we're square rooting a bunch of positive numbers - we know the result has to be positive. We probably 'created' the extraneous root when we squared both sides of the equation.Huh? I can't let this one by as it contradicts any and all maths I ever studied. Minus times minus gives plus - the root was there all the time and is not extraneous.
BTW check out Ted's link page and you'll see that the value of the second solution is what my original Google (
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibnat.html) mentioned as the 'closely related value which we write as phi with a small "p" ', and is the reciprocal of Phi as well as (Phi-1).
Which means (I think, my head is now hurting as it's a long time since I did any of this!
) that small-p phi is actually really the same ratio anyway. (Value n of the series):(value n+1) rather than (value n):(value n-1). So there's not even any reason to eliminate this solution from the quadratic equation.
..I was trying to write this post some 16 hours ago and at that point it killed not only my brain but my computer too... I was also planning to address FF's later concern about a negative value for the side of a triangle. I don't have a problem with this, or see that it is complicated. Assume that the side in question lies along the x-axis of a graph and has one end at the point of crossing the y-axis (0,0). The difference between positive and negative value for the length of the side is whether the triangle then lies to the right or the left of the y-axis. Put another way, which end of that side you have arbitrarily decided to define as (0,0).
I see positive and negative as directional rather than absolute-value (consider also time and latitude as examples where we apply not just distance but also direction from an arbitrary point), but I think I'd better stop there as 1) I am out of my depth and 2) it is hard to type with a kitten on the keyboard!!!
Bel, I apologise for hijacking your thread in the first place!!
