The relation R={(x,y):x,y∈Z and x+y is even} is:
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Step-by-Step Solution
Step 1: Check for Reflexivity
A relation R on a set A is reflexive if (x,x)∈R for every x∈A. In this case, we need to check if x + x is always even for any integer x. The sum x + x simplifies to 2x.
Step 2: Evaluate Reflexivity
Since 2x is a multiple of 2, it is always an even number for any integer x. Therefore, (x,x)∈R for all x∈Z, which means the relation R is reflexive.
Step 3: Check for Symmetry
A relation R is symmetric if whenever (x,y)∈R, then (y,x)∈R. This means if x + y is even, we need to check if y + x is also even. Since addition of integers is commutative, x+y=y+x.
Step 4: Evaluate Symmetry
Since x+y=y+x, if x + y is even, then y + x must also be even. Thus, (y,x)∈R whenever (x,y)∈R, which means the relation R is symmetric.
Step 5: Check for Transitivity
A relation R is transitive if whenever (x,y)∈R and (y,z)∈R, then (x,z)∈R. This means if x + y is even and y + z is even, we need to check if x + z is also even.
Step 6: Evaluate Transitivity
If x + y is even, then x and y must have the same parity (both even or both odd). Similarly, if y + z is even, then y and z must have the same parity. If x and y have the same parity, and y and z have the same parity, it implies that x and z must also have the same parity. When x and z have the same parity, their sum x + z will always be even. For example, if x is odd and y is odd, x+y is even. If y is odd and z is odd, y+z is even. Then x is odd and z is odd, so x+z is even. Thus, the relation R is transitive.
Step 7: Conclusion
Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.