Is there a simple method for square root of big.Rat?

I need to find the square root of a big.Rat. Is there a way to do it without losing (already existing) accuracy?

For example, I could convert the numerator and denominator into floats, get the square root, and then convert it back…

func ratSquareRoot(num *big.Rat) *big.Rat {
    f, exact := num.Float64() //Yuck! Floats!
    squareRoot := math.Sqrt(f)
    var accuracy int64 = 10 ^ 15 //Significant digits of precision for float64
    return big.NewRat(int64(squareRoot*float64(accuracy)), accuracy) 
    // ^ This is now totally worthless. And also probably not simplified very well.

}

…but that would eliminate all of the accuracy of using a rational. Is there a better way of doing this?

>Solution :

The big.Float type has a .Sqrt(x) operation, and handles defining explicitly the precision you aim for. I’d try to use that and convert the result back to a Rat with the same operations in your question, only manipulating big.Int values.