SOLUTION: Let U = {x ∈ Z | 0 ≤ x ≤ 20}. Also, let A = set of all even integers; B= set of all prime numbers; C = set of all numbers coprime with 2; and D={y ∈ U | y = 2z + 3, for

Algebra ->  sets and operations -> SOLUTION: Let U = {x ∈ Z | 0 ≤ x ≤ 20}. Also, let A = set of all even integers; B= set of all prime numbers; C = set of all numbers coprime with 2; and D={y ∈ U | y = 2z + 3, for       Log On


   



Question 1167141: Let U = {x ∈ Z | 0 ≤ x ≤ 20}. Also, let A = set of all even integers;
B= set of all prime numbers; C = set of all numbers coprime with 2; and
D={y ∈ U | y = 2z + 3, for some z∈Z}.
Represent sets A, B, C, D using set-roster notation.

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
U = {x ∈ Z | 0 ≤ x ≤ 20} = {0,1,2,3,...,20};

A = set of all even integers = ...-6%2C-4%2C-2%2C0%2C2%2C4%2C6%2C...;

B= set of all prime numbers = 2%2C3%2C5%2C7%2C11%2C13%2C17%2C19%2C...;

C = set of all numbers coprime with 2 = 3%2C5%2C7%2C9%2C11%2C...;

{y ∈ U | y= 2z + 3, for some z∈Z} 

To do the last one:

0%3C=2z+%2B+3%3C=20

Solve for z in the middle.
Subtract 3 from all three sides:

-3%3C=2z%3C=17

Divide all three sides by 2

-1.5%3C=z%3C=8.5

Since z is an integer,

-1%3C=z%3C=8 = {-1,0,1,2,3,...,8}, substitute each 
in 2z+3:

{y ∈ U | y = 2z + 3, for some z ∈ Z} = {1,3,5,7,...,19}

Edwin