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Computer Science. LESSON ON Number Base Subtraction and Simple Equations. Objective. In this lesson 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!.

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## Computer Science

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### Computer Science

LESSON ON

Number Base Subtraction and Simple Equations

John Owen, Rockport Fulton HS

Objective
• In this lesson 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!

John Owen, Rockport Fulton HS

Review Base Ten Subtraction
• In Base 10 subtraction, you use a very simple process.
• Look at this problem:

48

-37

= 11

John Owen, Rockport Fulton HS

Review Base Ten Subtraction

48

-37

= 11

• Each column is subtracted to get an answer of 11…pretty easy, huh?

John Owen, Rockport Fulton HS

Subtraction, Base 10
• Now look at this problem:

63

-37

• In this problem, you need to borrow.

John Owen, Rockport Fulton HS

Subtraction, Base 10

513

63

-37

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

John Owen, Rockport Fulton HS

Subtraction, Base 10

513

63

-37

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

John Owen, Rockport Fulton HS

Subtraction, Base 10

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.

John Owen, Rockport Fulton HS

Subtraction, Base 10

513

63

-37

26

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

John Owen, Rockport Fulton HS

Subtraction, Base 8
• Now let’s try base eight:

63

-37

• Again, in this problem, you need to borrow.

John Owen, Rockport Fulton HS

Subtraction, Base 8

511

63

-37

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

John Owen, Rockport Fulton HS

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.

John Owen, Rockport Fulton HS

Subtraction, Base 8

511

63

-37

24

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

John Owen, Rockport Fulton HS

Subtraction, Base 16

Now base 16:

519

63

-37

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

John Owen, Rockport Fulton HS

Subtraction, Base 16

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.

John Owen, Rockport Fulton HS

Subtraction, Base 16

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.

John Owen, Rockport Fulton HS

Subtraction, Base 16

Here’s another example in base 16

D6

-3B

• How is this one solved? Try it.

John Owen, Rockport Fulton HS

Subtraction, Base 16

C22

D6

-3B

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

John Owen, Rockport Fulton HS

Subtraction, Base 16

C22

D6

-3B

9B

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

John Owen, Rockport Fulton HS

Subtraction, Base 2

Now base 2:

11

- 1

10

• This one is easy…answer is 10

John Owen, Rockport Fulton HS

Subtraction, Base 2

Another in base 2:

02

110

- 1

• Here we need to borrow from the twos place…

John Owen, Rockport Fulton HS

Subtraction, Base 2

02

110

- 1

101

John Owen, Rockport Fulton HS

Subtraction, Base 2

Still another in base 2:

02

110

- 11

1

Now borrow again…

John Owen, Rockport Fulton HS

Subtraction, Base 2

2

100

- 11

11

Final answer is 11, base 2

John Owen, Rockport Fulton HS

Simple Equations

Here an equation to solve (base 10):

x + 6 = 14

John Owen, Rockport Fulton HS

Simple Equations

Solution…subtract 6 from both sides

x + 6 = 14

-6 -6

x = 8

John Owen, Rockport Fulton HS

Simple Equations

Now do it in base 8:

x + 6 = 14

John Owen, Rockport Fulton HS

Simple Equations

Solution…subtract 6 from both sides

x + 6 = 14

-6 -6

x = ?

John Owen, Rockport Fulton HS

Simple Equations

12

x + 6 = 14

-6 -6

x = 6

John Owen, Rockport Fulton HS

Simple Equations

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

x + 2A = F3

John Owen, Rockport Fulton HS

Simple Equations

Solution?

x + 2A = F3

John Owen, Rockport Fulton HS

Simple Equations

Subtract 2A from both sides:

E19

x + 2A = F3

- 2A-2A

x =C9

John Owen, Rockport Fulton HS

Exercises
• Now try these exercises
• 12 - 12 =
• 78 - 68 =
• F16 - A16 =
• 158 - 68 =
• 4916 - 2B16 =

John Owen, Rockport Fulton HS

Exercises
• 738 - 348 =
• 3E16 – 2F16 =
• 1012 - 102 =
• 11012 - 112 =
• 10102 - 1112 =
• 7168 - 3648 =
• 7768 + 3378 =

John Owen, Rockport Fulton HS

Exercises

Now let’s mix it up a bit!

• AE16 + 768 = _________8
• 2348 + 110110112 = _________16
• 101102 - F16 + 768 = _________10
• 38 + 3910 - 1101012 = _________16
• 11112 - F16 + 1510 = _________16

John Owen, Rockport Fulton HS

Exercises

And finally, some equations

• x16 + 7616 = AB16
• x2 - 10112 = 1012
• x8 + 568 = 728
• x2 + 2510 = 1F16
• x8 + 3748 - 65568 = BAD16
• 378 + X16 = 110111102

John Owen, Rockport Fulton HS