Subtraction from a number, means to take one number from another and look at what is left over. We always either take the lesser number from the greater, so that there is some remainder, for example three from five leaves two, or we subtract equal from equal, where the remainder is nothing; for example when we subtract three from three. It is not possible to take a greater from a lesser number, for it is not possible to take away what is not there.

If you now want to perform a subtraction, you proceed as follows:

Write the digits in turn as many and whichever you like. Below it write the same number of digits or less, but not more. If the number below which is being subtracted has fewer digits then no other condition need be stated. But if it has the same number of digits as the top number then make sure the last digit of the top row is greater than the last corresponding digit of the bottom row. As stated previously, when I speak of the 'last digit', I mean the one on our left hand side. We need this digit to be greater than the bottom one to ensure that the whole number (on top) is greater than the whole number (on the bottom). This being so, even if the rest of the digits in the bottom number are greater than the digits in the top number, taken one at a time, nevertheless the top number is greater and in no way smaller than the bottom one, that is, than the number being subtracted.

Suppose then that the numbers are written with digits corresponding, units with units, tens with tens and so on in turn. If the first digit^{22} on the bottom row is less than the first digit of the top, then take the lesser from the greater and write the number remaining above the first digit on the first row. If it is equal, then subtract equal from equal and, as stated above, write the remaining zero above. If it is greater, since we cannot take the greater from the lesser, borrow a unit which signifies ten, from the next digit after it, that is from the second digit of the bottom row. (This digit after the first place occupies the second decadic position.) Having added this ten to the smaller number in the top row, subtract the larger from that total and write the remainder again above the smaller digit. If you can take the second digit on the bottom line, after adding the unit to it - for this unit is again thought of as a unit in respect to the number in the corresponding column, but as ten in respect to the preceeding column, since every number is taken to be ten times the number in the preceeding column, viz; ten compared to the unit, a hundred to ten, a thousand to a hundred and so on - if then you can take away this second digit, after adding to it the unit, then take it away and write the remainder, if there is any, above that second digit on the top row, but if not then write 0. Again, if the second digit on the bottom row, with the unit, is greater than the second digit on the top, again borrow a unit, that is a ten, from the third column on the bottom and add the ten to the second digit on the top and take away the second digit on the bottom, with the unit, and again write the remainder above the second digit on the top. Proceeding thus to the end, you will have the desired result. For the number remaining after you take the whole of the lesser number from the whole of the greater is precisely the number written above the top row.

So that what is being said might become clearer to you in an example, let it be as follows:^{23}

I wish to take 3 from 2 but I can't, since 3 is greater than 2. I add a unit to the 4 next to the 3. I regard the unit as a ten^{24} and I say 10 and 2 (make 12). I subtract 3 from 12 leaving 9. I write this number above the 2. Again I wish to take 4, along with the unit, from 1 , but am unable. I add a unit to the 8 after the 4 and, regarding this as ten, I say ten and 1 is 11. \[\begin{array} {|ccccc|} \hline 5 & 4 & 6 & 1 & 2 \\ 1 & 8 & 7 & 6 & 9 \\ \hline 5 & 4 & 6 & 1 & 2 \\ 3 & 5 & 8 & 4 & 3 \\ 1 & 1 & 1 & 1 & \\ \hline \end{array}\]

I take from 11 the 4 with the unit, that is 5, leaving 6 and I write this above the 1. Using this method I reach the last column and since I can subtract 3 with a unit from 5, I subtract from it 4, that is 3 with a unit, leaving a unit. This number I write above the 5.

With this in mind another example is given, in order to show that if the remainder is zero, it is still written. Beginning in the third place in the accompanying example,

\[\begin{array} {|ccccc|} \hline 3 & 1 & 2 & 5 & 6 \\ 0 & 7 & 0 & 8 & 8 \\ \hline 3 & 1 & 2 & 5 & 6 \\ 2 & 4 & 1 & 6 & 8 \\ 1 & 1 & 1 & 1 & \\ \hline \end{array}\]

after adding the unit to the digit thus making 2, I can subtract this completely from 2 and nothing is left. I write 0 above the two. I continue in the same way to the end of the diagram.

A further example will show how the calculation is carried out when there are fewer digits on the bottom line.^{25}

\[\begin{array} {|ccccc|} \hline 6 & 4 & 5 & 4 & 3 & 2 \\ 6 & 3 & 8 & 6 & 7 & 4 \\ \hline 6 & 4 & 5 & 4 & 3 & 2 \\ & & 6 & 7 & 5 & 8 \\ & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}\]

Going to the 4th place, since I am unable to take the 6 with the unit added, that is 7, from the 5 I write a unit next to the units under the 4 and I regard this as 10. Having added this to the 5, I proceed as previously explained. After this I take the unit from the 4 leaving 3 and I write this above the 4. If there are two, three or more digits less in the bottom than the top^{26} a unit is not written below those digits from there to the end along the top row, but the digits are written as they are, each one written above itself in turn after those digits written above and included with the remainders.^{27}

If you wish to have a check, proceed as follows. Suppose in the last example, we say 8 and 4 is 12 and take the unit from it leaving 2. Write this above the 4. Again, 5 and 7 is 12, add in the unit you carried and we see 13, take away 10 and again carry the unit leaving 3. Write this above the 7. Again 7 and 6 is 13, adding in the unit makes 14. Then take away 10 and carry the unit leaving 4. Write this above the 6. Again 6 and 8 is 14 and adding in the unit makes 15. Take 10 amd again carry the unit leaving 5. Write this above the 8. Then since we have nothing further from the bottom line of our initial two rows, take the next digit in turn of those left, that is 3, and again add in the unit which you held making 4. Write this above the 3. Now since 6 has nothing to combine with^{28} nor is there a unit being carried to add to it, place this 6 as it is above itself. Now the topmost row turns out to be the same as the top row of the two initial rows, and the subtraction was performed correctly. Indeed this is always the method by which the check is made, that is, to add the lower of the two initial rows to the row of remainders. Check that the row resulting from this is the same as the top row of the initial two. If so, then the subtraction has been correctly done.