Mathematical Systems - Modular Arithmetic, Applications of Modular Arithmetic, Introduction to Group PDF

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Modular Arithmetic, Applications of Modular Arithmetic, Introduction to Group Theory...

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MODULE 10 MATHEMATICAL SYSTEMS (Modular Arithmetic, Applications of Modular Arithmetic, Introduction to Group Theory)

III. ACTIVITIES TOPIC 1 Modular Arithmetic 1. Calculate the following using a 12-hour clock a. 5

8 = 1

b. 10

11 = 9

c.

6= 4

10

d. 1

4= 9

2. Determine whether the congruence is true or false. Justify your answer. a. 5 ≡ 20 mod 4 false not a congruence because (5 – 20)/4 does not divide or when you divided 5 by 4, the quotient will be 1 remainder 1 not 20. b. 100 ≡ 4 mod 8 is true congruence because (100 – 4)/8 does divide or when you divided 100 by 8, the quotient will be 12 remainder 4, that means the congruence is true. c. 25 ≡ 85 mod 12 is true congruence because (25 – 85)/12 = -5 does divide. d. 7 ≡ 21 mod 3 false not a congruence because (7 – 21)/3 does not divide or when you divided 7 by 3, the quotient will be 2 remainder 1 not 21. 3. Perform the modular arithmetic a. (42 + 35) mod 3 Answer: 42 + 35 = 77 divided by 3 = 25 remainder 2, therefore the answer is (42 + 35) mod 3 = 2 b. (19 – 6) mod 5 Answer: 19 – 6 = 13 divided by 5 = 2 remainder 3, therefore the answer is (19 – 6) mod 5 = 3 https://docs.google.com/forms/u/0/d/e/1FAIpQLSfcjKQ_Je5W2NQs5XEaEx8QaU3xfrBvUe f2iTkd4AP1v7muIw/formResponse https://docs.google.com/forms/d/e/1FAIpQLSfcjKQ_Je5W2NQs5XEaEx8QaU3xfrBvUef2iT kd4AP1v7muIw/viewscore? viewscore=AE0zAgAoHiX6XMwskrsxFX3eKv17HgkfBCmD3blH1A9PZsHBf8wc5I0tD0pFwN0AA c. (5 • 12) mod 4 Answer: 5 x 12 = 60 divided by 4 = 15, therefore the answer is (5 • 12) mod 4 = 0

d. (26 • 11) mod 15 Answer:

26 x 11 = 286 divided by 15 = 19 remainder 1, therefore the answer is (26 • 11) mod 15 = 1

4. Find all whole number solutions of the congruence equation. a. (5x + 4) ≡ 2 mod 8 Solution: x=0

5(0) + 4 ≢ 2 mod 8

not a solution

x=1

5(1) + 4 ≢ 2 mod 8

not a solution

x=2

5(2) + 4 ≢ 2 mod 8

not a solution

x=3

5(3) + 4 ≢ 2 mod 8

not a solution

x=4

5(4) + 4 ≢ 2 mod 8

not a solution

x=5

5(5) + 4 ≢ 2 mod 8

not a solution

x=6

5(6) + 4 ≡ 2 mod 8

A solution

x=7

5(7) + 4 ≢ 2 mod 8

not a solution

x=8

5(8) + 4 ≢ 2 mod 8

not a solution

x=9

5(9) + 4 ≢ 2 mod 8

not a solution

The solution between 0 and 9 is 6. b. (3x + 12) ≡ 7 mod 10 Solution: x=0

3(0) + 12 ≢ 7 mod 10

not a solution

x=1

3(1) + 12 ≢ 7 mod 10

not a solution

x =2

3(2) + 12 ≢ 7 mod 10

not a solution

x=3

3(3) + 12 ≢ 7 mod 10

not a solution

x=4

3(4) + 12 ≢ 7 mod 10

not a solution

x=5

3(5) + 12 ≡ 7 mod 10

A solution

x=6

3(6) + 12 ≢ 7 mod 10

not a solution

x=7

3(7) + 12 ≢ 7 mod 10

not a solution

x=8

3(8) + 12 ≢ 7 mod 10

not a solution

x=9

3(9) + 12 ≢ 7 mod 10

not a solution

The solution between 0 and 9 is 5.

