Lemma2.2(Y+9k)mod 9= Ymod 9e.g.8Claim:(20 + 17) mod 9 = ((20 mod 9) + (17 mod 9)) mod 9Why is it correct?ByEuclid'sDivisionTheorem,wecanwrite20asfollows.20 = 9q1 + r1where qi and ri are some unique integers.= 9q1 + (20 mod 9)By Euclid's Division Theorem, we can write 17 as follows.17 = 992 + r2whereqz and rz are some unique integers= 9q2 + (17 mod 9)Consider (20 + 17) mod 9= {[9q1 + (20 mod 9)] + [9q2 + (17 mod 9)]) mod 9= (9q1 + (20 mod 9) + 9q2 + (17 mod 9)) mod 9= [(20 mod 9) + (17 mod 9) + 9q1 + 9q2] mod 9= [ (20 mod 9) + (17 mod 9) + 9(q1 + q2) mod 9(by Lemma 2.2)= [ (20 mod 9) + (17 mod 9)] mod 9
e.g.8 Claim: (20 + 17) mod 9 = ((20 mod 9) + (17 mod 9)) mod 9 Why is it correct? By Euclid’s Division Theorem, we can write 20 as follows. 20 = 9q1 + r1 where q1 and r1 are some unique integers. = 9q1 + (20 mod 9) By Euclid’s Division Theorem, we can write 17 as follows. 17 = 9q2 + r2 where q2 and r2 are some unique integers. = 9q2 + (17 mod 9) Consider (20 + 17) mod 9 = {[9q1 + (20 mod 9)] + [9q2 + (17 mod 9)]} mod 9 = {9q1 + (20 mod 9) + 9q2 + (17 mod 9)} mod 9 = [ (20 mod 9) + (17 mod 9) + 9q1 + 9q2 ] mod 9 = [ (20 mod 9) + (17 mod 9) + 9(q1 + q2 )] mod 9 = [ (20 mod 9) + (17 mod 9)] mod 9 Lemma 2.2 (Y + 9k) mod 9 = Y mod 9 (by Lemma 2.2)
e.g.9 (Page 14)E.g. 0 +5 2 = 2(0 + 2) mod 5 = 2 mod 5 = 2 E.g., 1:52 = 2(1 2) mod 5 = 2 mod 5 = 2
e.g.9 (Page 14) ◼ E.g. 0 +5 2 = 2 ◼ E.g., 1. 52 = 2 (0 + 2) mod 5 = 2 mod 5 = 2 (1.2) mod 5 = 2 mod 5 = 2