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.

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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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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”
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.
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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
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
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
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