TOPIC 2 Group Theory 5. Does the set { , 1, 2, 3, 4} with operation multiplication module 5 form a group? TOPIC 3 Arithmetical Operations with Complex Numbers 6. Perform as indicated in each item: a. ADD: (12 + 8i) + (6 – 4i) Answer: (12 + 8i) + (6 – 4i) = (12 + 6) + (8 – 4)i = 18 + 4i

Thus, (12 + 8i) + (6 – 4i) = 18 + 4i b. ADD: (6 + √-144) + (3 − √-81) Answer; Step 1 (6 + √-144) = (6 + 12i) (3 − √-81) = (3 − 9i) Step 2 (6 + 12i) + (3 − 9i) = (6 + 3) + (12i − 9i) = 9 + 3i Thus, (6 + √-144) + (3 − √-81) = 9 + 3i c. SUBTRACT: (25 + 18i) – (14 – 6i) Answer: 25 + 18i − 14 + 6i = (25 – 14) + (18i + 6i) = 11 + 24i Thus, (25 + 18i) – (14 – 6i) = 11 + 24i d. SUBTRACT: (22 − √-49) – (12 − √-16) Answer: Step 1 (22 − √-49) = (22 – 7i) (12 − √-16) = (12 – 4i) Step 2 22 – 7i − (12 – 4i) = 22 – 7i – 12 + 4i = (22 – 12) + (-7i + 4i) = 10 – 3i Thus, (22 − √-49) – (12 − √-16) = 10 – 3i e. MULTIPLY: (8 + 5i) • (6 + 2i) Answer: 8(6 + 2i) + 5i(6 + 2i) = (8 x 6) + (8 x 2i) + (5i x 6) + (5i x 2i) = 48 + 16i + 30i + 10i² = -1 then, = 48 + 16i + 30i + 10(-1) = 48 + 46i – 10 = 38 + 46i Thus, (8 + 5i) • (6 + 2i) = 38 + 46i

f.

MULTIPLY: (6 + √-25) • (2 − √-121) Answer: Step 1 (6 + √-25) = (6 + 5i) (2 − √-121) = (2 – 11i) Step 2 (6 + 5i) • (2 – 11i) = 6(2 – 11i) + 5i(2 – 11i) = (6 x 2) − (6 x 11i) + [(5i x 2) – (5i x 11i)] = 12 – 66i + 10i + 55i² = 12 – 66i + 10i + 55(-1) = 12 – 56i – 55

= −43 – 56i Thus, (6 + √-25) • (2 − √-121) = −43 – 56i TOPIC 4 Cryptology 7. Use a cyclical alphabetic encrypting code that shifts the letters stated number of positions to decode the encrypted message for 8th positions: VWJWLG QA XMZNMKB. Answer: c ≡ (p + 8) mod 26 V : c ≡ (22 + 8) mod 26 = 30 mod 26 = 4. So V will be coded by D W : c ≡ (23 + 8) mod 26 = 31 mod 26 = 5. So W will be E J : c ≡ (10 + 8) mod 26 = 18 mod 26. So J will be R W : c ≡ (23 + 8) mod 26 = 31 mod 26 = 5. So W will be E L : c ≡ (12 + 8) mod 26 = 20 mod 26. So L will be T G : c ≡ (7 + 8) mod 26 = 13 mod 26. So G will be O Therefore, VWJWLG is decoded as DERETO Q : c ≡ (17 + 8) mod 26 = 25 mod 26. So Q will be Y A : c ≡ (1 + 8) mod 26 = 9 mod 26. So A will be I Therefore, QA is decoded as YI X : c ≡ (24 + 8) mod 26 = 32 mod 26 = 6. So X will be F M : c ≡ (13 + 8) mod 26 = 21 mod 26. So M will be U Z : c ≡ (0 + 8) mod 26 = 8 mod 26. So Z will be H N : c ≡ (14 + 8) mod 26 = 22 mod 26. So N will be V M : c ≡ (13 + 8) mod 26 = 21 mod 26. So M will be U K : c ≡ (11 + 8) mod 26 = 19 mod 26. So K will be S B : c ≡ (2 + 8) mod 26 = 10 mod 26. So B will be J. Therefore, XMZNMKB is decoded as FUHVUSJ

8. Use a cyclical alphabetic encrypting code to decode the encrypted message: AOB HVS HCFDSRCSG

9. Use the encrypting congruence c ≡ (3p + 2) mod 26 to code the message TOWER OF LONDON. Answer: c ≡ (3p + 2) T : c ≡ 3(20) + 2 = 62 mod 26 = 2 remainder 10. So T will be coded by J O : c ≡ 3(15) + 2 = 47 mod 26 = 21. So O will be U W : c ≡ 3(23) + 2 = 71 mod 26 = 2 remainder 19. So W will be S E : c ≡ 3(5) + 2 = 17 mod 26. So E will be Q R : c ≡ 3(18) + 2 = 56 mod 26 = 2 remainder 4. So R will be D Therefore, TOWER is coded as JUSQD O : c ≡ 3(15) + 2 = 47 mod 26 = 21. So O will be U F : c ≡ 3(6) + 2 = 20 mod 26. So F will be T Therefore, OF is coded as UT

