Section 1.1: Integer Operations and the Division Algorithm

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Presentation transcript:

Section 1.1: Integer Operations and the Division Algorithm MAT 320 Spring 2008 Dr. Hamblin

Addition “You have 4 marbles and then you get 7 more. How many marbles do you have now?” 4 11 7

Subtraction “If you have 9 toys and you give 4 of them away, how many do you have left?” 5 4 9

Multiplication “You have 4 packages of muffins, and each package has 3 muffins. How many total muffins do you have?” 4 12 3

Division “You have 12 cookies, and you want to distribute them equally to your 4 friends. How many cookies does each friend get?” 3 12

Examining Division 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 divided by 4 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 divided by 12 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.

Expressing the Answer As an Equation 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

3409 divided by 13 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.

How Division Works 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

Theorem 1.1: The Division Algorithm (aka The Remainder Theorem) 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

Ways to Find the Quotient and Remainder We’ve already talked about the repeated subtraction method Method 2: Guess and Check Fill 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: Calculator Type in a/b on your calculator. The number before the decimal point is q. Solve for r in the equation a = bq + r

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

-30 divided by 8 “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 divided by 8, continued -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

Caution! Negative numbers are tricky, be sure to always check your answer Be careful when using the calculator method Example: -41 divided by 7 The 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