Proving m + 1/m = Integer When m = 1

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Homework Help Overview

The discussion centers around proving that if \( m \) is a positive rational number, then \( m + \frac{1}{m} \) is an integer only when \( m = 1 \). Participants explore the implications of this statement and the conditions under which it holds true.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the representation of \( m \) as a fraction \( \frac{p}{q} \) and the resulting expression \( \frac{p}{q} + \frac{q}{p} \). Some question the validity of concluding that both terms are integers based solely on their sum being an integer.
  • There are attempts to clarify the conditions under which the sum \( \frac{a}{b} + \frac{b}{a} \) can be an integer, with discussions about divisibility and the implications of \( a \) and \( b \) being coprime.
  • Some participants express uncertainty about the correct interpretation of the problem, whether it is to show that \( m + \frac{1}{m} \) is an integer or that \( \frac{m + 1}{m} \) is an integer only if \( m = 1 \).

Discussion Status

The discussion is ongoing, with various interpretations and approaches being explored. Some participants have provided insights into the mathematical relationships involved, while others are still grappling with the proof and the conditions required for it to hold. There is no explicit consensus yet, but productive lines of reasoning are being developed.

Contextual Notes

Participants note that \( m \) must be a positive rational number and discuss the implications of this constraint on the proof. There is also mention of the need to consider cases where \( a \) and \( b \) are coprime, which adds complexity to the problem.

  • #31
I showed that my premise, that there is a lowest terms fraction satisfying a/b+b/a=n leads to the conclusion that fraction is, in fact, not in lowest terms. This is an impossible situation. Hence I must have been wrong in assuming such a fraction exists.
 
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  • #32
I think the initial assumption that \frac {a}{b} is in the lowest terms, hence they are co-prime (or rather no common factors), but you've proven that there exists a prime factor p that divides both of them, hence they are not co-prime which is a contradiction.

Anyway why do we even use co-prime when "no common factor" suffices?

EDIT: I can't seem to get the latex to work :(
 
Last edited:
  • #33
I usually say relatively prime when "no common factor" suffices. Just force of habit, I guess.
 
Last edited:
  • #34
What is RHS? You mentioned it in your proof.
 
  • #35
RHS='right hand side' (of the equation).
 

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