Let C be the circle relation defined on the set of real numbers. For every x, y ER, XCyx² + y² = 1. (a) Is C reflexive? Justify your answer. C is reflexive for every real number x, x C x. By definition of C this means that for every real number x, show this is the case. (x, x² + x²) = Since this does not equal✔ 1, C is not (b) Is C symmetric? Justify your answer. C is symmetric for all real numbers x and y, if x C y theny y² + x² = | This is true ✓ ✔ IS ✔ symmetric. (c) Is C transitive? C is transitive x² +2²✔✔ (x, y, z) = reflexive. Then x² + y²✔ ], y² +2²=✔ X = 1. This is false and x² + 2✔✔✔✔ 1. Thus, C is not ✓ ✓ CX✔✔ By definition of C, this means that for all real numbers x and y, if x² + y2 = | because, by the commutative property of addition, x2 + y² =✔✔✔✔✔ + x² for all Justify your answer. for all real numbers x, y, and z, if x C y and y C z then x C z. By definition of C this means that for all real numbers x, y, and z, if x2 + y2 = 1 and y2 + 2✔✔✔ = 1 then = 1. This is false . For example, let x, y, and z be the following numbers entered as a comma-separated list. ✓ ✓ Find an example x and x2 + x2 that ✔✔✔ transitive. then real numbers x and y. Thus, C ►
Let C be the circle relation defined on the set of real numbers. For every x, y ER, XCyx² + y² = 1. (a) Is C reflexive? Justify your answer. C is reflexive for every real number x, x C x. By definition of C this means that for every real number x, show this is the case. (x, x² + x²) = Since this does not equal✔ 1, C is not (b) Is C symmetric? Justify your answer. C is symmetric for all real numbers x and y, if x C y theny y² + x² = | This is true ✓ ✔ IS ✔ symmetric. (c) Is C transitive? C is transitive x² +2²✔✔ (x, y, z) = reflexive. Then x² + y²✔ ], y² +2²=✔ X = 1. This is false and x² + 2✔✔✔✔ 1. Thus, C is not ✓ ✓ CX✔✔ By definition of C, this means that for all real numbers x and y, if x² + y2 = | because, by the commutative property of addition, x2 + y² =✔✔✔✔✔ + x² for all Justify your answer. for all real numbers x, y, and z, if x C y and y C z then x C z. By definition of C this means that for all real numbers x, y, and z, if x2 + y2 = 1 and y2 + 2✔✔✔ = 1 then = 1. This is false . For example, let x, y, and z be the following numbers entered as a comma-separated list. ✓ ✓ Find an example x and x2 + x2 that ✔✔✔ transitive. then real numbers x and y. Thus, C ►
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Circle Relation on Real Numbers**
Let \( C \) be the circle relation defined on the set of real numbers.
For every \( x, y \in \mathbb{R} \), \( x \, C \, y \iff x^2 + y^2 = 1 \).
### (a) Is \( C \) reflexive? Justify your answer.
\( C \) is reflexive if and only if for every real number \( x, \, x \, C \, x \). By definition of \( C \), this means for every real number \( x \), \( x^2 + x^2 = 1 \).
\[
(x, x^2 + x^2) = (x, 2x^2)
\]
Since this **does not equal** 1, \( C \) is **not** reflexive.
### (b) Is \( C \) symmetric? Justify your answer.
\( C \) is symmetric if and only if for all real numbers \( x \) and \( y \), if \( x \, C \, y \) then \( y \, C \, x \).
By definition of \( C \), this means that for all real numbers \( x \) and \( y \), if \( x^2 + y^2 = 1 \) then \( y^2 + x^2 = 1 \).
This is **true** because, by the commutative property of addition, \( x^2 + y^2 = y^2 + x^2 \) for all real numbers \( x \) and \( y \). Thus, \( C \) is **symmetric**.
### (c) Is \( C \) transitive? Justify your answer.
\( C \) is transitive if and only if for all real numbers \( x, y, \) and \( z \), if \( x \, C \, y \) and \( y \, C \, z \) then \( x \, C \, z \).
By definition of \( C \), this means that for all real numbers \( x, y, \) and \( z \), if \( x^2 + y^2 = 1 \) and \( y^2 + z^2 = 1 \) then \( x^2 + z^2 = 1 \).
This](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa0e38307-1ade-44bc-b712-aaeda4c58098%2Fb34491fe-2550-4d0f-acc3-03abfcdcfc46%2Fzc8y92_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Circle Relation on Real Numbers**
Let \( C \) be the circle relation defined on the set of real numbers.
For every \( x, y \in \mathbb{R} \), \( x \, C \, y \iff x^2 + y^2 = 1 \).
### (a) Is \( C \) reflexive? Justify your answer.
\( C \) is reflexive if and only if for every real number \( x, \, x \, C \, x \). By definition of \( C \), this means for every real number \( x \), \( x^2 + x^2 = 1 \).
\[
(x, x^2 + x^2) = (x, 2x^2)
\]
Since this **does not equal** 1, \( C \) is **not** reflexive.
### (b) Is \( C \) symmetric? Justify your answer.
\( C \) is symmetric if and only if for all real numbers \( x \) and \( y \), if \( x \, C \, y \) then \( y \, C \, x \).
By definition of \( C \), this means that for all real numbers \( x \) and \( y \), if \( x^2 + y^2 = 1 \) then \( y^2 + x^2 = 1 \).
This is **true** because, by the commutative property of addition, \( x^2 + y^2 = y^2 + x^2 \) for all real numbers \( x \) and \( y \). Thus, \( C \) is **symmetric**.
### (c) Is \( C \) transitive? Justify your answer.
\( C \) is transitive if and only if for all real numbers \( x, y, \) and \( z \), if \( x \, C \, y \) and \( y \, C \, z \) then \( x \, C \, z \).
By definition of \( C \), this means that for all real numbers \( x, y, \) and \( z \), if \( x^2 + y^2 = 1 \) and \( y^2 + z^2 = 1 \) then \( x^2 + z^2 = 1 \).
This
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