Let the number A be a part of BC, and another number D be the same part of another number EF that A is of BC.
I say that the sum of A and D is also the same part of the sum of BC and EF that A is of BC.
Since, whatever part A is of BC, D is also the same part of EF, therefore, there are as many numbers equal to D in EF as there are in BC equal to A.
Divide BC into the numbers equal to A, namely BG and GC, and EF into the numbers equal to D, namely EH and HF. Then the multitude of BG and GC equals the multitude of EH and HF.
And, since BG equals A, and EH equals D, therefore the sum of BG and EH also equals the sum of A and D. For the same reason the sum of GC and HF also equals the sum of A and D.
Therefore there are as many numbers in BC and EF equal to A and D as there are in BC equal to A. Therefore, the sum of BC and EF is the same multiple of the sum of A and D that BC is of A. Therefore, the sum of A and D is the same part of the sum of BC and EF that A is of BC.
Therefore, if a number is part of a number, and another is the same part of another, then the sum is also the same part of the sum that the one is of the one.
If a number a is one nth of a number b, and if d is one nth of e, then a + d is one nth of b + e. As a single algebraic equation this says
The sample value taken for 1/n in the proof is 1/2.
Although this proposition is only stated for the sum of two numbers, it is used for sums of arbitrary size.
Suppose that a is one nth of b and d is one nth of e. Then b is a sum of n a’s while e is a sum of n d’s. Therefore, b + e is a sum of n (a + d)’s. Therefore, a + d is one nth of b + e.
Note that there are no justifications that we can give for the steps in the proof. Even the definition of number VII.Def.2 is not relevant. The argument is completely based in principles of informal addition, namely unrestricted commutativity and associativity of addition where the number of terms n is unrestricted.
In the middle of the proof it is shown that n(a + b) = na + nb. In particular, when b = a, that says, n(2a) = 2(na). As the proposition is used for more than 2 terms, and the proof works as well for m terms, it can be used to show n(ma) = m(na), which is done in VII.16, the proposition preceding the statement for commutativity of multiplication of numbers.
Thus, the foundations of formal number theory in the Elements are principles of informal number theory.