# Solving Linear Equation over a Field

A set is a well-defined collection of distinct things or objects, such as animals, plants, vowels (a, e, i, o, u), swearwords, etc.

By well-defined, we mean that there is a rule that enables us to determine whether a given object is an element of the set or not.

A field is a set F together with two binary operations on F called addition and multiplication. These operations are required to satisfy the following properties, referred to as field axioms:

1. Associativity of addition and multiplication: a + (b + c) = (a + b) + c, a ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c.
2. Commutativity of addition and multiplication: a + b = b + a, a ⋅ b = b ⋅ a.
3. Additive and multiplicative identity: there exist two different elements 0 and 1 in F such that a + 0 = a and a ⋅ 1 = a.
4. Additive and multiplicative inverses: for every element a in F, there exists an element in F, denoted as −a, called the additive inverse of a, such that a + (−a) = 0; and another element in F, denoted by a-1 or 1/a, called the multiplicative inverse of a, such that a ⋅ a-1 = 1.
5. Distributivity of multiplication over addition: a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c).

Examples: ℚ, ℝ, and ℂ are fields, but ℕ (2 - 3 ∉ ℕ), ℤ (3 does not have a multiplicative inverse), and ℤn are not fields. 2 and 4 do not have multiplicative inverses in ℤ6, there is no an element a ∈ ℤ6 such that a·2≡1 (mod 6).

n is a field if and only if n is prime.

Fields are the environment where linear equations can be solved. Let F be a field, ax + b = c is a linear equation with variable x and a, b, c ∈ F (elements in the field are called scalars). A solution is any value or assignment to the variable x with makes the equation true or produces a true statement.

For linear equations ax + b = c, there is one and only one solution in any field. ax + b = c ⇒ [Additive inverses] (ax + b) + (-b) = c -b ⇒ [Associativity of addition] ax + (b - b) = c -b ⇒ [Additive identity] ax = c - b ⇒ [Multiplicative inverses] a-1(ax) = a-1(c -b) ⇒ [Associativity of multiplication] (a-1a)x = c-ba ⇒ [Multiplicative identity] x = c-ba

Solve the following equations:

1. 4x + 3 = 5 over ℚ. 4x + 3 = 5 ⇒ 4x = 5 - 3 = 2 ⇒ 4x = 2 ⇒ x = 12.
2. 4x + 7 = 21 over ℝ ⇒ 4x = 21 -7 = 14 ⇒ x = 14/4 = 7/2 = 3.5.
3. (2+4i)x + (1+5i) = 2+2i over ℂ ⇒ (2+4i)x = 2+2i-(1+5i) = 1 -3i ⇒ (2+4i)x = 1 -3i ⇒ x = $\frac{1-3i}{2(1+2i)}=\frac{1-3i}{2(1+2i)}(\frac{1-2i}{1-2i})=\frac{1-3i-2i-6}{2(1+2i-2i+4)}=\frac{-5}{10}+\frac{-5i}{10}=\frac{-1}{2}-\frac{i}{2}$
4. 4x + 1 = 3 over ℤ5 ↭ 4x + 1 ≡ 3 (mod 5) ⇒ 4x ≡ 2 (mod 5) ⇒ x ≡ 2/4 (mod 5) ≡ 1/2 (mod 5) ≡ (1 + 5)/2 (mod 5) ≡ 3 (mod 5). 3 is the solution over ℤ5.
5. 4x + 7 = 5 over ℤ9 ↭ 4x + 7 ≡ 5 (mod 9) ⇒ 4x ≡ -2 (mod 9) ⇒ 4x ≡ 7 (mod 9) ⇒ x ≡ 7/4 (mod 9) ≡ (7+9)/4 (mod 9) ≡ 4(mod 9). The solution is 4.
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