. AE16 + 768 = _________8

15. 2348 + 110110112 = _________16

16. 101102 - F16 + 768 = _________10

17. 38 + 3910 - 1101012 = _________16

18. 11112 - F16 + 1510 = _________16


Exercises

And finally, some equations19. x16 + 7616 = AB16

20. x2 - 10112 = 1012

21. x8 + 568 = 728

22. x2 + 2510 = 1F16

23. x8 + 3748 - 65568 = BAD16

24. 378 + X16 = 110111102


ANSWERS (JUMBLED)35

37

69

110

177

332

354

1010

0

1

5

7

11

11

14

19

1335

10000

14037

1E

1F

BF

F

F

155

Representing Fractional Numbers

Computers store fractional numbers Negative and positive

Storage technique based on floating-point notation Example: 1.345E+5 1.345 = mantissa, E = exponent, + 5 moves

decimal IEEE-754 specification

Uses binary mantissas and exponents Implementation details are part of

advanced study

Conversion to Decimal (dn…..d2 d1 d0 . d-1 d-2 ……) R = ( )10

Examples

Example

Part5:Data Representation in Binary

Binary values map to two states On or off

Bit Each 1 and 0 (on and off ) in a computer

Byte Group of 8 bits

Word Collection of bytes (typically 4 bytes)

Nibble Half a byte or 4 bits

Connecting with Computer Science, 2e 159

Representing Whole Numbers

Whole numbers (integer numbers) Stored in fixed number of bits 2010 stored as 16-bit integer 0000011111011010

Equivalent hex value: 07DA Signed numbers stored with twos

complement Leftmost bit reserved for sign

1 = negative and 0 = positive If positive: leave as is If negative: perform twos complement

Reverse bit pattern and add 1 to number using binary addition

160

161

Figure 7-5, Storing numbers in a twos complement 8-bit field

Representing Whole Numbers (cont’d.)

One’s Complement

Way to represent negative values. Change every one in binary to zero

and every zero to 1

Example

Convert (1010011)2 one’s complement. (0101100) 1’s

Convert (1000010011) which is one’s complement to its binary value. (0111101100)2

Two’s complement It is a way to represent a number with

both value and sign. The most important property is that the

Most Significant Bit has a negative weight.

To convert from two’s complement to decimal is an easy way. The only difference is that during computation the weight of the last bit is negative.

Example to convert from two’s complement to decimal.

Converting from decimal number to two’s complement We have two cases.

If it is a positive decimal number. It is the same as normal binary except that you add 0 as MSB to the binary number.

If it is negative decimal number. Treat the number as if it is positive. Convert it to binary number. Put Zeroes to the left of the number. Convert the zero to one and one to zero for

all digits Add one to the whole number.

Subtraction using two’s complement Subtraction process requires two operands:

minuend and subtrahend, normally the subtrahend is greater than the minuend. Make sure you have more digits that

accommodate the value of the numbers by adding zeros to the left of the two numbers.

Convert the subtrahend to two’s complement. Add the two numbers. If the number of digits of the result exceeds

numbers of digits both numbers. Cancel that digit.

Example of Two’s complement subtraction. Subtract 1001 from 101.

Number of digits ( we need 5 digits for those numbers) so , we add trailing zeros

Minuend becomes 00101 and the subtrahend becomes 01001

Convert the subtrahend to two’s complement So it 01001 becomes 10111

Add both numbers: 00101 to 10111 00101 +10111 = 11100

As the number of digits of the result is that same as the number of the digits of both number. This is the result.

(Note that as the last bit of the result is 1, it means the number is negative , to find its value. Follow the rules to convert from two’s complement to decimal)

Example of Two’s complement subtraction. Subtract 110 from 1011.

Number of digits ( we need 5 digits for those numbers) so , we add trailing zeros

Minuend becomes 01011 and the subtrahend becomes 00110

Convert the subtrahend to two’s complement So it 00110 becomes 11010

Add both numbers: 01011 to 11010 01011+ 11010 = 100101

Because we have six digits, the left most digits is discarded and keep the number of the digits to 5. the result will b 00101

(Note that as the last bit of the result is 0, it means the number is positive, to find its value. Follow the rules to convert from two’s complement to decimal)


Exercises 1-5

Convert each of these to base 2

1. 528

2. 2738

3. 6178

4. 44728

5. 35028


Exercises 6 - 10


6. 67316

7. 2A516

8. DEB16

9. 50C16

10. 293716


Exercises 11 - 15


11. 4510

12. 3610

13. 1710

14. 7210

15. 5710


Exercises 16 - 20


16. 1100112

17. 11110012

18. 100101102

19. 110110112

20. 1101001012


Exercises 21 - 25


21. 45316

22. D1016

23. 72916

24. BCEF16

25. 4A616


Exercises 26 - 30


26. 4610

27. 8910

28. 7010

29. 12010

30. 27310


Exercises 31 - 35


31. 1011012

32. 11100012

33. 110011012

34. 10001001012

35. 11000010012


Exercises 36 - 40


36. 468

37. 2768

38. 7258

39. 56248

40. 70138


Exercises 41 - 45


41. 7410

42. 15610

43. 16810

44. 41510

45. 55010


ANSWERS…JUMBLED!

26

56

63

71

106

131

170

171

225

6420

10001

100100

101010

101101

111001

136357

10010000

10111011

226

226

309

333

421

645

2123

2246

3451

110001111

1010100101

10100001100

11001110011

11101000010

100100111010

110111101011

10100100110111

19F

1D5

2D

4A

9C

A8

B94

BE

CD

EOB

computer science introduction to the number base unit adapted from slides by john owen computer...

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