In a comment on my “Gilded Ratios” essay fluffy wondered about a variation on the Golden and Golden-like ratios. What’s interesting about the Golden Ratio and similar numbers is that their reciprocal — one divided by them — is a whole number less than the original number. That is, 1 divided by 1.618(etc) is 0.618(etc), which is 1 less than the original number. 1 divided by 2.414(etc) is 0.414(etc), exactly 2 less than the original 2.414(etc). 1 divided by 3.302(etc) is 0.302(etc), exactly 3 less than the original 3.302(etc).
fluffy wondered about a variation. Is there some number x that’s exactly 2 less than 2 divided by x? Or a (presumably) differently number that’s exactly 3 less than 3 divided by it? Yes, there is.
Let me call the whole number difference — the 1 or 2 or 3 or so on, referred to above — by the name b. And let me call the other number — the one that’s b less than b divided by it — by the name x. Then a number x, for which b divided by x is exactly b less than itself, makes true the equation . This is slightly different from the equation used last time, but not very different. Multiply both sides by x, which we know not to be zero, and we get a polynomial.
Yes, quadratic formula, I see you waving your hand in the back there. And you’re right. There are two x’s that will make that equation true. The positive one is . The negative one you get by changing the + sign, just before the square root, to a – sign, but who cares about that root? Here’s the first several of the (positive) silver-leaf ratios:
|Some More Numbers With Cute Reciprocals|
Looking over those hypnotic rows of digits past the decimal inspires thoughts. The part beyond the decimal keeps rising, closer and closer to 1. Does it ever get past 1? That is, might (say) the silver-leaf number that’s 2,038 more than its reciprocal be 2,039.11111 (or something)?
No, it never does. There are a couple of ways to prove that, if you feel like. We can take the approach that’s easiest (to my eyes) to imagine. It takes a little algebraic grinding to complete. That is to look for the smallest number b for which the silver-leaf number, , is larger than . Follow that out and you realize that it’s any value of b for which 0 is greater than 4. Logically, therefore, we need to take b into a private room and have a serious talk about its job performance, what with it not existing.
A harder proof to imagine working out, but that takes no symbol manipulation, comes from thinking about these reciprocals. Let’s imagine we had some b for which its corresponding silver-leaf number x is more than b + 1. Then, x – b has to be greater than 1. But if x is greater than 1, then its reciprocal has to be less than 1. We again have to talk with b about how its nonexistence is keeping it from doing its job.
Are there other proofs? Most likely. I was satisfied by this point, and resolved not to work on it more until the shower. Updates after breakfast, I suppose.