For all m,n define a relation R to be x + y is even - show R is on equivalence relation? Solution An equivalence relation is a relation that is reflexive, symmetric, and transitive. Reflexive: We must show that xRx for all x in Z. Now, consider that x+x=2x. As this is 2 times some integer, we see that x+x is divisible by 2, meaning that it is even. Therefore, xRx, and the relation is reflexive. Symmetric: We must show that xRy => yRx. Now, suppose that xRy. Then x+y is an even number. However, x+y=y+x, so y+x must also be even, which implies that yRx, so the relation is symmetric. Transitive: We must show that if xRy and yRz, then xRz. Now, xRy => x+y is an even number. Also, yRz => y+z is even. Now, the sum of two even numbers is even, so (x+y)+(y+z)=x+2y+z is also even. Furthermore, 2y must be even (it\'s divisible by 2), and the difference of two even numbers is even, so (x+2y+z)-2y=x+z is even. This implies that xRz, so the relation is transitive. Thus, the relation R is an equivalence relation..

1. For all m,n define a relation R to be x + y is even - show R is on equivalence relation?
Solution
An equivalence relation is a relation that is reflexive, symmetric, and transitive.
Reflexive:
We must show that xRx for all x in Z. Now, consider that x+x=2x. As this is 2 times some
integer, we see that x+x is divisible by 2, meaning that it is even. Therefore, xRx, and the
relation is reflexive.
Symmetric:
We must show that xRy => yRx. Now, suppose that xRy. Then x+y is an even number.
However, x+y=y+x, so y+x must also be even, which implies that yRx, so the relation is
symmetric.
Transitive:
We must show that if xRy and yRz, then xRz. Now, xRy => x+y is an even number. Also, yRz
=> y+z is even. Now, the sum of two even numbers is even, so (x+y)+(y+z)=x+2y+z is also
even. Furthermore, 2y must be even (it's divisible by 2), and the difference of two even numbers
is even, so (x+2y+z)-2y=x+z is even. This implies that xRz, so the relation is transitive.
Thus, the relation R is an equivalence relation.