Final answer:
The given relation f is not a function because there is an integer, such as 6, which is divisible by both 2 and 3, resulting in two different outputs (6,1) and (6,5). This violates the definition of a function, where each input must correspond to exactly one output.
Step-by-step explanation:
To determine if a relation is a function, we must check whether each input (in this case, each integer in Z) corresponds to exactly one output. The given relation f is defined as f ={(x,1):2 divides x} ∪ {(x,5):3 divides x}. This means that for every integer x that is divisible by 2, we associate it with the number 1, and for every integer x that is divisible by 3, we associate it with the number 5.
In order to prove this is not a function, we need to find an integer x that is both divisible by 2 and 3. Such numbers are called multiples of 6. For example, if we take x = 6, which is the smallest positive integer divisible by both 2 and 3, we get two pairs: (6,1) and (6,5). This shows that our x = 6 corresponds to more than one output, which violates the definition of a function. Therefore, the given relation is not a function.