An IEEE754 64-bit double can represent any 32-bit integer, simply because it has 53-odd^(a) bits available for precision and the 32-bit integer only needs, well, 32 :-)

It would be plausible for a (non IEEE754 double precision) 64-bit floating point number to have less than 32 bits of precision. That would allow truly huge numbers (due to the exponent) but at the cost of precision.

The bottom line is that, provided there are more bits of precision in the mantissa of the floating point number than there are in the integer (and enough bits in the exponent to scale it), then it can be represented without loss of precision.

^(a) Technically, the 53rd bit of precision is an implied 1 at the start of the sequence so the amount of "variablity" may only be 52 bits. Whether it's 52 or 53, it's still enough bits to represent every 32-bit integer.

xs:double and xs:float are defined by IEEE rules. See

3.3.5 double

The ·value space· of double contains the non-zero numbers m × 2e , where m is an integer whose absolute value is less than 253, and e is an integer between −1074 and 971, inclusive.

By contrast, the limits for xs:integer and xs:decimal are implementation-defined.

Section 3.3.3: [Definition:] decimal represents a subset of the real numbers, which can be represented by decimal numerals. The ·value space· of decimal is the set of numbers that can be obtained by dividing an integer by a non-negative power of ten, i.e., expressible as i / 10n where i and n are integers and n ≥ 0.

Section 5.4: All ·minimally conforming· processors must support decimal values whose absolute value can be expressed as i / 10^k, where i and k are nonnegative integers such that i < 1016 and k ≤ 16 (i.e., those expressible with sixteen total digits).