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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?