I am trying to work on the following problem and I am willing to transfer universities to find a good advisor:
We do not need to take the whole of the real numbers to do engineering. Since engineers only care about the first n decimal places and treat everything after that as noise, if every term in an equation has units of meters, replace them with units of centimeters, millimeters, decimeters, etc, effectively multiply everything by 10^n for what ever n value is important. Then go and use the ring of integers).
(update)I took some liberties when posting the problem that I did not think would be relevant but have clearly shown to be vital based on the comments I have received. Because engineers have the option to work in which ever set of units are desired ie, centimetres, milimeters, nanometres, etc, seconds, miliseconds, nanoseconds, etc. any problem that is not dimensionless to start with can be written as a problem in the integers with some rounding. As far as engineers are concerned, anything over n decimal places is literately noise from their equipment and not meaningful information. It does not matters that noise1+noise2+noise3+...etc > threshold for not noise. Engineers live with this all the time, of course it would be nice to be able to find out the approximate size of the error term. Does this make the initial equation not in a ring? Yes, but the equation that the engineers consider to be a good enough approximation will be in the ring of integers. Just like when we approximate functions with taylor series there will be error associated with this method. Have I figured out the exact extend to the error yet, no. As important as this is that is something that I need to work on. Yes, the value of n for an engineer will change if they multiply 2 numbers together with differing values n1 and n2 needed to make all significant figures into integers, but if the dimensions of the numbers provide a useful tool to avoid this pitfall. If I multiply x in meters by y in seconds before switch everything from meters to nanometres and seconds to nanoseconds, then the conversion from meters to nanometers is multiplication by 10^9, and the conversion from seconds to nanoseconds is again multiplication by 10^9. So the product x*y would have to be multiplied by 10^18 in the conversion. Similarly if I wanted to convert the volume to cubic nanometers from cubic meters I would have to multiply by 10^27. (end update)
Call S the result of truncating the the field of real numbers to n decimal places (with n being a value relevant to the engineering problem in question), and then multiplying by 10^n. S will be a ring of integers with the usual operations of addition and multiplication.
Let R be the group ring by taking S , and crossing it with the free abelian group, on 7 generators, Z^7. IE with operations performed as follows 2seconds*4meters/second = 8 meters, 2seconds+4meters/second= 2seconds+4meters/second . This group ring is close enough of an approximation to the engineering approximations used that a solution found in the group ring would be meaningful to an engineer.
Now the free abelian group has all its subgroups as normal subgroups. Take the homomorphism f, from R to R with the free abelian group on 6 generators Z^6, defined by f:(S, x1, x2, … x7) → (S, x1, x2, … x6, 0). Because this is well defined by the first isomorphism theorem for rings, it shows that (0, 0, 0, … x7) is an ideal for R. Similarly, g:(S, x1, x2, … x7) → (0, x1, x2, … x6, 0), and h:(S, x1, x2, … x7) → ( S mod r, x1 mod a1, x2 mod a2 , … , x7mod a7 ) show that ( S, 0, 0,…,0, a7x7) and ( r S, a1x1, a2x2,… a7x7) and must also be an ideals of R. Therefore the Chinese Remainder theorems can be used on these ideals of R.
Further, the Chinese remainder theorem can be used on polynomials of this group ring R[x], with x not only having values in the integers but also values in terms of meters, seconds, etc.
Dear David,
You stated many assumptions to create a new field of numbers.
I think you need to take care of many points.
Assume that a and b are real numbers defined up to n decimals in the proposed truncated field.
BUT ab yields a new different number with more decimals exceeding n.
If you want to truncate the result into n decimals by rounding, then you will face anew obstruction.
You can find ab = ac; a is not zero and b is different c. You lose the basic property of your proposed field.
Ex: 0.10000001 x 0.0000000001= 0.10000001 x 0.0000000002
but 0.0000000001 is not equal to 0.0000000002
moreover 0.10000001 x 0.0000000001 = 0 after rounding process but neither of the two numbers is zero.
Best wishes
I will update the initial question to rephrase what I am doing.I took some liberties when posting the problem that I did not think would be relevant but have clearly shown to be vital based on the comments I have received. Because engineers have the option to work in which ever set of units are desired ie, centimetres, milimeters, nanometres, etc, seconds, miliseconds, nanoseconds, etc. any problem that is not dimensionless to start with can be written as a problem in the integers with some rounding. As far as engineers are concerned, anything over n decimal places is literately noise from their equipment and not meaningful information. It does not matters that noise1+noise2+noise3+...etc > threshold for not noise. Engineers live with this all the time, of course it would be nice to be able to find out the approximate size of the error term. Does this make the initial equation not in a ring? Yes, but the equation that the engineers consider to be a good enough approximation will be in the ring of integers. Just like when we approximate functions with taylor series there will be error associated with this method. Have I figured out the exact extend to the error yet, no. As important as this is that is something that I need to work on. Yes, the value of n for an engineer will change if they multiply 2 numbers together with differing values n1 and n2 needed to make all significant figures into integers, but if the dimensions of the numbers provide a useful tool to avoid this pitfall. If I multiply x in meters by y in seconds before switch everything from meters to nanometres and seconds to nanoseconds, then the conversion from meters to nanometers is multiplication by 10^9, and the conversion from seconds to nanoseconds is again multiplication by 10^9. So the product x*y would have to be multiplied by 10^18 in the conversion. Similarly if I wanted to convert the volume to cubic nanometers from cubic meters I would have to multiply by 10^27.
I agree that most cases this will not work. However, I think it will work for a handful of cases and here is why: I think the scaling aspect of multiplication is dealt with the by using the correct choice of dimensions before using the rounding. If I need the first 8 digits and I convert something from meters to nanometers before rounding off at the ninth digit there should be no problem as far as I can tell. All further multiplication will be done in nanormeters so that significant digits will not be lost, even if the goal is to find the volume in cubic nanormeters. The engineers will have no problem converting back to meters from nanometers.
Some care will have to be taken with terms that are ratios but I can say much about the ratios in the equation based on dimensional analysis. There will have to be judicious choices made in each equation about what units everything is in and what items are actually variables. However, dimensional analysis allows us to rewrite the equation and combined variables. I am not saying that this will work for each and every equation but I would be fine if there were a handful of equations to which this could be applied with painstaking care. I agree that there is not an homorphism from the original problem to the new group-ring and that addition done before rounding would have different results than addition done after rounding. You seem to think there is a problem with this set-up and that any approximation of the problem in this fashion would have an error that is either huge or highly variant and hard to calculate depending on the problem. I would like to understand better why this won't work.
I really liked your approach towards a better understanding of the problem @ David Infortunio!!!! Philosophically speaking, this should the approach towards constructing a meaningful problem!!
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