Proving transitivity, stuck at at algebra

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Homework Help Overview

The discussion revolves around the properties of a relation defined on integers, specifically focusing on proving symmetry and transitivity. The original poster is exploring the implications of the relation defined by the condition 5|2a+3b.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to manipulate the equation 2a + 3b = 5m to explore its implications for symmetry. Some participants question the introduction of a third variable, c, in the context of transitivity, while others suggest focusing on symmetry instead.

Discussion Status

Participants are actively engaging with the problem, offering hints and exploring different approaches. There is a recognition of the need to clarify the definitions and properties being examined, particularly regarding symmetry and transitivity.

Contextual Notes

There appears to be some confusion regarding the properties being proven, with the original poster initially intending to show transitivity but later clarifying a focus on symmetry. The discussion reflects an ongoing exploration of algebraic manipulation and its implications for the defined relation.

barylwires
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1. aRb on Z if 5|2a+3b
2. Since 5|2a+3b, 2a+3b=5m, so 2a=5m-3b.
3. Consider 3a+2b=2a+2b+a=5m-3b+2b+a=5m-b+a. I can't show that 5 divides that, but there is another way to wrangle that into a better form. I need help seeing how that works. Thanks.
 
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For transitivity you are given aRb and bRc and need to show aRc. You haven't even introduced c yet, nor written in terms of a and c what you have to show.
 


Sorry. I'm trying to show symmetry, not transitivity.
 
Hint: What happens if you subtract 5a and 5b from your equation 2a + 3b = 5m?
 
Good things happen. I think I can write: Assume aRb. Then 5|2a+3b iff 5|-(2a+3b) iff 5|-2a-3b iff 5|-3b-2a+5b+5a iff 5|2b+3a. Now aRb implies bRa, thus R is symmetrical.

I'll sleep on that. Many thanks!
 

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