Functional Equation: A, B, C Answers

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Discussion Overview

The discussion revolves around a functional equation involving a function \( f: \mathbb{N} \rightarrow \mathbb{N} \) that satisfies specific properties related to coprime natural numbers and prime numbers. Participants explore the implications of these properties on the values of \( f \) at various points, particularly \( f(1) \), \( f(2) \), and other primes, while attempting to solve for specific expressions involving \( f \).

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • Some participants assert that since \( 1 \) is coprime to every natural number, it follows that \( f(1) = 0 \).
  • It is proposed that for any prime \( c > 2 \), the relationship \( f(2c) = f(2) + f(c) \) holds, leading to the conclusion that \( f(c) = f(2) \).
  • A participant notes that this reasoning allows for answering the posed questions (a), (b), and (c) in terms of \( f(2) \), but they express uncertainty about how to evaluate \( f(2) \).
  • Another participant points out a potential contradiction arising from setting \( c = 2 \) and \( d = 3 \), leading to the conclusion that \( f(2) = 0 \) and consequently \( f(p) = 0 \) for every prime \( p \).
  • There is a suggestion that if the derived conclusion does not lead to a contradiction, it could provide a full solution, although it raises questions about the completeness of the original problem statement.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the functional equation, particularly regarding the value of \( f(2) \) and its consequences for other primes. The discussion remains unresolved as participants explore these implications without reaching a consensus.

Contextual Notes

There are indications of missing conditions or assumptions in the original problem statement that could affect the conclusions drawn by participants.

juantheron
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http://latex.codecogs.com/gif.latex?\hspace{-20}$%20A%20function%20$f:\mathbb{N}%20\rightarrow%20\mathbb{N}$%20and%20satisfies%20$f(ab)%20=%20f(a)+f(b)$.\\%20Where%20$a$%20and%20$b$%20are%20Coprime%20Natural%20no.\\%20and%20$f(c+d)%20=%20f(c)+f(d)\forall$%20prime%20no.%20$c$%20and%20$d$.%20Then\\%20(a)%20The%20value%20of%20$f(1)+f(2)+f(3)%20=%20$\\%20(b)%20$\frac{f(5)+f(7)}{f(4)}%20=$\\%20(c)%20$f(9)-f(6)+f(3)%20=%20$
 
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jacks said:
http://latex.codecogs.com/gif.latex?\hspace{-20}$%20A%20function%20$f:\mathbb{N}%20\rightarrow%20\mathbb{N}$%20and%20satisfies%20$f(ab)%20=%20f(a)+f(b)$.\\%20Where%20$a$%20and%20$b$%20are%20Coprime%20Natural%20no.\\%20and%20$f(c+d)%20=%20f(c)+f(d)\forall$%20prime%20no.%20$c$%20and%20$d$.%20Then\\%20(a)%20The%20value%20of%20$f(1)+f(2)+f(3)%20=%20$\\%20(b)%20$\frac{f(5)+f(7)}{f(4)}%20=$\\%20(c)%20$f(9)-f(6)+f(3)%20=%20$

Since \(1\) is coprime to every natural \(f(1)=0\).

Also since for any prime \(c>2\) we have \(f(2c)=f(2)+f(c)\) and \(f(2c)=f(c+c)=f(c)+f(c)\) we conclude that \(f(c)=f(2)\), which is sufficient to allow us to answer (a), (b) and (c) in terms of \(f(2)\).

At present I don't see any means of evaluating \(f(2)\).CB
 
Last edited:
CaptainBlack said:
Since \(1\) is coprime to every natural \(f(1)=0\).

Also since for any prime \(c>2\) we have \(f(2c)=f(2)+f(c)\) and \(f(2c)=f(c+c)=f(c)+f(c)\) we conclude that \(f(c)=f(2)\), which is sufficient to allow us to answer (a), (b) and (c) in terms of \(f(2)\).

CB
But that leads to something strange if you put $c=2$ and $d=3$, because it then follows that $f(2) = f(5) = f(2+3) = f(2) + f(3) = 2f(2).$ Thus $f(2)=0$ and hence $f(p)=0$ for every prime $p.$
 
Opalg said:
But that leads to something strange if you put $c=2$ and $d=3$, because it then follows that $f(2) = f(5) = f(2+3) = f(2) + f(3) = 2f(2).$ Thus $f(2)=0$ and hence $f(p)=0$ for every prime $p.$

If that does not entail a contradiction then that gives us a full solution, including undefined for (b), it also answers the implied question I added to my post between you starting to reply and my seeing your reply :) . Alternativly there is a condition missing from the statement of the question.

CB
 

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