I don't know why I was using S('') anyway; `cf_q([0, 1, 1, 7, [24, 8]]).equals((sqrt(3) + 2)/7)` works fine too.  Avoids the issue of sympify, but still doesn't use ==.  The other thing I was thinking of was doing `.as_numer_denom` to reduce it to a form I could compare better.  What's the least of the evils?