# Why does the “borrow” algorithm for subtraction work? [duplicate]

I would like to understand how the borrow algorithm for simple subtraction ( base 10, the one we learn when we are kids ) works, does anyone have any material on this? Thank you.

## marked as duplicate by MJD, Community♦Sep 7 '16 at 18:40

• Perhaps you first should ask why the subtraction algorithm itself works? Why is $392-141$ the same as $100\cdot(3-1)+10\cdot(9-4)+1\cdot(2-1)$? Once you understand that, seeing how borrowing fits into this is not too hard. – Arthur Sep 7 '16 at 17:16

Let's consider the borrowing in this example:

$$\begin{array}{rrrr} &3&4&2\\ -& &5&8\\\hline \end{array}$$

We can't take 8 from 2, so we "borrow" one of the 4 tens from 342, and make it into 10 units instead, and add it to the 2 units we already have. That leaves 3 tens in the tens' column, and 12 units in the units' column:

$$\require{cancel} \begin{array}{rrrr} \quad &3&\color{darkred}{\cancelto{3}4}&\color{darkred}{{}^1}2\\ -& &5&8\\\hline \end{array}$$

What we did here is like making change: We had 4 dimes and 2 pennies, but that was not convenient because we wanted to take away 8 pennies. So we changed one of the dimes into pennies, and then we had 3 dimes and 12 pennies, and it was easy to take away 8 pennies. This "borrowing" trick is just like making change.

Then we take 8 units from 12 units, leaving 4 units: $$\require{cancel} \begin{array}{rrrr} \quad &3&{\cancelto{3}4}&{{}^1}2\\ -& &5&8\\\hline &&&\color{darkred}{4}\end{array}$$

Now we want to subtract 5 tens from 3 tens, but, again, we can't. So we do almost the same as before, but this time we "borrow" one of the hundreds from the 3 hundreds in the hundreds' column, and turn it into 10 tens, adding it to the 3 tens that are already in the tens' column:

$$\require{cancel} \begin{array}{rrrr} \quad &\color{darkred}{\cancelto23}&\cancelto{\color{darkred}{1}3}4&{}^12\\ -& &5&8\\\hline &&&4 \end{array}$$

Again, this is like making change: we had 3 dollars and 3 dimes, and we changed one dollar into ten dimes, leaving 2 dollars and 13 dimes. That is allowed in arithmetic for the same reason it is allowed in money dealings: it doesn't change the actual amount!

Anyway subtracting 5 tens from 13 tens we have 8 tens:

$$\require{cancel} \begin{array}{rrrr} \quad &\cancelto23&\cancelto{13}4&{}^12\\ -& &5&8\\\hline &&\color{darkred}{8}&4 \end{array}$$

and then 2 hundreds minus no hundreds is 2 hundreds:

$$\require{cancel} \begin{array}{rrrr} \quad &\cancelto23&\cancelto{13}4&{}^12\\ -& &5&8\\\hline &\color{darkred}{2}&8&4 \end{array}$$