Proving "[a]_n ∩ [b]_n is Empty or Equals [b]_n

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Homework Help Overview

The discussion revolves around proving a statement related to the intersection of two sets defined by modular arithmetic, specifically [a]_{n} and [b]_{n}. The original poster is attempting to establish conditions under which the intersection is either empty or the two sets are equal.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to assume the existence of an element in the intersection of the two sets and to derive implications from that assumption. There are questions about the relationships between the elements and their modular properties, particularly concerning the implications of the equations derived from the assumptions.

Discussion Status

Participants are exploring the implications of their assumptions and questioning the relationships between the elements involved. There is a lack of consensus on the next steps, with some participants expressing uncertainty about how to proceed with the reasoning related to the differences between a and b.

Contextual Notes

There is a focus on the modular properties of the elements and the implications of their relationships, with some participants noting potential confusion regarding the assumptions made in the problem statement.

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Homework Statement


Prove that either [tex][a]_{n}[/tex][tex]\cap[/tex][tex]<b>_{n}</b>[/tex]=empty set or [tex][a]_{n}[/tex]=[tex]<b>_{n}</b>[/tex].

Homework Equations


The Attempt at a Solution


I want to assume there is an element x in [tex][a]_{n}[/tex][tex]\cap[/tex][tex]<b>_{n}</b>[/tex] and show this implies [tex][a]_{n}[/tex]=[tex]<b>_{n}</b>[/tex].
This tells me x is in [tex][a]_{n}[/tex] and [tex]<b>_{n}</b>[/tex].
That's where I get stuck.
 
Last edited:
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sorry, I meant to show this implies [a]=.
 
If x=a mod n then n|x-a. x=b mod n means n|x-b.

If x|x-a and n|x-b... what can you say about a-b?
 
x-a=x-b
x-x=a-b
0=a-b
b=a
 
x-a is probably not equal to x-b
 
Ok then I'm not really sure where to go with a-b then.
 

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