The addition of the units’ line, or 8 + 4 + 3 + 7 + 6 + 5, gives 33, that is, 3 tens and 3 units. Put 3 in the units’ place, and add together the line of tens, taking in at the beginning the 3 tens which were created by the addition of the units’ line. That is, find 3 + 0 + 7 + 2 + 3 + 0, which gives 15 for the number of tens; that is, 1 hundred and 5 tens. Add the line of hundreds together, taking care to add the 1 hundred which arose in the addition of the line of tens; that is, find 1 + 0 + 4 + 7 + 8, which gives exactly 20 hundreds, or 2 thousands and no hundreds. Put a cipher in the hundreds’ place (because, if you do not, the next figure will be taken for hundreds instead of thousands), and add the figures in the thousands’ line together, remembering the 2 thousands which arose from the hundreds’ line; that is, find 2 + 2 + 1 + 9 + 1, which gives 15 thousands, or 1 ten thousand and 5 thousand. Write 5 under the line of thousands, and collect the figures in the line of tens of thousands, remembering the ten thousand which arose out of the thousands’ line; that is, find 1 + 1, or 2 ten thousands. Write 2 under the ten thousands’ line, and the operation is completed.

34. As an exercise in addition, you may satisfy yourself that what I now say of the following square is correct. The numbers in every row, whether reckoned upright, or from right to left, or from corner to corner, when added together give the number 24156.

2016 4212 1656 3852 1296 3492 936 3132 576 2772 216
252 20524248 16923888 13323528 9723168 6122412
2448 2882088 42841728 39241368 35641008 2808 648
684 2484 324 21244320 17643960 14043204 10442844
2880 7202520 3602160 43561800 36001440 32401080
1116 2916 756 2556 396 21963996 18363636 14763276
3312 11522952 7922592 362232 40321872 36721512
1548 33481188 2988 432 2628 72 22684068 19083708
3744 15843384 8283024 4682664 1082304 41041944
1980 37801224 3420 864 3060 504 2700 144 23404140
4176 16203816 12603456 9003096 5402736 1802376

35. If two numbers must be added together, it will not alter the sum if you take away a part of one, provided you put on as much to the other. It is plain that you will not alter the whole number of a collection of pebbles in two baskets by taking any number out of one, and putting them into the other. Thus, 15 + 7 is the same as 12 + 10, since 12 is 3 less than 15, and 10 is three more than 7. This was the principle upon which the whole of the process in (29) was conducted.

36. Let a and b stand for two numbers, as in (24). It is impossible to tell what their sum will be until the numbers themselves are known. In the mean while a + b stands for this sum. To say, in algebraical language, that the sum of a and b is not altered by adding c to a, provided we take away c from b, we have the following equation:

(a + c) + (b - c) = a + b;

which may be written without brackets, thus,

a + c + b - c = a + b.

For the meaning of these two equations will appear to be the same, on consideration.