L : c ≡ 3(12) + 2 = 38 mod 26 = 12. So L will be L O : c ≡ 3(15) + 2 = 47 mod 26 = 21. So O will be U N : c ≡ 3(14) + 2 = 44 mod 26 = 18. So N will be R D : c ≡ 3(4) + 2 = 14 mod 26 . So D will be N O : c ≡ 3(15) + 2 = 47 mod 26 = 21. So O will be U N : c ≡ 3(14) + 2 = 44 mod 26 = 18. So N will be R Therefore, LONDON is coded as LURNUR

10. Decode the message LOFT JGMK LBS MNWMK that was encrypted using the congruence c ≡ (3p + 4) mod 26 Answer: c ≡ (3p + 4) mod 26 L : c ≡ 3(12) + 4 = 40 mod 26 = 14. So, L will be N O : c ≡ 3(15) + 4 = 49 mod 26 = 23. So, O will be W F : c ≡ 3(6) + 4 = 22 mod 26. So, F will be V T : c ≡ 3(20) + 4 = 64 mod 26 = 2 remainder 12. So, T will be L Therefore, LOFT is decoded as NWVL J : c ≡ 3(10) + 4 = 34 mod 26 = 8. So, J will be H G : c ≡ 3(7) + 4 = 25 mod 26. So, G will be Y M : c ≡ 3(13) + 4 = 43 mod 26 = 17. So, M will be Q K : c ≡ 3(11) + 4 = 37 mod 26 = 11. So, K will be K Therefore, JGMK is decoded as HYQK L : c ≡ 3(12) + 4 = 40 mod 26 = 14. So, L will be N B : c ≡ 3(2) + 4 =10 mod 26. So, B will be J S : c ≡ 3(19) + 4 = 61 mod 26 = 2 remainder 9. So, S will be I Therefore, LBS is decoded as NJI M : c ≡ 3(13) + 4 = 43 mod 26 = 17. So, M will be Q N : c ≡ 3(14) + 4 = 46 mod 26 = 20. So, N will be T W : c ≡ 3(23) + 4 =73 mod 26 = 2 remainder 21. So, W will be U M : c ≡ 3(13) + 4 = 43 mod 26 = 17. So, M will be Q K : c ≡ 3(11) + 4 = 37 mod 26 = 11. So, K will be K Therefore, MNWMK is decoded as QTUQK

IV. ASSESSMENT TOPIC 1 Modular Arithmetic 1. Calculate the following using a 12-hour clock a. 11

5=4

b. 5

11 = 6

c. 8

8= 4

d. 5

5 = 12

2. Determine whether the congruence is true or false. Justify your answer. a. 21 ≡ 45 mod 6 is true congruence because (21 – 45)/6 is -4 does divide. e. 5 ≡ 20 mod 4 false not a congruence because (5 – 20)/4 does not divide or when you divided 5 by 4, the quotient will be 1 remainder 1 not 20. b. 72 ≡ 30 mod 5 false not a congruence because (72 – 30)/5 does not divide or when you divided 72 by 5, the quotient will be 8 remainder 32 not 30. c. 18 ≡ 60 mod 7 is true congruence because (18 – 60)/7 is -6 does divide.

3. Perform the modular arithmetic a. (9 + 15) mod 7 Answer: 9 + 15 = 24 divided by 7 = 3 remainder 3, therefore the answer is (9 + 15) mod 7 = 3 b. (8 – 15) mod 12 Answer: 8 – 15 = -7 divided by 12 = does not divide c. (6 • 8) mod 9 Answer: 6 x 8 = 48 divided by 9 = 5 remainder 3, therefore the answer is (6 • 8) mod 9 = 3 d. (14 • 20) mod 8 Answer: 14 x 20 = 280 divided by 8 = 35, therefore the answer is (14 • 20) mod 8 = 0

4. Find all whole number solutions of the congruence equation. a. (3x + 12) ≡ 7 mod 10 Solution: x=0 x=1 x=2 x=3 x=4 x=5 x=6 x=7 x=8 x=9

3(0) + 12 ≢ 7 mod 10 3(1) + 12 ≢ 7 mod 10 3(2) + 12 ≢ 7 mod 10 3(3) + 12 ≢ 7 mod 10 3(4) + 12 ≢ 7 mod 10 3(5) + 12 ≡ 7 mod 10 3(6) + 12 ≢ 7 mod 10 3(7) + 12 ≢ 7 mod 10 3(8) + 12 ≢ 7 mod 10 3(9) + 12 ≢ 7 mod 10

The solution between 0 and 9 is 5. b. 3x ≡ 8 mod 11 Solution:

not a solution not a solution not a solution not a solution not a solution A solution not a solution not a solution not a solution not a solution...