Let R be a relation on the set of positive integers, Z+, defined by xRy <--> x+y is even?

1. R is reflexive

Notice that x + x = 2x.

2. R is symmetric

Assume that x + y = 2q for some integer q. Since addition is commutative, then y + x = 2q.

3. R is transitive

Assume that x + y = 2k for some integer k and assume that y + z = 2m for some integer m. Adding the two equations yield

x + y + y + z= 2k + 2m

x + 2y + z= 2k + 2m

x + z= 2k + 2m - 2y

x + z= 2(k + m - y)

Therefore, R is an equivalence relation.

commencing with a million, 2, ___, there are 8 solutions because of the fact z ought to be 3, 4, 5, 6, 7, 8, 9, or 10 commencing with a million, 3, ___, there are 7 all a thank you to a million, 9, 10 with a million answer. 8+7+..+a million=36 Then 2, 3, __ = 6 procedures; 2, 4, __ = 5 procedures, and so on for 6+..+a million=21 Then 3,4, ___ has 4 procedures 3,5,__ has 3, and so on 4+..+a million=10 Then 4,5 __ has 2 procedures, 4,6,__ has a million way so 2+a million=3 No procedures have x = 5 or larger. So upload all of them up.