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

[Dick]

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.

 

[truewt]

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 :(

 

[Dick]

I usually say relatively prime when "no common factor" suffices. Just force of habit, I guess.

 

[disregardthat]

What is RHS? You mentioned it in your proof.

 

[Dick]

RHS='right hand side' (of the equation).