Chapter 5: Equivalence Relations Do The Following Pro

Chapter 5 Equivalence Relationsdo The Following Pro

Chapter 5: Equivalence Relations Do the following problems involving equivalence relations, each is worth 8 points. Note that "iff" stands for "if and only if". Also, to show "whether or not" means to give a proof for it if it is true and to give a counter example if it is false. Finally a R b means "a is related to b", i.e. (a,b)∈R. Let R be the relation on socks in your drawer defined by the following property: a R b iff a and b have the same number of stripes. Show whether or not R is reflexive.

The relation R on socks is defined based on the number of stripes, which introduces an equivalence relation because it partitions the socks into classes characterized by the quantity of stripes. To verify whether R is reflexive, we examine whether every sock is related to itself. Since any sock has a number of stripes that is equal to itself, for any sock a, a R a because a and a definitely have the same number of stripes. Therefore, R is reflexive.

Next, we assess if R is symmetric. Suppose a R b, meaning sock a and sock b have the same number of stripes. Since having the same number of stripes is a symmetric property, this implies b R a. Therefore, R is symmetric.

Lastly, we consider transitivity. Assume a R b and b R c, i.e., socks a and b share the same number of stripes, and socks b and c do as well. Since both a and c have the same number of stripes as b, it follows that a and c have the same number of stripes. Thus, a R c, and R is transitive.

In conclusion, R is an equivalence relation because it is reflexive, symmetric, and transitive. The equivalence classes under R are determined by the number of stripes; each class contains socks sharing the same number of stripes. The number of such classes depends on the maximum number of stripes in your drawer; if, for example, the maximum is n, then there are n+1 classes (assuming zero stripes as a class). If the number of stripes can be arbitrarily large or unknown, the classes are countably infinite.

On the relation R defined on natural numbers by a R b iff a+b is even

This relation groups natural numbers based on the parity of their sum. To verify reflexivity, take any natural number a. Since a+a=2a, which is even, a R a holds for all a, confirming that R is reflexive.

Symmetric property asserts that if a R b, then b R a. Given that a+b is even, then by commutativity, b+a is also even, so b R a. Hence, R is symmetric.

For transitivity, assume a R b and b R c. That is, a+b and b+c are both even. When the sum of two numbers is even, the parity of the addends must be the same. Since a+b is even, a and b share the same parity; similarly, b and c share the same parity. Therefore, a and c must also share the same parity, making a+c even, and thus a R c. Therefore, R is transitive.

As R is reflexive, symmetric, and transitive, it is an equivalence relation. The equivalence classes correspond to the parity of the natural numbers: one class with even numbers and another with odd numbers, making two classes in total.

On relation R defined by f(a)=f(b), where f(n) is the number of 1's in the binary representation of n

Reflexivity holds since for any number a, f(a)=f(a), so a R a.

Symmetry holds because if f(a)=f(b), then f(b)=f(a), so b R a.

Transitivity is also satisfied since if f(a)=f(b) and f(b)=f(c), then f(a)=f(c), hence a R c.

This relation R partitions the natural numbers into classes based on the number of 1's in their binary representation. Since the number of 1's can be zero or any non-negative integer, there are countably infinite equivalence classes, each characterized by a specific number of 1's.

On functions f,g mapping natural numbers to natural numbers, with the relation defined by: f R g iff for all n, f(n) ≤ g(n)

Reflexivity is evident because for any function f, f(n) ≤ f(n) for all n, so f R f.

Symmetry fails because if f(n) ≤ g(n) for all n, it does not necessarily mean g(n) ≤ f(n). Therefore, f R g does not imply g R f in general.

Transitivity holds: if f R g and g R h, then f(n) ≤ g(n) and g(n) ≤ h(n), thus f(n) ≤ h(n), so f R h.

Since the relation is reflexive and transitive but not symmetric, it is a partial order, not an equivalence relation.

On relation R on natural numbers defined by a R b iff a % 7 = b % 7

Reflexivity: For any a, a % 7 = a % 7, so a R a.

Symmetry: If a R b, then a % 7 = b % 7, which implies b % 7 = a % 7, so b R a.

Transitivity: If a R b and b R c, then a % 7 = b % 7 and b % 7 = c % 7, hence a % 7 = c % 7, so a R c.

Therefore, R is an equivalence relation. The equivalence classes are determined by the remainders when divided by 7. There are exactly 7 classes, corresponding to remainders 0 through 6.

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