Finding quadratic irrationals from their decimal form

[rabbit boy]

Okay, this is something that's been bugging me for a while. A lot of the polynomials with higher powers can't be solved using algebraic methods and must be estimated. So my idea was to take these answers and find a way to convert them into an exact form, such as quadratic irrationals/surds.

I do not know if all zeroes of polynomials can be represented as quadratic irrationals. However, I did find out that you can use continued fractions to get a quadratic irrational solution from its decimal value to its $\dfrac{a + b\sqrt{c}}{d}$ form.

The problem I'm having with continued fractions is that you have to keep going until you find where they repeat, for many quadratic irrationals there are dozens of numbers in the series, and I don't have enough accuracy in my calculations to find where it repeats.

And then of course there's guess and check, but I don't know a way of doing it efficiently yet.

So what should I do?

 

[wywong]

No matter how many decimal places you got, there are in theory infinite number of irrationals that fit, and a lot of them not in the form of surds. So I think it will to futile to look for such a way.

 

[rabbit boy]

No matter how many decimal places you got, there are in theory infinite number of irrationals that fit, and a lot of them not in the form of surds. So I think it will to futile to look for such a way.

I already know it's possible to find what the quadratic irrational is with continued fractions. I made a thread about it on another forum.

http://bbs.zoklet.net/showthread.php?t=11555

 

[HallsofIvy]

But that was not your question. You asked about using continued fractions to get exact solutions to higher degree equations. "I do not know if all zeroes of polynomials can be represented as quadratic irrationals." No, they cannot. All quadratic irrationals can be represented as surds. Not all zeroes of polynomials can be. Further, it is at least theorectically possible to represent all continued fractions as surds so it is not possible to represent all solutions to polynomial equations as continued fractions either.

 

[confinement]

I once read that if we could invert the function x^5 + x = y to be x = f(y) then we could solve all 5 degree polynomials using algebraic operations together with the function f.

 

[lurflurf]

I once read that if we could invert the function x^5 + x = y to be x = f(y) then we could solve all 5 degree polynomials using algebraic operations together with the function f.

What you read is true, so what? All polynomials can be solved in any algebraically complete field such as the complex numbers. The question is what operations are needed to effect the solution. Field operations and root extractions suffice for at most quartic equations. A possible additional operation to allow solution of quintics in the bring radical you mention.

 

[rabbit boy]

But that was not your question. You asked about using continued fractions to get exact solutions to higher degree equations. "I do not know if all zeroes of polynomials can be represented as quadratic irrationals." No, they cannot. All quadratic irrationals can be represented as surds. Not all zeroes of polynomials can be. Further, it is at least theorectically possible to represent all continued fractions as surds so it is not possible to represent all solutions to polynomial equations as continued fractions either.

My question is, for zeroes that can be represented in the form $\dfrac{a + b\sqrt{c}}{d}$, what is a good way to convert them from decimal form to that?

I already know how to do it with continued fractions, but sometimes it requires too many digits of precision. so I'm probably looking for a different way to do it.

I'm sorry for not being clear about that.

And thank you, you two, for answering whether all of them can be represented in that form. I was just a bit grumpy yesterday.

 

[lurflurf]

My question is, for zeroes that can be represented in the form $\dfrac{a + b\sqrt{c}}{d}$, what is a good way to convert them from decimal form to that?

I think in general it would be hard.