1. Let f: X –→ Y and define x~y if f (x) =f(y). Show that - is an equivalence relation on X. Definition: For a set X, we say it is an equivalence relation if: a) Reflexivity: x~x for all x E X. b) Symmetry: if x~y, then y~x c) Transitivity: if x~y and y~z, then x~z. Proof: a) Reflexive: f (x) = f(x) for all x E X. Therefore, x~x for all x E X. b) Symmetric: Let x~y Then, f(x) = f(y) Then, f(y) = f(x) Therefore: y~x. c) Transitive: Let x~y and y~z Then, f(x) = f(y) and f(y) = f(z) Then, f(x) = f(z) %3D Therefore: x~z. In conclusion, ~ on set X is an equivalence relation because all three properties were tisfied to represent an equivalence relation on X.
1. Let f: X –→ Y and define x~y if f (x) =f(y). Show that - is an equivalence relation on X. Definition: For a set X, we say it is an equivalence relation if: a) Reflexivity: x~x for all x E X. b) Symmetry: if x~y, then y~x c) Transitivity: if x~y and y~z, then x~z. Proof: a) Reflexive: f (x) = f(x) for all x E X. Therefore, x~x for all x E X. b) Symmetric: Let x~y Then, f(x) = f(y) Then, f(y) = f(x) Therefore: y~x. c) Transitive: Let x~y and y~z Then, f(x) = f(y) and f(y) = f(z) Then, f(x) = f(z) %3D Therefore: x~z. In conclusion, ~ on set X is an equivalence relation because all three properties were tisfied to represent an equivalence relation on X.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can you please look over my proof and make suggestions/corrections?
Thank you,
Transcribed Image Text:1. Let f: X → Y and define x~y if f(x) = f(y). Show that - is an equivalence relation
on X.
Definition: For a set X, we say it is an equivalence relation if:
a) Reflexivity: x~x for all x E X.
b) Symmetry: if x~y, then y~x
c) Transitivity: if x~y and y~z, then x~z.
Proof:
a) Reflexive: f (x) = f (x) for all x E X.
Therefore, x~x for all x E X.
b) Symmetric: Let x~y
Then, f(x) = f(y)
Then, f (y) = f(x)
Therefore: y~x.
c) Transitive: Let x~y and y~z
Then, f(x) = f (y) and f(y) = f(z)
Then, f(x) = f(z)
Therefore: x~z.
In conclusion,
on set X is an equivalence relation because all three properties were
satisfied to represent an equivalence relation on X.
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