Division & Divisibility
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Division & Divisibility. Division. a divides b if a is not zero there is a m such that a.m = b “a is a factor of b” “b is a multiple of a” a|b. Division. If a|b and a|c then a|(b+c) “ If a divides b and a divides c then a divides b plus c ”. a|b  a.x = b a|c  a.y = c

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Division

  • a divides b if

    • a is not zero

    • there is a m such that a.m = b

  • “a is a factor of b”

  • “b is a multiple of a”

  • a|b


Division

  • If a|b and a|c then a|(b+c)

  • “If a divides b and a divides c then a divides b plus c”

  • a|b  a.x = b

  • a|c  a.y = c

  • b+c = a.x + a.y

  • = a(x + y)

  • and that is divisible by a


Division

  • a|b  a.m = b

  • b.c = a.m.c

  • which is divisible by a


Division

  • a|b  a.x = b

  • b|c  b.y = c

  • c = a.x.y

  • and that is divisible by a


Division

Theorem 1 (page 202, 6th ed, page 154, 5th ed)


The Division Algorithm (aint no algorithm)

dividend

divisor

remainder

quotient

  • a is an integer and d is a positive integer

    • there exists unique integers q and r,

    • 0  r  d

    • a = d.q. + r

a divided by d = q remainder r

NOTE: remainder r is positive and divisor d is positive


Division

  • a = d.q + r and 0 <= r < d

  • a = -11 and d = 3 and 0 <= r < 3

    • -11 = 3q + r

    • q = -4 and r = 1

  • a = d.q + r and 0 <= r < d

  • a = -63 and d = 20 and 0 <= r <= 20

    • -63 = 20q + r

    • q = -4 and r = 17

  • a = d.q + r and 0 <= r < d

  • a = -25 and d = 15 and 0 <= r < 15

    • -25 = 15.q + r

    • q = -2 and r = 10


Division

  • a = d.q + r and 0 <= r < d

  • a = -11 and d = 3 and 0 <= r < 3

    • -11 = 3q + r

    • q = -4 and r = 1

Troubled by this?

Did you expect q = -3 and r = -2?

What if 3 of you went to a café and got a bill for £11?

Would you each put £3 down and then leg it?

Or £4 each and leave £1 tip?


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