### Computer Number Bases Lesson 2

```Computer Science
LESSON 2 ON
Number Bases
John Owen, Rockport Fulton HS
1
Objective

In the last lesson you learned
by the computer, which were
 Base Two – binary
 Base Eight – octal
John Owen, Rockport Fulton HS
2
Base Conversion

You also learned how to convert
from the decimal (base ten) system
to each of the new bases…binary,
John Owen, Rockport Fulton HS
3
Other conversions

Now you will learn other
conversions among these four
number systems, specifically:
Binary to Decimal
 Octal to Decimal

John Owen, Rockport Fulton HS
4
Other conversions

As well as
Binary to Octal
 Octal to Binary

John Owen, Rockport Fulton HS
5
Other conversions

And finally

John Owen, Rockport Fulton HS
6
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.
John Owen, Rockport Fulton HS
7
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.
John Owen, Rockport Fulton HS
8
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.
John Owen, Rockport Fulton HS
9
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.
John Owen, Rockport Fulton HS
10
Binary to Decimal

Can you figure the decimal values
for these binary values?
11
 101
 110
 1111
 11011

John Owen, Rockport Fulton HS
11
Binary to Decimal

11 is 3 in base ten
 101 is 5
 110 is 6
 1111 is 15
 11011 is 27

John Owen, Rockport Fulton HS
12
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.
Technically this is the 80 place.
John Owen, Rockport Fulton HS
13
Octal to Decimal


The 2nd digit from the right, 111, is
in the “eights” place, the “base”
place, and is worth 8.
Technically this is the 81 place.
John Owen, Rockport Fulton HS
14
Octal to Decimal


The 3rd digit from the right, 111, is
in the “sixty-fours” place, the “base
squared” place, and is worth 64.
Technically this is the 82 place.
John Owen, Rockport Fulton HS
15
Octal to Decimal

The total value of this octal
number, 111, is 64+8+1, or
seventy-three.
John Owen, Rockport Fulton HS
16
Octal to Decimal

Can you figure the value for these
octal values?
21
 156
 270
 1164
 2105

John Owen, Rockport Fulton HS
17
Octal to Decimal

21 is 17 in base 10
 156 is 110
 270 is 184
 1164 is 628
 2105 is 1093

John Owen, Rockport Fulton HS
18



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.
Technically this is the 160 place.
John Owen, Rockport Fulton HS
19


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.
John Owen, Rockport Fulton HS
20


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.
John Owen, Rockport Fulton HS
21

The total value of this hexadecimal
number, 111, is 256+16+1, or two
hundred seventy-three.
John Owen, Rockport Fulton HS
22

Can you figure the value for these
2A
 15F
 A7C
 11BE
 A10D

John Owen, Rockport Fulton HS
23

2A is 42 in base 10
 15F is 351
 A7C is 2684
 11BE is 4542
 A10D is 41229

John Owen, Rockport Fulton HS
24
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.
John Owen, Rockport Fulton HS
25
Binary to Octal








000 base 2 equals 0 base 8
0012 = 18
0102 = 28
0112 = 38
1002 = 48
1012 = 58
1102 = 68
1112 = 78
John Owen, Rockport Fulton HS
26
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
John Owen, Rockport Fulton HS
27
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
John Owen, Rockport Fulton HS
28
Binary to Octal

1111002 = 748
 1001012 = 458
 1110012 = 718
 11001012 = 1458
 1100111012 = 6358

John Owen, Rockport Fulton HS
29


The conversion between binary and
Since 2 to the power of 4 equals
16, it takes 4 base 2 digits to
combine to make a base 16 digit.
John Owen, Rockport Fulton HS
30








0000 base 2 equals 0 base 8
00012 = 116
00102 = 216
00112 = 316
01002 = 416
01012 = 516
01102 = 616
01112 = 716
John Owen, Rockport Fulton HS
31








10002
10012
10102
10112
11002
11012
11102
11112
=
=
=
=
=
=
=
=
816
916
A16
B16
C16
D16
E16
F16
John Owen, Rockport Fulton HS
32


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
John Owen, Rockport Fulton HS
33

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
John Owen, Rockport Fulton HS
34

110111002 = DC16
 101101012 = B516
 100110012 = 9916
 1101101012 = 1B516
 1 1010 0101 11012 = 1A5D16

John Owen, Rockport Fulton HS
35
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
John Owen, Rockport Fulton HS
36



Binary is the inverse of Binary to
For each “hex” digit, translate it
into the equivalent four-digit binary
group.
For example, 45 base 16 equals
01000101 base 2
John Owen, Rockport Fulton HS
37
Binary Exercises

Convert each of these to binary:
638
 12316
 758
 A2D16
 218
 3FF16

John Owen, Rockport Fulton HS
38
Binary Exercises

638 = 1100112
 12316 = 1001000112 (drop leading 0s)
 758 = 1111012
 A2D16 = 1100001011012
 218 = 100012
 3FF16 = 11111111112

John Owen, Rockport Fulton HS
39



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.
John Owen, Rockport Fulton HS
40

For example:
4A316
 = 0100 1010 00112
 = 010 010 100 0112
 = 22438

John Owen, Rockport Fulton HS
41



Converting from Octal to
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.
John Owen, Rockport Fulton HS
42

For example:
3718
 = 011 111 0012
 = 1111 10012
 = F98

John Owen, Rockport Fulton HS
43

Convert each of these:
638 = ________16
 12316 = ________8
 758 = ________16
 A2D16 = ________8
 218 = ________16
 3FF16 = ________8

John Owen, Rockport Fulton HS
44

638 = 3316
 12316 = 4438
 758 = 3D16
 A2D16 = 50558
 218 = 1116
 3FF16 = 17778

John Owen, Rockport Fulton HS
45
Number Base Conversion
Summary


Now you know twelve different
number base conversions among
the four different bases (2,8,10,
and 16)
With practice you will be able to do
these quickly and accurately, to the
point of doing many of them in
John Owen, Rockport Fulton HS
46
Practice



Now it is time to practice.
Go to the Number Base Exercises
slide show to find some excellent
practice problems.
Good luck and have fun!
John Owen, Rockport Fulton HS
47
```
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}