The phrase "a sum of unit" is not correct in written English.
It seems to be a misphrasing; the correct expression would be "a sum of units."
Example: "The total cost is calculated as a sum of units sold during the month."
Alternatives: "a total of units" or "an aggregate of units."
First, we approximate the fraction of the requested bandwidth over the available bandwidth per slot, by means of an Egyptian Fraction [20, 21], that is, a sum of distinct unit fractions.
In both cases, a sum of basic units, no matter how many, is insufficient to produce a significant difference.
By the above theorem, we know that (S=S_1times S_2), where (S_1) is a Boolean ring and each element of the ring (S_2) is a sum of two units.
Then (R=R_1times R_2), where (R_1) is a Boolean ring and each element of the ring (R_2) is a sum of two units.
Furthermore, it is known that if R is an abelian regular self-injective ring then (R=R_1times R_2), where (R_1) is a Boolean ring and each element of ring (R_2) is a sum of two units [21].
Vámos in [29] proved that every element of a right self-injective ring R is a sum of two units if R has no non-zero corner ring which is Boolean.
If you flipped a switch to turn 50 of those strings on with some unit value, leaving the rest off, and joined them at the end and saw the combined final result of the power, you would have completed something similar to an addition operation leading to a sum of 50 units.
Using ideas that were implicit in [20], it was shown in [11] that this ring decomposition can be streamlined further and shown that if R is a von Neumann regular right self-injective ring then (R=R_1times R_2), where (R_1) is an abelian regular self-injective ring and each element of the ring (R_2) is a sum of two units.
That is, a (overline{Delta m/z}) value of 58.042 Th may as well suggest the occurrence of a propylene oxide unit instead of a sum of two different units—and propylene units are known to exist for homologue series [52].
A primary score was calculated from a sum of the "story unit scores".
