LESSON ON Number Base Subtraction and Simple Equations Computer Science LESSON ON Number Base Subtraction and Simple Equations John Owen, Rockport Fulton HS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

John Owen, Rockport Fulton HS Subtraction, Base 16 C22 D6 -3B 9B 22 minus B (11) is B. C minus 3 is 9. Answer is 9B John Owen, Rockport Fulton HS

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

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

John Owen, Rockport Fulton HS Subtraction, Base 2 02 110 - 1 101 Subtract to get the answer. John Owen, Rockport Fulton HS

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

John Owen, Rockport Fulton HS Subtraction, Base 2 2 100 - 11 11 Final answer is 11, base 2 John Owen, Rockport Fulton HS

John Owen, Rockport Fulton HS Simple Equations Here an equation to solve (base 10): x + 6 = 14 John Owen, Rockport Fulton HS

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

John Owen, Rockport Fulton HS Simple Equations Now do it in base 8: x + 6 = 14 John Owen, Rockport Fulton HS

John Owen, Rockport Fulton HS Simple Equations Solution…subtract 6 from both sides x + 6 = 14 -6 -6 x = ? John Owen, Rockport Fulton HS

John Owen, Rockport Fulton HS Simple Equations Answer is 6, base 8 12 x + 6 = 14 -6 -6 x = 6 John Owen, Rockport Fulton HS

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

John Owen, Rockport Fulton HS Simple Equations Solution? x + 2A = F3 John Owen, Rockport Fulton HS

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

John Owen, Rockport Fulton HS Exercises Now try these exercises 12 - 12 = 78 - 68 = F16 - A16 = 158 - 68 = 4916 - 2B16 = CC16 - AD16 = John Owen, Rockport Fulton HS

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

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

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

John Owen, Rockport Fulton HS ANSWERS (JUMBLED) 1 5 7 11 14 19 35 37 69 110 177 332 354 1010 1335 10000 14037 1E 1F BF F John Owen, Rockport Fulton HS