Consider the relation R on the set of real numbers defined by R = {(x, y) | x^2

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Question: Consider the relation R on the set of real numbers defined by R = {(x, y) | x^2 + y^2 = 1}. What type of relation is R?

Options:

Correct Answer: Symmetric

Solution:

R is symmetric because if (x,y) is in R, then (y,x) is also in R. It is not reflexive or transitive.

Consider the relation R on the set of real numbers defined by R = {(x, y) | x^2 + y^2 = 1}. What type of relation is R?

Consider the relation R on the set of real numbers defined by R = {(x, y) | x^2 + y^2 = 1}. What type of relation is R?

Step 1: Understand the relation R. It consists of pairs (x, y) where the sum of the squares of x and y equals 1. This describes a circle with a radius of 1 centered at the origin (0, 0).
Step 2: Check if R is symmetric. A relation is symmetric if whenever (x, y) is in R, then (y, x) is also in R. For example, if (0.6, 0.8) is in R (since 0.6^2 + 0.8^2 = 1), then (0.8, 0.6) is also in R (since 0.8^2 + 0.6^2 = 1). Therefore, R is symmetric.
Step 3: Check if R is reflexive. A relation is reflexive if every element is related to itself, meaning (x, x) must be in R for all x. However, for (x, x) to be in R, x^2 + x^2 = 1 must hold, which is not true for all real numbers. Therefore, R is not reflexive.
Step 4: Check if R is transitive. A relation is transitive if whenever (x, y) is in R and (y, z) is in R, then (x, z) must also be in R. For example, if (0.6, 0.8) and (0.8, -0.6) are in R, (0.6, -0.6) is not in R because 0.6^2 + (-0.6)^2 does not equal 1. Therefore, R is not transitive.

Symmetric Relation – A relation R is symmetric if for every (x, y) in R, (y, x) is also in R.
Reflexive Relation – A relation R is reflexive if for every x in the set, (x, x) is in R.
Transitive Relation – A relation R is transitive if whenever (x, y) and (y, z) are in R, then (x, z) is also in R.