How can mathematical induction be used to prove a modular arithmetic problem?

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Discussion Overview

The discussion revolves around the use of mathematical induction to prove a statement related to modular arithmetic involving a set of natural numbers and their sum modulo a given integer. Participants are exploring the formulation and implications of the problem, as well as the specifics of the modulus operation.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents a modular arithmetic problem involving a set of natural numbers and seeks to prove a specific equality using mathematical induction.
  • Another participant expresses uncertainty about the interpretation of the modulus operation in the context of the problem.
  • A third participant attempts to clarify the modulus operation, suggesting a nested application of the modulus function with a consistent divisor 'm'.
  • A later reply acknowledges the clarification and suggests that the problem is suitable for mathematical induction, noting that the base case for n=1 is trivial.
  • Participants encourage sharing thoughts and approaches to solving the problem, emphasizing collaborative exploration.

Areas of Agreement / Disagreement

There is no consensus on the proof method yet, as participants are still discussing the formulation of the problem and the specifics of the modulus operation. Multiple interpretations of the modulus term exist, indicating some disagreement or uncertainty.

Contextual Notes

Participants have not fully resolved the assumptions regarding the modulus operation, and there are unresolved details about the application of mathematical induction in this context.

Bipolarity
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I'm trying to prove something in modular arithmetic that I came upon across my studies in comp sci. Consider a set of natural numbers {n_{1},n_{2},n_{3},...n_{k}}

Consider two more natural numbers m and p such that

(\sum^{k}_{i=1}n_{i} ) \ mod \ m = p

Now prove that

((((n_{1} \ mod \ m + n_{2}) \ mod \ m + n_{3}) \ mod \ m + n_{4}) \ mod \ m + ... + n_{k}) \ mod \ m = p

All help would be appreciated.

BiP
 
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Hey bipolarity.

I'm just unsure about what the modulus term is. For example is it (a mod (m + n2)) or is it a mod m + n2 etc.
 
chiro said:
I'm just unsure about what the modulus term is. For example is it (a mod (m + n2)) or is it a mod m + n2 etc.
I think he means:

mod_m( ... (mod_m( mod_m(n_{1}) + n_{2})) ... + n_{k})
 
Sorry! Thanks rcgldr for clarifying what I meant.
I meant the divisor in the modulus operator is always 'm'.

So basically you are adding the next number, then taking the remainder after dividing by m. Then again you are adding the next number, then taking the remainder after dividing by m. You do this until you run out of numbers, at which point you finally take the remainder after dividing by m. The final answer should be 'p'. But I'm trying to prove it.

BiP
 
Bipolarity said:
Sorry! Thanks rcgldr for clarifying what I meant.
I meant the divisor in the modulus operator is always 'm'.

So basically you are adding the next number, then taking the remainder after dividing by m. Then again you are adding the next number, then taking the remainder after dividing by m. You do this until you run out of numbers, at which point you finally take the remainder after dividing by m. The final answer should be 'p'. But I'm trying to prove it.

BiP

Thanks Bipolarity.

This looks like a good candidate for mathematical induction. The first case is trivial because n1 MOD m and n1 MOD m are the same so proven for n = 1.

I think I know how you could do it but I'll ask you for your thoughts and any work that you have for solving the problem in your own terms (we encourage that highly here on PF).
 

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