Expand the square and simplify: a2+45+a+(6a−b)5=51⇒a2+a−6+(6a−b)5=0
So you need a2+a−6=0 and 6a−b=0 by comparing the rational and irrational parts of both sides of the equation. Tripped up plenty of people in the exam, took me some time to work it out and the only reason I did was because I'd looked up complex numbers for fun and was used to dealing with imaginary and real parts, this was analogous.
Expand the square and simplify: a2+45+a+(6a−b)5=51⇒a2+a−6+(6a−b)5=0
So you need a2+a−6=0 and 6a−b=0 by comparing the rational and irrational parts of both sides of the equation. Tripped up plenty of people in the exam, took me some time to work it out and the only reason I did was because I'd looked up complex numbers for fun and was used to dealing with imaginary and real parts, this was analogous.
Could there be an irrational number that fits a and b?
If you remove the condition that a and b are rational, I don't think the equation is solvable by conventional means or at all. It's like saying x+iy=a+ib and letting a,b,x,y∈C.
If you remove the condition that a and b are rational, I don't think the equation is solvable by conventional means or at all. It's like saying x+iy=a+ib and letting a,b,x,y∈C.
D'you have a better way to justify taking the positive root?
I'm not very familiar with the theory of continued fractions, so maybe it's done with a more general argument there, but if I had to do it here, I would replace the continued fraction with an equivalent recursive series, then show that the series is decreasing and bounded below by some positive number. That proves than the sequence converges to a +ve limit.
That would then be enough to justify taking the +ve root, and the trick that you've used above (which implicitly assumes convergence)
Maybe someone like Gregorius knows how this is done more generally?
[edit: Have I tagged him properly? If not, something is effed up on this site]
I'm not very familiar with the theory of continued fractions, so maybe it's done with a more general argument there, but if I had to do it here, I would replace the continued fraction with an equivalent recursive series, then show that the series is decreasing and bounded below by some positive number. That proves than the sequence converges to a +ve limit.
Ah, thanks. I'll go have a read up on that in a sec because I'm not quite sure what you mean by an equivalent recursive series. How would you go about constructing that recursive series?
I'm not very familiar with the theory of continued fractions, so maybe it's done with a more general argument there, but if I had to do it here, I would replace the continued fraction with an equivalent recursive series, then show that the series is decreasing and bounded below by some positive number. That proves than the sequence converges to a +ve limit.
That would then be enough to justify taking the +ve root, and the trick that you've used above (which implicitly assumes convergence)
Maybe someone like Gregorius knows how this is done more generally?
[edit: Have I tagged him properly? If not, something is effed up on this site]
Only here for a couple of minutes, so haven't read the question properly! But what I think you're looking for is the idea of a convergent to a continued fraction.
Ah, thanks. I'll go have a read up on that in a sec because I'm not quite sure what you mean by an equivalent recursive series. How would you go about constructing that recursive series?