Understanding the Addition Theorem for Modular Operations

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Discussion Overview

The discussion revolves around the Addition Theorem for modular operations, specifically the proof of the theorem which states that for three integers x, y, and d (where d > 0), the equation (x+y)%d = (x%d + y%d) %d holds. Participants are exploring the steps in the proof and clarifying the reasoning behind certain transitions in the mathematical argument.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions how the proof transitions from (q(1)d + r(1) + q(2)d + r(2))%d to (r(1) + r(2))%d, seeking clarification on the treatment of the terms involving q(1)d and q(2)d.
  • Another participant suggests that the reason for ignoring q(1)d and q(2)d is that they are zero modulo d, implying that these terms do not affect the result of the modular operation.
  • Another participant counters that since d is greater than zero, q(1)d and q(2)d should not be considered zero, indicating a potential misunderstanding of the modular arithmetic involved.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of the terms q(1)d and q(2)d in the proof. There is no consensus on whether these terms can be disregarded in the context of the modular operation.

Contextual Notes

Participants reference the definition of modular arithmetic and its implications, but there are unresolved assumptions regarding the treatment of terms involving d in the proof.

Who May Find This Useful

This discussion may be useful for students and individuals interested in modular arithmetic, particularly those seeking clarification on the Addition Theorem and its proof.

ajbiol
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Hello all,

I was wondering if someone can explain to me a step in a proof given to me by my professor in regards to a modular operation theorem.

Addition theorem: Given three integers x, y, d (d > 0), (x+y)%d = (x%d + y%d) %d

Proof:
Let x = q(1)d + r(1) and y = q(2)d + r(2).
We have (x+y)%d = (q(1)d + r(1) + q(2)d + r(2)) %d
= (r(1) + r(2)) %d
Therefore: (x+y)%d = (x%d + y%d) %d

I don't get how my professor jumped from (q(1)d + r(1) + q(2)d + r(2))%d to (r(1) + r(2))%d.

Is there a specific reason for why we just ignore the product of q(1)d and q(2)d?

Thank you in advance.
 
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Is there a specific reason for why we just ignore the product of q(1)d and q(2)d?
The Big Idea is that d is zero modulo d, so q(1)d is zero because you're multiplying q(1) by zero.

As for the technical detail, doesn't the equality follow directly from the definition of x%d? If you don't think so, then please state what definition you are using, and apply that definition to the two sides of that equation.
 
d has to be greater than zero in our given. so i don't believe q(1)d and q(2)d are negated because d is zero.

ajbiol said:
Addition theorem: Given three integers x, y, d (d > 0), (x+y)%d = (x%d + y%d) %d
 
ajbiol said:
d has to be greater than zero in our given. so i don't believe q(1)d and q(2)d are negated because d is zero.
d is zero modulo d.
 
oh i get it! ty ty!
 

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