Equivalence Relations in Mathematics: Understanding Transitivity and Symmetry

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Homework Help Overview

The discussion revolves around equivalence relations in mathematics, specifically focusing on the properties of transitivity and symmetry as they relate to the equation a/b = c/d. Participants are exploring the definitions and implications of these properties within the context of equivalence relations.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to determine whether the relation defined by a/b = c/d is transitive and symmetric. There are questions about the definitions of reflexivity, symmetry, and transitivity, as well as the implications of these properties for specific pairs of values.

Discussion Status

Some participants are providing guidance on the definitions of the properties of equivalence relations, while others are questioning the original poster's understanding of symmetry and transitivity. There is an ongoing exploration of how these properties apply to specific examples, and clarification is being sought regarding the nature of equivalence classes.

Contextual Notes

Participants are discussing the requirement for properties to hold for all pairs of values rather than individual values, which has led to some confusion. The original poster's understanding of the equivalence relation is being challenged, particularly regarding the conditions for symmetry.

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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