Re: [tuning-math] Proofs Concerning Ratios of Minimal Numerator/Denominator

Very nice, thank you Ryan! -Carl

At 04:39 PM 2013/06/06, you wrote:
>Conjecture 1:
>Let A and B be real numbers such that B > A >= 1 and B-A<1. Let C be
>the real interval [A,B]. Then there exists exactly one rational
>number in C with minimal numerator.
>
>Proof:
>Suppose that there were at least two ratios of minimal numerator
>within the interval. Then we can state that:
>
>N/D > N/(D+1)
>N/D > (N-1)/D > N/(D+1)
>
>This violates our original assumption and we therefore conclude that
>the interval C must have exactly one ratio with minimal numerator.
>
>
>
>Conjecture 2:
>There exists exactly one rational number in C with minimal denominator.
>
>Proof:
>Suppose that there were at least two ratios of minimal denominator
>within the interval. Then we can state that:
>
>(N+1)/D > N/D
>(N+1)/D > (N-A)/(D-1) > N/D
>
>N/D > A > (N+1-D)/D
>N/D > A > (N+1)/D - 1
>
>Substitute E=A+1.
>(N+D)/D > E > (N+1)/D
>
>Because N+D, N+1, N and D are all pairwise coprime with each other,
>and since N > D, we know that there must be an integer solution for E
>which satisfies the above inequalities. This means that a rational
>number with a smaller denominator must exist within the interval C,
>violating our original assumption and concluding the proof.
>
>
>
>Conjecture 3:
>The minimal ratios mentioned in the previous two conjectures (the
>ratio of minimal denominator and minimal numerator) are in fact
>equivalent, and not distinct.
>
>Proof:
>Suppose that the two ratios were distinct. Let us call the ratio of
>minimal denominator (N+A)/D, and the ratio of minimal numerator
>N/(D+B). This gives us the following inequalities:
>
>B > 0
>A > 0
>(N+A)/D > N/(D+B)
>(N+A)/D > N/D > N/(D+B)
>
>This violates the conclusions of the first two conjectures, meaning
>that it also violates our original assumption. Therefore the ratios
>are equivalent, and not distinct.
>