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Computer Science LESSON ON Number Base Subtraction and Simple Equations John Owen, Rockport Fulton HS 1 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 2 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 3 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 4 Subtraction, Base 10 Now look at this problem: 63 -37 In this problem, you need to borrow. John Owen, Rockport Fulton HS 5 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 6 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 7 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 8 Subtraction, Base 10 513 63 -37 26 Then you subtract the two columns with a result of 26. John Owen, Rockport Fulton HS 9 Subtraction, Base 8 Now let’s try base eight: 63 -37 Again, in this problem, you need to borrow. John Owen, Rockport Fulton HS 10 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 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. John Owen, Rockport Fulton HS 12 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 13 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 14 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 15 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 16 Subtraction, Base 16 Here’s another example in base 16 D6 -3B How is this one solved? Try it. John Owen, Rockport Fulton HS 17 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 18 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 19 Subtraction, Base 2 Now base 2: 11 - 1 10 This one is easy…answer is 10 John Owen, Rockport Fulton HS 20 Subtraction, Base 2 Another in base 2: 02 110 - 1 Here we need to borrow from the twos place… John Owen, Rockport Fulton HS 21 Subtraction, Base 2 02 110 - 1 101 Subtract to get the answer. John Owen, Rockport Fulton HS 22 Subtraction, Base 2 Still another in base 2: 02 110 - 11 1 Now borrow again… John Owen, Rockport Fulton HS 23 Subtraction, Base 2 2 100 - 11 11 Final answer is 11, base 2 John Owen, Rockport Fulton HS 24 Simple Equations Here an equation to solve (base 10): x + 6 = 14 John Owen, Rockport Fulton HS 25 Simple Equations Solution…subtract 6 from both sides x + 6 = 14 -6 -6 x = 8 John Owen, Rockport Fulton HS 26 Simple Equations Now do it in base 8: x + 6 = 14 John Owen, Rockport Fulton HS 27 Simple Equations Solution…subtract 6 from both sides x + 6 = 14 -6 -6 x = ? John Owen, Rockport Fulton HS 28 Simple Equations Answer is 6, base 8 12 x + 6 = 14 -6 -6 x = 6 John Owen, Rockport Fulton HS 29 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 30 Simple Equations Solution? x + 2A = F3 John Owen, Rockport Fulton HS 31 Simple Equations Subtract 2A from both sides: E19 x + 2A = F3 - 2A -2A x = C9 John Owen, Rockport Fulton HS 32 Exercises 1. 2. 3. 4. 5. 6. Now try these exercises 12 - 12 = 78 - 68 = F16 - A16 = 158 - 68 = 4916 - 2B16 = CC16 - AD16 = John Owen, Rockport Fulton HS 33 Exercises 7. 8. 9. 10. 11. 12. 13. 738 - 348 = 3E16 – 2F16 = 1012 - 102 = 11012 - 112 = 10102 - 1112 = 7168 - 3648 = 7768 + 3378 = John Owen, Rockport Fulton HS 34 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 John Owen, Rockport Fulton HS 35 Exercises And finally, some equations 19. x16 + 7616 = AB16 20. x2 - 10112 = 1012 21. x8 + 568 = 728 22. x2 + 2510 = 1F16 23. x8 + 3748 - 65568 = BAD16 24. 378 + X16 = 110111102 John Owen, Rockport Fulton HS 36 ANSWERS (JUMBLED) 0 1 5 7 11 11 14 19 35 37 69 110 177 332 354 1010 1335 10000 14037 1E 1F BF F F John Owen, Rockport Fulton HS 37