## Return to Answer

7 fixed rollback error

Similarly, every numerical simulation in physics (or chemistry, biology, physiology, or medicine) always has to use finite precision representation of values, such that there is a limit to the largest and smallest integer represented by a fixed number of bits, and such that there is a limited amount of "floating-point-precision" available in dividing the bits of a floating-point representation of a real number into a set number of bits for the mantissa and a set number of bits for the exponent.

For example, assuming that $d=64$-bits are used to represent "real numbers" as floating point numbers in computations, $m=48$ bits may be allocated to the mantissa, allowing the numerator to be $2^{48}$ yielding approximately $14$ digits of base-ten specificity to the numerator; this leaves $d-m=16$ bits to the exponent which may be signed (+/-) yielding a range of -32768 to +32767.

In this case, the floating point number is in the range $n\times 2^{d-m}$, where $-(2^{47} \le n \le +(2^{47}-1)$, and ${-32768} \le d \le {32768}$.

6 Rollback to Revision 4

As Colin Tan said, "[using] only countably many symbols, then there will always be a countable models." Whereas the field $\mathbb{Q}$ of rational numbers uses a finite-number of symbols for an uncountable number of rational numbers. I misunderstood and misapplied a concept.

The field generated by rational numbers is quite different from the "approximation space" rendered by using a finite number of bits interpreted as a floating-point number low-precision approximation to real numbers. I'm editing my answer to point out my misunderstanding. @Hans-Stricker, I've fixed my error by pointing it out, but leaving it up (below the ruled line) so that some other bit-flipper like me will see why {0,1}$^n \times${0,1}$^n$ is not equivalent to $\mathbb{Q}$

below this is my original (erroneous) answer

If the total number of bits is $d$, the number of bits allocated to the exponent, $m$, may be decreased while simultaneously increasing the number of bits, $d-m$, allocated to the mantissa, increasing the "precision" of the numerator while decreasing the range over $\mathbb{R}$ spanned by this particular approximating set of {0,1}$^m \times${0,1}$^{d-m}$ (which is not equivalent to $\mathbb{Q}$, as I erroneously stated originally)

Thus every numerical simulation is already, in a way, based on $\mathbb{Q}^d$ when models of $d$-dimensional systems are created and iterated using Euler or Runge-Kutta of whatever order.

5 remove text

As Colin Tan said, "[using] only countably many symbols, then there will always be a countable models." Whereas the field $\mathbb{Q}$ of rational numbers uses a finite-number of symbols for an uncountable number of rational numbers. I misunderstood and misapplied a concept.

The field generated by rational numbers is quite different from the "approximation space" rendered by using a finite number of bits interpreted as a floating-point number low-precision approximation to real numbers. I'm editing my answer to point out my misunderstanding. @Hans-Stricker, I've fixed my error by pointing it out, but leaving it up (below the ruled line) so that some other bit-flipper like me will see why {0,1}$^n \times${0,1}$^n$ is not equivalent to $\mathbb{Q}$

below this is my original (erroneous) answer

If the total number of bits is $d$, the number of bits allocated to the exponent, $m$, may be decreased while simultaneously increasing the number of bits, $d-m$, allocated to the mantissa, increasing the "precision" of the numerator while decreasing the range over $\mathbb{R}$ spanned by this particular approximating set of {0,1}$^m \times${0,1}$^{d-m}$ (which is not equivalent to $\mathbb{Q}$, as I erroneously stated originally)

Thus every numerical simulation is already, in a way, based on $\mathbb{Q}^d$ when models of $d$-dimensional systems are created and iterated using Euler or Runge-Kutta of whatever order.

4 removed extra erroneous text
3 pointed out error of approximation vs. finite field
2 shifting decimal points
1