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The Division Algorithm

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1 The Division Algorithm
In the context of division, 203 is called the dividend, 8 the divisor, 25 the quotient and 3 the remainder. The relationship between these four quantities can also be expressed as This theorem is referred to as the division algorithm or the division identity. Notation: The greatest common divisor (gcd) of a and b can be denoted by (a,b) when the context is unambiguous.

2 Theorem: If a = bq + r, then (a,b) = (b,r)
Proof The repeated application of the above theorem until the gcd is identified is known as the Euclidean Algorithm.

3

4 Find the greatest common divisor of 203 and 8 using the Euclidean algorithm.
Thus the gcd of 203 and 8 is 1. Note that when the gcd of two numbers is 1, then the numbers are said to be co-prime. Find the greatest common divisor of 132 and 424 using the Euclidean algorithm. Thus the gcd of 132 and 424 is 4.

5 A Consequence of the Euclidean Algorithm
Express the gcd of 132 and 424 in the form 132s+424t The gcd is identified here.

6 D The gcd is identified here. C B A

7 Show that 280 and 117 are relatively prime.
Find a and b such that 280a + 117b = 1 Hence the gcd of 280 and 117 is 1 so 280 and 117 are relatively prime.

8 Page 145 Exercise 4 Questions 1(a), (c), (e), (g), (i)
Page 147 Exercise 5 Questions 1 to 4 TJ Exercise 7 and 8

9 The Division Algorithm and Number Bases
We take it for granted that a number can be expressed as a sum of multiples of powers of 10. Perhaps it is because we have 10 fingers that this system developed. There is evidence that the Babylonians used a different base: 60. Computers work using a logic which calculates using 2 as a base. Computer programmers use 8 and 16 frequently. Consider the number 48. What would this number be in base 5? We don’t really need the subscript for base 10. In base 16 we have to invent new digits. A = 10, B = 11, C = 12 all the way to F = 15.

10 (First convert to base 10 number)

11 Page 151 Exercise 7

12 Proof by Induction The statement is true for n = 1.
Assume the statement is true for n = k. Consider n = (k + 1) The statement is true for n = 1 and if true for n = k is true for n = k + 1. Therefore, by induction the statement is true for all n ≥ 1.

13 We have established: You will establish In the next exercise Between them we can establish a formula for any series whose nth term is a cubic expression. You have to remember the above formula NOTE:

14 When n = 1, Assume the statement is true for n = k. Consider n = k + 1. The statement is true for n = 1 and if true for n = k then it is true for n = k + 1. Therefore by induction the statement is true for all n ≥ 1.

15 Welcome to the end of the Advanced Higher Course
Page 141 Exercise 3A – As many as possible. TJ Exercise 5. Welcome to the end of the Advanced Higher Course


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