Divisibility Proof (Abstract Algebra)

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Homework Help Overview

The discussion revolves around a proof in abstract algebra concerning divisibility. The original poster is tasked with proving that if 5 divides a natural number \( a \) and 15 divides the expression \( 2ax + r \), then 5 must also divide \( r \).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the definition of divisibility and the necessary form to demonstrate that \( 5|r \). There are attempts to manipulate expressions involving \( a \) and \( r \) to connect the given conditions to the desired conclusion.

Discussion Status

Participants are actively engaging with the problem, providing hints and suggestions for manipulating the expressions. There is acknowledgment of progress in the proof, with one participant confirming that the steps taken make sense, indicating a productive direction in the discussion.

Contextual Notes

There is an emphasis on the importance of following through the proof process rather than simply arriving at the conclusion, reflecting the forum's learning-first approach.

ETuten
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Homework Statement



let a belong to N and x,r belong to Z use the definition of divisibility along with the axioms of Integers to prove that IF 5|a and 15|(2ax+r) then 5|r

Homework Equations

How do I continue the proof??



The Attempt at a Solution

So I have: let a belong to N and x,r belong to Z. Assume 5|a and 15|(2ax+r). Then there is s,t belonging to Z such that a=5s and (2ax+r)=3(5T).
 
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How does the definition apply to 5|r? Can you get the form required to show it?
 
It is an if then proof with 5|r being what you are trying to prove.
 
Use the fact that 5|a to substitute.
 
ETuten said:
It is an if then proof with 5|r being what you are trying to prove.

Yes of course but you must end up with your next to last step being the defining form of 5|r and the last step being "thus 5|r".

You almost have it. Play with what you have and what you need and see if you can connect the two. I can tell you but the point of you going through the proof is you going through the proof.
 
I now have this so
r=15t-2ax=
15t-2(5s)x=
5(3T)-5(2s)x=
5(3t-2sx)=r
since 3t-2sx belongs to Z we have 5|r

does that make sense??
 
ETuten said:
I now have this so
r=15t-2ax=
15t-2(5s)x=
5(3T)-5(2s)x=
5(3t-2sx)=r
since 3t-2sx belongs to Z we have 5|r

does that make sense??

Looks good.
 

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