Relations

To solve the given problem, we need to determine the equivalence class of the point using the relation and are at the same distance from the origin .

1. First, we calculate the distance of the point from the origin. The distance from the origin is given by the formula: . For the point :

1. The equivalence class of a point under this relation consists of all points that are equidistant from the origin as the given point. Therefore, we need all points such that the distance from the origin is :

1. Squaring both sides, we obtain:

1. Thus, the set of all points satisfying this equation is:

Therefore, the equivalence class of the point is the set of points such that . This corresponds to the correct answer:

Step 1: Divisibility conditions
We are given two sets and . We need to find the number of elements in the relation where divides and divides .
Step 2: Divisibility for dividing
For each , there are 2 elements in that satisfy the divisibility condition.
Step 3: Divisibility for dividing
For each , there are 2 elements in that satisfy the divisibility condition.
Step 4: Total number of relations
Each element in has 2 choices for divisibility with elements in , so the total number of relations is: Thus, the number of elements in the relation is 36.

To determine if the given relation on is reflexive, symmetric, and/or transitive, we analyze each property in the context of the definition of : if and only if is divisible by 5.

1. Reflexivity:
   • A relation is reflexive if for all .
   • Substitute for : we get .
   • Since 0 is divisible by 5, the relation is reflexive.
2. Symmetry:
   • A relation is symmetric if whenever , then .
   • If is divisible by 5, is also divisible by 5 (because a number divisible by 5 is also divisible by 5).
   • Hence, the relation is symmetric.
3. Transitivity:
   • A relation is transitive if whenever and , then .
   • Let and , both divisible by 5. For transitivity, should be divisible by 5.
   • However, knowing and are divisible by 5 doesn't ensure will also be divisible by 5.
   • For example, if , then , , but , which is divisible by 5.
   • Instead, if and other choices remain 0, then it's not guaranteed that is not divisible by 5, specifically.
   • Thus, the relation may not be transitive.

Considering these observations, the correct answer is: Reflexive and symmetric but not transitive.

The correct answer is 13.
R = {(1, 2), (2, 3), (2, 4)}
for reflexive, we need to add,
(1, 1), (2, 2), (3, 3), (4, 4)
for symmetric
if (1, 2) ∈ R then (2, 1) ∈ R
if (2, 3) ∈ R then (3, 2) ∈ R
if (2, 4) ∈ R then (4, 2) ∈ R
So set becomes
{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (2, 3), (3, 2), (2, 4), (4, 2)}
for transitive As (1, 2) ∈ R (2, 3) ∈ R
then (1, 3) ∈ R then (3, 1) ∈ R (for symmetric)
& (1, 2) ∈ R (2, 4) ∈ R
then (1, 4) ∈ R
then (4, 1) ∈ R (for symmetric)
(3, 2) ∈ R (2, 4) ∈ R
then (3, 4) ∈ R then (4, 3) ∈ R (for symmetric)
so set S = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (2, 3), (3, 2), (2, 4), (4, 2), (1, 3), (3, 1), (1, 4), (4, 1), (3, 4), (4, 3)}
so 13 new elements are added
⇒ n = 13

To determine if the relation on the interval given by if and only if is an equivalence relation, we must check if it satisfies the properties of reflexivity, symmetry, and transitivity.

• Reflexivity: A relation is reflexive if every element is related to itself. Here, for , we need . We use the trigonometric identity , which simplifies to . Therefore, is reflexive.
• Symmetry: A relation is symmetric if for any elements and , whenever , then . Here, if , then for , we need . Since and , we have , thus symmetry holds.
• Transitivity: A relation is transitive if whenever and , then . Assuming and , then . Therefore, . Simplifying gives , which holds as . Hence, transitivity is satisfied.

Since satisfies reflexivity, symmetry, and transitivity, is indeed an equivalence relation.

We are given the set , and the relations . The remaining elements are: We need to determine the number of relations on the set containing at most 6 elements, which are reflexive and transitive but not symmetric.
Step 1: If the relation contains exactly 4 elements
The reflexive pairs must be included. We are required to select one additional element from the remaining ones: Only 1 way is possible: We choose the pair , which makes the relation transitive.
Thus, for 4 elements, there is exactly 1 way to form the relation.
Step 2: If the relation contains exactly 5 elements
Again, we must include . Now, we must choose 2 additional elements from the remaining pairs: The possible choices are: and , or and .
Thus, for 5 elements, there are 2 ways to form the relation.
Step 3: If the relation contains exactly 6 elements
In this case, we must include all reflexive pairs and all the other pairs: There are 3 possible ways to form a 6-element relation: 1. 2. 3.
Thus, for 6 elements, there are 3 ways to form the relation.
Final Answer: Total number of ways = .