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JA8. Algebraic Structures

Task 1

Choose a pair of two-digit numbers and find their greatest common divisor. Show the steps.

According to (CueMath, n.d), to find the greatest common divisor of two numbers (a and b):

  • Find the divisors of ‘a’ and ‘b’.
  • List the common divisors of ‘a’ and ‘b’.
  • Now find the divisor which is the highest among the common divisors.

I chose numbers 112 and 44.

  • Divisors of 112: 1, 2, 4, 7, 8, 14, 16, 28, 56, 112.
  • Devisors of 44: 1, 2, 4, 11, 22, 44.
  • Common divisors: 1, 2, 4.
  • Thus, the greatest common divisor of 112 and 44 is 4.

Task 2

Choose any one modular arithmetic operation (addition modulo +_n or multiplication modulo n X _n ) from the set of positive integers ℤ (for example, ℤ5= {0,1,2,3,4} do not use the same example with your choice of n). Find the identities and inverses of all elements with respect to the operation you choose. If they do not exist, explain why. You can use an operation table to explain your reasoning.

I will choose ℤ3= {0,1,2} and the operation of multiplication modulo *_3.

* (mod 3) 0 1 2
0 0 0 0
1 0 1 2
2 0 2 1
  • Identity: ∃e ∈ G,∀a ∈ G, a∘e = e∘a = a, by observing the table:
    • 0∘1 = 0 (identity of 0 is 1)
    • 1∘1 = 1 (identity of 1 is 1)
    • 2∘1 = 2 (identity of 2 is 1)
    • The identity of ℤ3 is 1.
  • Inverse: ∀a ∈ G,∃a−1 ∈ G, a∘a−1 = a−1∘a = e, we find e=1 from previous question; thus, we need to find a−1 such that a∘a−1 = 1, by observing the table:
    • 0∘0 = 0 (0 has no inverse, as 0 ≠ e)
    • 1∘1 = 1 (inverse of 1 is 1)
    • 2∘2 = 1 (inverse of 2 is 2)
    • 2 and 1 have inverses, but 0 does not have an inverse.

Task 3

Find and explain solution for the equation x^2+ 5x +1 = 0 in ℤ3, the integers modulo 3.

The ℤ3 = {0, 1, 2} and the equation x^2+ 5x +1 = 0 can be rewritten as x^2+ 5x +1 ≡ 0 (mod 3) and the solution of

0 mod 3 = 0.

Let’s discuss each case separately:

  • For x = 0:
    • f(0) = (0)^2 + 5(0) + 1 = 0 + 0 + 1 = 1
    • f(0) mod 3 = 1 mod 3 = 2
    • f(0) mod 3 ≢ 0 mod 3 thus x = 0 is not a solution.
  • For x = 1:
    • f(1) = (1)^2 + 5(1) + 1 = 1 + 5 + 1 = 7 and
    • f(1) mod 3 = 7 mod 3 = 1
    • f(1) mod 3 ≢ 0 mod 3, thus x = 1 is not a solution.
  • For x = 2:
    • f(2) = (2)^2 + 5(2) + 1 = 4 + 10 + 1 = 15 and
    • f(2) mod 3 = 15 mod 3 = 0
    • f(2) mod 3 ≡ 0 mod 3 , thus x = 2 is a solution.
x f(x) f(x) mod 3 f(x) mod 3 ≡ 0 mod 3
0 1 1 No
1 7 1 No
2 15 0 Yes

The solution for the equation x^2+ 5x +1 = 0 in ℤ3 is x = 2.

References