def _binary_float_to_ratio(x):
    """x -> (top, bot), a pair of ints s.t. x = top/bot.

    The conversion is done exactly, without rounding.
    bot > 0 guaranteed.
    Some form of binary fp is assumed.
    Pass NaNs or infinities at your own risk.

    >>> _binary_float_to_ratio(10.0)
    (10, 1)
    >>> _binary_float_to_ratio(0.0)
    (0, 1)
    >>> _binary_float_to_ratio(-.25)
    (-1, 4)
    """

    if x == 0:
        return 0, 1
    f, e = math.frexp(x)
    signbit = 1
    if f < 0:
        f = -f
        signbit = -1
    assert 0.5 <= f < 1.0
    # x = signbit * f * 2**e exactly

    # Suck up CHUNK bits at a time; 28 is enough so that we suck
    # up all bits in 2 iterations for all known binary double-
    # precision formats, and small enough to fit in an int.
    CHUNK = 28
    top = 0
    # invariant: x = signbit * (top + f) * 2**e exactly
    while f:
        f = math.ldexp(f, CHUNK)
        digit = int(f)
        assert digit >> CHUNK == 0
        top = (top << CHUNK) | digit
        f = f - digit
        assert 0.0 <= f < 1.0
        e = e - CHUNK
    assert top

    # reduce to lowest terms
    while not top&1:
        top >>= 1
        e += 1

    # Add in the sign bit.
    top = signbit * top

    # now x = top * 2**e exactly; fold in 2**e
    if e>0:
        return (top * 2**e, 1)
    else:
        return (top, 2 ** -e)


import math

for c in (math.pi, 7, 7.5, 0.875):
    n, d = _binary_float_to_ratio(c)
    print (n, d), (n+0.0)/d, c