Suppose R is the relation on N where aRb means that a ends in the same digit in which b ends. Determine whether R is an equivalence relation on N. And,
Suppose that R and S are equivalence relations on a set A. Prove that the R �¿ S is also an equivalence relation on A.
To determine whether a relation is an equivalence relation, we need to check if it satisfies three conditions: reflexivity, symmetry, and transitivity. Let's go through each condition for the relation R on N.
1. Reflexivity: For R to be reflexive, every element a in N should be related to itself. In this case, we need to check if every natural number ends with the same digit as itself.
Since every natural number ends in a certain digit, it is clear that every number is related to itself. For example, 42 ends in 2, and 42 is related to itself. Therefore, R is reflexive.
2. Symmetry: For R to be symmetric, if a is related to b, then b should also be related to a. In this case, if a ends in the same digit as b, then b should also end in the same digit as a.
Let's consider two natural numbers, a and b, such that aRb (a ends in the same digit as b). By the definition of R, this means a and b have the same last digit. Therefore, b and a will also have the same last digit. So, if aRb, then bRa. R is symmetric.
3. Transitivity: For R to be transitive, if a is related to b and b is related to c, then a should also be related to c. In this case, if a ends in the same digit as b, and b ends in the same digit as c, then a should also end in the same digit as c.
Let's consider three natural numbers, a, b, and c, such that aRb (a ends in the same digit as b) and bRc (b ends in the same digit as c). By the definition of R, this means a, b, and c all have the same last digit. Therefore, a also ends in the same digit as c. So, if aRb and bRc, then aRc. R is transitive.
Since R satisfies all three conditions (reflexivity, symmetry, and transitivity), we can conclude that R is an equivalence relation on N.
Moving on to the second part of your question:
To prove that the composition of two equivalence relations, R and S, denoted as R �¿ S, is also an equivalence relation on a set A, we need to show that it satisfies the same three conditions.
1. Reflexivity: To prove reflexivity for R �¿ S, we need to show that every element a in A is related to itself under R �¿ S. Since R and S are equivalence relations, we know that each element a is related to itself under both R and S. Therefore, a is related to itself under their composition R �¿ S. R �¿ S is reflexive.
2. Symmetry: To prove symmetry for R �¿ S, we need to show that if a is related to b under R �¿ S, then b is related to a as well. Let's assume a is related to b under R �¿ S. This means that there exists an element c in A such that (a, c) is in R and (c, b) is in S. Since R and S are equivalence relations, we know that both (c, a) and (b, c) are also in R and S, respectively. Therefore, b is related to a under R �¿ S. R �¿ S is symmetric.
3. Transitivity: To prove transitivity for R �¿ S, we need to show that if a is related to b and b is related to c under R �¿ S, then a is related to c as well. Let's assume a is related to b and b is related to c under R �¿ S. This means that there exist elements x and y in A such that (a, x) is in R, (x, b) is in S, (b, y) is in R, and (y, c) is in S. Since R and S are equivalence relations, we know that (x, a) and (c, y) are also in R and S, respectively. Therefore, a is related to c under R �¿ S. R �¿ S is transitive.
Since R �¿ S satisfies all three conditions (reflexivity, symmetry, and transitivity), we can conclude that R �¿ S is also an equivalence relation on A.