Section 1.1: Integer Operations and the Division Algorithm

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Section 1.1: Integer Operations and the Division Algorithm

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Section 1.1: Integer Operations and the Division Algorithm

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Section 1.1: Integer Operations and the Division Algorithm

MAT 320 Spring 2008

Dr. Hamblin

- “You have 4 marbles and then you get 7 more. How many marbles do you have now?”

4

11

7

- “If you have 9 toys and you give 4 of them away, how many do you have left?”

5

4

9

- “You have 4 packages of muffins, and each package has 3 muffins. How many total muffins do you have?”

4

12

3

- “You have 12 cookies, and you want to distribute them equally to your 4 friends. How many cookies does each friend get?”

3

12

- As you can see, division is the most complex of the four operations
- Just as multiplication is repeated addition, division can be thought of as repeated subtraction

- 28 – 4 = 24
- 24 – 4 = 20
- 20 – 4 = 16
- 16 – 4 = 12
- 12 – 4 = 8
- 8 – 4 = 4
- 4 – 4 = 0
- Once we reach 0, we stop. We subtracted seven 4’s, so 28 divided by 4 is 7.

- 92 – 12 = 80
- 80 – 12 = 68
- 68 – 12 = 56
- 56 – 12 = 44
- 44 – 12 = 32
- 32 – 12 = 20
- 20 – 12 = 8
- We don’t have enough to subtract another 12, so we stop and say that 92 divided by 12 is 7, remainder 8.

- Since 28 divided by 4 “comes out evenly,” we say that 28 is divisible by 4, and we write 28 = 4 · 7.
- However, 92 divided by 12 did not “come out evenly,” since 92 12 · 7. In fact, 12 · 7 is exactly 8 less than 92, so we can say that 92 = 12 · 7 + 8.

remainder

dividend

quotient

divisor

- Subtracting 13 one at a time would take a while
- 3409 – 100 · 13 = 2109
- 2109 – 100 · 13 = 809
- 809 – 50 · 13 = 159
- 159 – 10 · 13 = 29
- 29 – 13 = 19
- 19 – 13 = 3
- So 3409 divided by 13 is 262 remainder 3.
- All in all, we subtracted 262 13’s, so we could write 3409 – 262 · 13 = 3, or 3409 = 13 · 262 + 3.

- Start with dividend a and divisor b (“a divided by b”)
- Repeatedly subtract b from a until the result is less than a (but not less than 0)
- The number of times you need to subtract b is called the quotient q, and the remaining number is called the remainder r
- Once this is done, a = bq + r will be true

- Let a and b be integers with b > 0. Then there exist unique integers q and r, with 0 r < b and a = bq + r.
- This just says what we’ve already talked about, in formal language

- We’ve already talked about the repeated subtraction method
- Method 2: Guess and CheckFill in whatever number you want for q, and solve for r. If r is between 0 and b, you’re done. If r is too big, increase q. If r is negative, decrease q.
- Method 3: CalculatorType in a/b on your calculator. The number before the decimal point is q. Solve for r in the equation a = bq + r

- Notice that in the Division Algorithm, b must be positive, but a can be negative
- How do we handle that?

- “You owe me 30 dollars. How many 8 dollar payments do you need to make to pay off this debt?”
- Instead of subtracting 8 from -30 (which would just increase our debt), we add 8 repeatedly

- -30 + 8 = -22
- -22 + 8 = -14
- -14 + 8 = -6 (debt not paid off yet!)
- -6 + 8 = 2
- So we made 4 payments and had 2 dollars left over
- -30 divided by 8 is -4, remainder 2
- Check: -30 = 8 · (-4) + 2

- Negative numbers are tricky, be sure to always check your answer
- Be careful when using the calculator method
- Example: -41 divided by 7The calculator gives -5.857…, but if we plug in q = -5, we get r = -6, which is not a valid remainder
- The correct answer is q = -6, r = 1