Let A be the set of all ordered pairs of positive integers and R be the relation defined on A where (a,b)R(c,d) means that b-a=d-c.
a)Show that R is an equivalence relation.
b)Find [(3, 5)] and [(7, 1)].
For R to be an equivalence relation, it must satisfy all the following conditions:
1) Reflexivity, i.e. (a,b)R(a,b) ∀(a,b).
2) Symmetry, i.e.
(a,b)R(c,d) => (c,d)R(a,b)
3) Transistivity, i.e.
(a,b)R(c,d) and (c,d)R(e,f) => (a,b)R(e,f).
Use the definition of R to show that all three conditions are satisfied and hence R is an equivalence relation.
To show that relation R is an equivalence relation, we need to prove three properties: reflexivity, symmetry, and transitivity.
a) Reflexivity:
To show that R is reflexive, we need to demonstrate that for all elements (a, b) in A, (a, b) R (a, b).
In this case, we have (a, b) R (a, b) if and only if b - a = b - a, which is true for any pair of positive integers. Therefore, R is reflexive.
b) Symmetry:
To show that R is symmetric, we need to prove that for any elements (a, b) and (c, d) in A, if (a, b) R (c, d), then (c, d) R (a, b).
Let's assume (a, b) R (c, d), which means b - a = d - c.
Rearranging this equation, we have: d - c = b - a.
This implies that (c, d) R (a, b). Therefore, R is symmetric.
c) Transitivity:
To show that R is transitive, we need to prove that for any elements (a, b), (c, d), and (e, f) in A, if (a, b) R (c, d) and (c, d) R (e, f), then (a, b) R (e, f).
Assume (a, b) R (c, d), which implies b - a = d - c.
Similarly, assume (c, d) R (e, f), which gives us d - c = f - e.
Substituting d - c with f - e in the first equation, we get b - a = f - e.
Therefore, (a, b) R (e, f). Thus, R is transitive.
Since R satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation.
b) To find [(3, 5)], we need to determine the equivalence class of (3, 5) under the relation R. The equivalence class [(3, 5)] is the set of all ordered pairs (a, b) that are related to (3, 5) by R, meaning b - a = 5 - 3.
Let's solve this equation: b - a = 5 - 3.
This simplifies to: b - a = 2.
To find all possible solutions, we can assign a value to a and calculate b using the equation b = a + 2.
For example, if we choose a = 1, then b = 1 + 2 = 3. Therefore, (1, 3) is an element of [(3, 5)].
Similarly, if we assign a = 2, then b = 2 + 2 = 4. Thus, (2, 4) is another element of [(3, 5)].
Similarly, (3, 5) itself is also an element of [(3, 5)].
Therefore, [(3, 5)] = {(1, 3), (2, 4), (3, 5)}.
To find [(7, 1)], we follow the same process.
We solve the equation b - a = 1 - 7, which simplifies to b - a = -6.
This implies b = a - 6.
By assigning different values to a, we can find all possible elements of [(7, 1)].
For example, if we choose a = 8, then b = 8 - 6 = 2. Thus, (8, 2) is an element of [(7, 1)].
Similarly, (7, 1) itself is also an element of [(7, 1)].
Therefore, [(7, 1)] = {(8, 2), (7, 1)}.