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I've seen this argument several times and always get lost at around the same point: > But that means that our candidate solution cannot be the quadratic formul
by alecbenzer 9y ago
I've seen this argument several times and always get lost at around the same point:
> But that means that our candidate solution cannot be the quadratic formula! If it were, then x_1 and x_2 would have ended up swapped, too.
Why?
- jacksnipe 9y agoBecause by definition, if we have found the right solution, then x_1 = r_1 and x_2 = r_2.
- alecbenzer 9y agoI don't see why that's true by definition. We're looking for a function f st. f(a,b,c) = {r1, r2} iff 0 = a·r1^2 + b·r1 + c = a·r2^2 + b·r2 + c. Right? So as long as the correct two roots are produced, why is the order that they are produced in relevant?
- akalin 9y agoBecause we start off with the candidate solution matching the real roots, and we continuously move the roots to swap them. At every point of the movement, the candidate solution match the real roots (including order) by assumption, and so if at the end of the movement the candidate solution returns to its initial state, i.e. it's equal to the swapped roots, then that would contradict continuity. Does that make sense?