Law - e.g.1 - e.g.2

e.g.5 (Page 11) Illustration of Lemma 2.2E.g.,i = 11n = 511 mod 5 = 1(11 + 5) mod 5 = 16 mod 5 = 1(11 + (-1)x5) mod 5 = 6 mod 5 = 1(11 + 2x5) mod 5 = 21 mod 5 = 1(11 + (-2)x5) mod 5 = 1 mod 5 = 1(11 + 3x5) mod 5 = 26 mod 5 = 1(11 +(-3)x5)mod5=-4mod 5= 1(11 + k.5) mod 5 = 1

e.g.5 (Page 11) ◼ Illustration of Lemma 2.2 ◼ E.g., i = 11 n = 5 11 mod 5 = 1 (11 + 5) mod 5 = 16 mod 5 = 1 (11 + 2x5) mod 5 = 21 mod 5 = 1 (11 + 3x5) mod 5 = 26 mod 5 = 1 (11 + k.5) mod 5 = 1 (11 + (-1)x5) mod 5 = 6 mod 5 = 1 (11 + (-2)x5) mod 5 = 1 mod 5 = 1 (11 + (-3)x5) mod 5 = -4 mod 5 = 1

e.g.6 (Page 11)Prove that11 mod 5 = (11 + k5) mod 5for all integers kLetr= 11mod5ByEuclid'sDivisionTheorem,wecanwrite11=5g+rwhere qand r are twounique integers and 0 ≤r<5Consider 11 + k·5= (5q + r) + k-5=5g+r+k5=5g+k5+r= 5(g+ k)+ rBy the definition of Euclid's division theorem, (11 +k·5)mod 5 = r

e.g.6 (Page 11) ◼ Prove that 11 mod 5 = (11 + k.5) mod 5 for all integers k Let r = 11 mod 5 By Euclid’s Division Theorem, we can write 11 = 5q + r where q and r are two unique integers and 0  r < 5 Consider 11 + k.5 = (5q + r) + k.5 = 5q + r + k.5 = 5q + k.5 + r = 5(q + k) + r By the definition of Euclid’s division theorem, (11 + k.5) mod 5 = r

e.g.7 (Page 12) Illustration of Lemma 2.3 E.g.,(2 + 8) mod 9 = (2 + (8 mod 9)) mod 9= (2 mod 9) + 8) mod 9= (2 mod 9) + (8 mod 9) mod 9(2·8) mod 9 = (2 : (8 mod 9) mod 9= (2 mod 9) - 8) mod 9= (2 mod 9) : (8 mod 9) mod 9

e.g.7 (Page 12) ◼ Illustration of Lemma 2.3 ◼ E.g., (2 + 8) mod 9 = (2 + (8 mod 9)) mod 9 = ((2 mod 9) + 8) mod 9 = ((2 mod 9) + (8 mod 9)) mod 9 (2.8) mod 9 = (2 . (8 mod 9)) mod 9 = ((2 mod 9) . 8) mod 9 = ((2 mod 9) . (8 mod 9)) mod 9

e.g.8 (Page 12)2x8mod9=16mod9=7((2 mod 9) (8 mod 9)) mod 9 = 2 x 8 mod 9 = 16 mod 9 = 7Conclusion:2 x 8 mod 9 = ((2 mod 9) (8 mod 9)) mod 9(2 + 8) mod 9 = 1(2 mod 9) + (8 mod 9)) mod 9 = (2 + 8) mod 9 = 10 mod 9 = 1Conclusion:(2 + 8) mod 9 = ((2 mod 9) + (8 mod 9)) mod 9

e.g.8 (Page 12) 2 x 8 mod 9 = 16 mod 9 = 7 ((2 mod 9) (8 mod 9)) mod 9 = 2 x 8 mod 9 = 16 mod 9 = 7 (2 + 8) mod 9 = 1 ((2 mod 9) + (8 mod 9)) mod 9 = (2 + 8) mod 9 = 10 mod 9 = 1 Conclusion: 2 x 8 mod 9 = ((2 mod 9) (8 mod 9)) mod 9 Conclusion: (2 + 8) mod 9 = ((2 mod 9) + (8 mod 9)) mod 9

e.g.8Claim:(20 + 17) mod 9 = ((20 mod 9) + (17 mod 9)) mod 9(2 + 8) mod 9 = 1(2 mod 9) + (8 mod 9)) mod 9 = (2 + 8) mod 9 = 10 mod 9 = 1Conclusion:(2 + 8) mod 9 = ((2 mod 9) + (8 mod 9)) mod 9

e.g.8 (2 + 8) mod 9 = 1 ((2 mod 9) + (8 mod 9)) mod 9 = (2 + 8) mod 9 = 10 mod 9 = 1 Conclusion: (2 + 8) mod 9 = ((2 mod 9) + (8 mod 9)) mod 9 Claim: (20 + 17) mod 9 = ((20 mod 9) + (17 mod 9)) mod 9