Step 1 of 1 A An equivalence relation R on a set S is one that satisfies these three properties for x, y,zeS. 1. (Reflexive) xRx. 2. (Symmetric). If xRy, then yRx. 3. (Transitive). If xRy and yRz, then xRz. Note that 120, and 021. So, in the relation, xRy in R if r2y, one is related to zero, but zero is not related to one; that is, the symmetry check fails. Therefore, the relation, xRy in R if x> y, is not an equivalence relation.
Step 1 of 1 A An equivalence relation R on a set S is one that satisfies these three properties for x, y,zeS. 1. (Reflexive) xRx. 2. (Symmetric). If xRy, then yRx. 3. (Transitive). If xRy and yRz, then xRz. Note that 120, and 021. So, in the relation, xRy in R if r2y, one is related to zero, but zero is not related to one; that is, the symmetry check fails. Therefore, the relation, xRy in R if x> y, is not an equivalence relation.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
I am confused about the concept. There are two solutions. I do not know which one makes sense. Please with explanation. Is it correct if I I under in this way:For equivalence relation,when checking three rules,does all elements should satisfy these, for example, the only way that x>=y and y>=x is they are equal,since x and y arr in R, they might not be equal, like when x=1 , y=0,the symmetric rule does not holds. When we check when a certain rule holds , does it hold if one case (x=y) can satisfy or it only holds only when for all random x and y which satisfy x >=y. Please also explains the partition in this situation, I do not know this concept well, what does it mean “ the partition arousing from each equivalence relations”

Transcribed Image Text:Step-by-step solution
Step 1 of 1
An equivalence relation R on a set S is one that satisfies these three properties for
x, y,ze S.
1. (Reflexive) xRx.
2. (Symmetric). If xRy, then yRx.
3. (Transitive). If xRy and yRz, then xRz.
Note that 120, and 021. So, in the relation, xRy in R if x2y, one is related to zero, but
zero is not related to one; that is, the symmetry check fails. Therefore, the relation, xRy in R if
x> y, is not an equivalence relation.
Comment
Was this solution helpful?
MacBook Pro
->
Search or type URL
@
#3
¥ $
2
3
4

Transcribed Image Text:Q30: Detelmie wheth a the given Yekation is cn equéralene belatin 2n the
Describe the
partition arising tom Pad equivdence relation.
re flative: xRY.
インィの
微
记
SYOmetric : xRY YR X.
→y>xの
Tewsitie: YRy, ソRマー→ オー
イラ Ond yフヌ
→ の
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images

Recommended textbooks for you

Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated

Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education

Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY

Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated

Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education

Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY

Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,

