1. Let T = {a, b, c, d}.
a. Find an example of a relation of this set that is transitive, but neither reflexive nor symmetric.
b. Find an example of a relation on this set that is both transitive and symmetric, but not reflexive.
To find an example of a relation on the set T = {a, b, c, d} that is transitive but neither reflexive nor symmetric, you can start by considering the transitivity requirement. Recall that a relation R on a set S is transitive if, for every distinct elements x, y, and z in S, if x is related to y and y is related to z, then x must be related to z.
a. Example of a transitive relation that is not reflexive or symmetric:
Let's define a relation R1 on T as follows:
R1 = {(a, b), (b, c), (a, c), (c, d)}
Explanation:
- R1 is not reflexive because it does not contain the pairs (a, a), (b, b), (c, c), and (d, d). Reflexivity means that every element in T should be related to itself.
- R1 is not symmetric because it contains the pair (a, b) but does not contain (b, a). Symmetry means that for every (x, y) in R1, (y, x) should also be in R1.
- R1 is transitive because for any two pairs (x, y) and (y, z) in R1, if x is related to y and y is related to z, then x is related to z. For example, (a, b) and (b, c) are in R1, and since a is related to b and b is related to c, we can conclude that a is also related to c.
b. Example of a relation that is transitive, symmetric, but not reflexive:
Let's define a relation R2 on T as follows:
R2 = {(a, a), (b, b), (c, c), (d, d), (a, b), (b, a)}
Explanation:
- R2 is not reflexive because it does not contain all the pairs (x, x) for every x in T. In this case, it's missing (c, c) and (d, d).
- R2 is symmetric because for every pair (x, y) in R2, (y, x) is also in R2. For example, (a, b) is in R2, and (b, a) is also present in R2. Symmetry means that the relation is bidirectional.
- R2 is transitive because every pair (x, y) and (y, z) in R2, if x is related to y and y is related to z, then x is related to z. For example, (a, b) and (b, a) are in R2, and since a is related to b and b is related to a, we can conclude that a is also related to a.