Let p be a prime number. For example, p could be 2,3,5,7, etc. We define a field Fp consisting of the elements {0, 1, 2, . . . , p − 1} with operations +, −, ×, ÷ considered only up to remainder when we divide by p. For example,in F5,2+4 = 6≡ 1,2−4 = −2 ≡ 3,and2×4 = 8 ≡3. SinceF5, 4 × 3 = 12 ≡ 2 (you can check that the multiplication of 4 with any other element in F5 is never 2), we say that 2 ÷ 4 ≡ 3 in F5. By row reduction, parametrize all solutions to the linear equations x + y + z = 3 3y + 2 = 4-2 x + y − z = 3 You should get a one parameter family of solutions (x, y, z), where each x, y, z ∈ F5.
Let p be a prime number. For example, p could be 2,3,5,7, etc. We define a field Fp consisting of the elements {0, 1, 2, . . . , p − 1} with operations +, −, ×, ÷ considered only up to remainder when we divide by p. For example,in F5,2+4 = 6≡ 1,2−4 = −2 ≡ 3,and2×4 = 8 ≡3. SinceF5, 4 × 3 = 12 ≡ 2 (you can check that the multiplication of 4 with any other element in F5 is never 2), we say that 2 ÷ 4 ≡ 3 in F5. By row reduction, parametrize all solutions to the linear equations x + y + z = 3 3y + 2 = 4-2 x + y − z = 3 You should get a one parameter family of solutions (x, y, z), where each x, y, z ∈ F5.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 87E
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Let p be a prime number. For example, p could be 2,3,5,7, etc. We define a field Fp consisting of the elements {0, 1, 2, . . . , p − 1} with operations +, −, ×, ÷ considered only up to remainder when we divide by p. For example,in F5,2+4 = 6≡ 1,2−4 = −2 ≡ 3,and2×4 = 8 ≡3. SinceF5, 4 × 3 = 12 ≡ 2 (you can check that the multiplication of 4 with any other element in F5 is never 2), we say that 2 ÷ 4 ≡ 3 in F5.
By row reduction, parametrize all solutions to the linear equations
x + y + z = 3
3y + 2 = 4
-2 x + y − z = 3
You should get a one parameter family of solutions (x, y, z), where each x, y, z ∈ F5.
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