Equivalence Relations in Mathematics: Understanding Transitivity and Symmetry

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SUMMARY

This discussion focuses on the properties of equivalence relations in mathematics, specifically transitivity and symmetry. The relation R is defined by the equation a/b = c/d, which is reflexive, symmetric, and transitive. The participants clarify that for R to be symmetric, the condition (a,b)R(c,d) must imply (c,d)R(a,b), and they highlight the importance of understanding equivalence classes, such as (1,2)R(2,4)R(3,6). The conversation emphasizes the necessity of demonstrating these properties through specific examples and definitions.

PREREQUISITES
  • Understanding of equivalence relations in mathematics
  • Familiarity with the concepts of reflexivity, symmetry, and transitivity
  • Basic knowledge of fractions and their equivalences
  • Ability to work with ordered pairs in mathematical contexts
NEXT STEPS
  • Study the formal definitions of equivalence relations in set theory
  • Learn about equivalence classes and their applications in mathematics
  • Explore examples of equivalence relations beyond fractions, such as modular arithmetic
  • Investigate the implications of equivalence relations in algebraic structures
USEFUL FOR

Students studying mathematics, particularly those focusing on abstract algebra and set theory, as well as educators teaching these concepts.

Kingyou123
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Homework Statement


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Homework Equations


a/b=c/d

The Attempt at a Solution


I started out with transitivity and I figure it's transitive because a/b=e/f = af=be=(a,b)R(e,f).
It can't be symmetric because ad does not equal bc, 1*4 does not equal 2*3
I'm I correct or completely wrong?
 
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Wrong. What does equivalence mean? Can you define all properties? (first in general)
 
You have to show that the relation R is:

1) Reflexive: (a,b)R(a,b)
2) Symmetric: (a,b)R(c,d) iff (c,d)R(a,b)
3) Transitive: If (a,b)R(c,d) and (c,d)R(e,f), then (a,b)R(e,f)

Can you work the rest out from here?
 
Kingyou123 said:

Homework Statement


View attachment 96365

Homework Equations


a/b=c/d

The Attempt at a Solution


I started out with transitivity and I figure it's transitive because a/b=e/f = af=be=(a,b)R(e,f).
Note that in the problem statement it says ##ad = bc \Leftrightarrow \frac a b = \frac c d##. IOW, these two equations are equivalent. You should not have the 2nd and 4th "equals" there.
Kingyou123 said:
It can't be symmetric because ad does not equal bc, 1*4 does not equal 2*3
You're missing the point. R would be symmetric if (a, b) R (c, d) implies that (c, d) R (a, b).

Think about what the relation as defined means, relative to the set of numbers in the problem. Do you understand why (1, 3) R (2, 6)?
Kingyou123 said:
I'm I correct or completely wrong?
 
Last edited:
TeethWhitener said:
You have to show that the relation R is:

1) Reflexive: (a,b)R(a,b)
2) Symmetric: (a,b)R(c,d) iff (c,d)R(a,b)
3) Transitive: If (a,b)R(c,d) and (c,d)R(e,f), then (a,b)R(e,f)

Can you work the rest out from here?
Symmetric= if (a,b)R(C,d) then (c,d)R(a,b) since ad=bc
Reflexive =since a/b=a/b
My confusion was with that I thought it had to be for all x values.
 
Kingyou123 said:
My confusion was with that I thought it had to be for all x values
Nope, all pairs of values.
 
TeethWhitener said:
Nope, all pairs of values.
Could you explain part b?
 
Kingyou123 said:
Could you explain part b?
For example 1/2 = 2/4 = 3/6, so (1,2)R(2,4) and (2,4)R(3,6). These three pairs are all equivalent and build together an equivalent class. One of them represents this class. But there are more classes.
 

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