Let R be the relation on ℤ+×ℤ+ defined by (a,b)R(c,d) if and only if a−2d=c−2b.
(a) prove that R is an equivalence relation
(b) list all elements of the equivalence class [(3,3)]
(c) find an equivalence class that has exactly 271 elements.
(d) is it true that for every positive integer n, there is an equivalence class that contains exactly n elements? explain
(a)
An equivalence relation must satisfy the following three requirements:
1. reflexivity
(a,b)R(c,d) is reflexive if
(a,b)R(a,b)is defined.
Since a-2d=c-2b ≡ a+2b=c+2d
so (a,b)R(a,b) is defined since a+2b=a+2b
Therefore R is reflexive.
2. symmetry
R is symmetric if
(a,b)R(c,d) <=> (c,d)R(a,b)
Since by definition,
a-2d=c-2b <=> c-2b=a-2d
R is symmetric
3. Transistivity
R is transitive if
(a,b)R(c,d) and (c,d)R(e,f) <=>(a,b)R(e,f)
We know that
a-2d=c-2b, and
c-2f=e-2d
Add preceding equations,
a-2d+c-2f=c-2b+e-2d
Simplify
a-2f=e=2b
<=>
(a,b)R(e,f)
Therefore R is transitive.
Since R satisfies all three requirements, R is an equivalence relation.
(b)
Elements of the equivalence class [(3,3)] consists of all elements (c,d) that satisfy (3,3)R(c,d), from which we can name the following set:
{(1,4),(3,3),(5,2),(7,1)}
(c) hints:
show that the number of elements of an equivalent class (a,b) has exactly
floor((a+2b-1)/2) elements. Floor(x) function return the greatest integer not exceeding x. Use mathematical induction if necessary.
Hence find the equivalent class that has 271 elements.
(d) hints:
Deduce from (c).
(a) To prove that R is an equivalence relation, we need to show that it satisfies three properties: reflexivity, symmetry, and transitivity.
1. Reflexivity: For any (a, b) in ℤ+ × ℤ+, we need to show that (a, b) R (a, b). According to the definition of R, this means a - 2b = a - 2b, which is true for any integers a and b. Therefore, R is reflexive.
2. Symmetry: We need to show that if (a, b) R (c, d), then (c, d) R (a, b). Suppose (a, b) R (c, d), which means a - 2d = c - 2b. We want to show that c - 2b = a - 2d. Rearranging the terms, we have a - c = 2b - 2d, which can be expressed as (c, d) R (a, b). Thus, R is symmetric.
3. Transitivity: We need to show that if (a, b) R (c, d) and (c, d) R (e, f), then (a, b) R (e, f). Suppose (a, b) R (c, d) and (c, d) R (e, f), which means a - 2d = c - 2b and c - 2f = e - 2d. Adding these two equations together, we get a - 2f = e - 2b. This implies (a, b) R (e, f), thus proving transitivity.
Since R satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation.
(b) To list all elements of the equivalence class [(3,3)], we need to find all pairs (a, b) such that (a, b) R (3, 3). According to the definition of R, this means a - 2(3) = 3 - 2b. Simplifying, we have a - 6 = 3 - 2b. Rearranging terms, we get a - 3 = 2 - 2b, or a - 3 = 2(1 - b). This implies that a - 3 must be an even number. So, all possible pairs (a, b) in the equivalence class [(3,3)] are:
(2, 2), (4, 1), (6, 0), (8, -1), (10, -2), ...
(c) To find an equivalence class that has exactly 271 elements, we need to find a pair (a, b) such that the number of solutions to the equation a - 2d = c - 2b is 271. The number of solutions to this equation can be found by considering the range of values for a and b.
Assuming positive integers for a and b, the range of a is from 1 to ∞ and the range of b is from 1 to ∞. For each value of a, we have a corresponding value of b that satisfies the equation. Therefore, the number of solutions is infinite, so there is no equivalence class that has exactly 271 elements.
(d) It is not true that for every positive integer n, there is an equivalence class that contains exactly n elements. As shown in part (c), the number of solutions to the equation a - 2d = c - 2b is infinite. Therefore, there are infinitely many elements in any non-trivial equivalence class.
(a) To prove that R is an equivalence relation, we need to show that it satisfies three properties: reflexivity, symmetry, and transitivity.
1. Reflexivity: For any (a, b) ∈ ℤ+ × ℤ+, we need to show that (a, b) R (a, b). To prove this, we observe that a - 2b = a - 2b, so the relation holds for every element (a, b) in the set. Therefore, R is reflexive.
2. Symmetry: For any (a, b) and (c, d) ∈ ℤ+ × ℤ+, if (a, b) R (c, d), then we need to show that (c, d) R (a, b). We have that a - 2d = c - 2b, and by rearranging the equation, we can obtain c - 2b = a - 2d. Thus, the relation is symmetric.
3. Transitivity: For any (a, b), (c, d), and (e, f) ∈ ℤ+ × ℤ+, if (a, b) R (c, d) and (c, d) R (e, f), then we need to show that (a, b) R (e, f). Assume that a - 2d = c - 2b and c - 2f = e - 2d. By adding these two equations, we have (a - 2d) + (c - 2f) = (c - 2b) + (e - 2d), simplifying to a - 2f = e - 2b. Therefore, (a, b) R (e, f), and the relation is transitive.
Since R satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation.
(b) To find the equivalence class [(3, 3)] for this relation, we need to determine all elements (a, b) such that (a, b) R (3, 3), i.e., a - 2(3) = 3 - 2(3) = 3 - 6 = -3. Solving this equation for a, we find that a = -3 + 2(3) = -3 + 6 = 3. Therefore, the equivalence class [(3, 3)] consists of all elements (3, b) for any positive integer b.
(c) To find an equivalence class with exactly 271 elements, we can consider the equation a - 2d = c - 2b, where we want the equivalence class to have 271 elements. We can set one parameter to any positive integer, say b = 1. Then, we can solve for a and c in terms of d:
a - 2d = c - 2(1)
a - c = 2d - 2
We need to find values of a and c that satisfy this equation and have a difference of exactly 2d - 2. We can choose a = 2d - 2 + 1 and c = 2d - 2 - 1, which gives a difference of 2d - 2 - (2d - 2) = 0. Therefore, for every positive integer d, the equivalence class [(2d - 2 + 1, 1)] has exactly 271 elements.
(d) It is true that for every positive integer n, there is an equivalence class that contains exactly n elements. We can choose any positive integer value for b and solve for a and c using the equation a - 2d = c - 2b. Suppose we set b = 1, then:
a - 2d = c - 2(1)
a - c = 2d - 2
By choosing a = 2d - 2 + 1 and c = 2d - 2 - 1, as we did in part (c), we can see that the difference between a and c is always 2. Therefore, by varying the values of d, we can create an equivalence class with any desired number of elements, including n for any positive integer n.