How to explain 1000 - 1 using the american method

Question

I have developed a small Android app. The app is a pocket calculator for addition and subtraction, in which the calculation is explained via both video and audio. If it is unclear please have a quick look at the video I made that explains it.
Of course I understand subtraction using the American method, but I am having a hard time explaining the idea of borrowing over more then one digit. For example 1000-1, already in the first step, I need to borrow a 1 from all the other digits. For me that is a recursion, but how can I explain that in my app? My target users are school kids and their parents.

Answer 1

The first three digits of $1000$ represent $100$ tens. If you regroup that as $99$ tens and $10$ units,then you can subtract $1$ unit from the $10$ units and get $99$ tens and $9$ units, i.e. $999$.

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Edited: I'm going to substantially expand on my answer above.

I think most school-aged children are taught to execute the algorithm by "borrowing" or "regrouping" from one digit to another, one at a time: from thousands to hundreds to tens to ones. I've illustrated this with the GIF below.

But the problem becomes much simpler if, instead, you aggregate the first three digits together (as I described in my response originally). Then the process looks like this:

Answer 2

I'll take a stab at it. What does $1000$ mean? It is a decimal representation: $$ 1000 = 1 \cdot 10^3+0 \cdot 10^2+0 \cdot 10^1+0 \cdot 1$$ This is hard to work with, however, if we notice $1000=999+1$ then we have: $$ 1000 = (9 \cdot 10^2+9 \cdot 10^1+9 \cdot 1)+1$$ Furthermore, if we are willing to violate the purist's form of the decimal representation then we can just as well write (this is "carrying the one"): $$ 1000 = 9 \cdot 10^2+9 \cdot 10^1+10 \cdot 1$$ Finally, to subtract $1$ we can subtract $$ 1 = 0\cdot 10^2+0 \cdot 10^1+1 \cdot 1$$ from $1000$ as follows: $$ \begin{align} 1000 &= 9 \cdot 10^2+9 \cdot 10^1+10 \cdot 1 \\ 1 &= 0\cdot 10^2+0 \cdot 10^1+1 \cdot 1 \\ \hline 1000-1 &= 9 \cdot 10^2+9 \cdot 10^1+9 \cdot 1 \end{align} $$ Which gives $1000-1=999$.

The procedure of carrying can always be understood in the way I indicate here. If we have to subtract too much from a particular digit then we can borrow from higher digits by taking one of the first larger digit and breaking it up for the smaller digits. Of course, there are more cases I have not discussed...

Maybe you can see how to twist my work here into something for your app?

A second view: my wife had this comment. How to explain to children? My way is wrong. Instead, when faced with the question $1000-1$ the story goes as follows:

1. We want to subtract $1$ from the one's place in $1000$ but the one's place has nothing so it has to borrow $1$ from the ten's place which is $10$ of the one's place.
2. Unfortunately, the ten's place also has nothing so it has to borrow $1$ from the hundred's place which counts as $10$ in the ten's place. But, the ten's place already gave $1$ to the one's place so it only has $9$.
3. Yet again, the hundred's place has nothing so it has to borrow $1$ from the thousand's place which gives $10$ hundreds. Again, the hundred's place already gave $1$ to the ten's place so only $9$ remains.
4. Finally, the thousands place only had $1$ and since it gave it up the hundreds there is nothing more to do.

Perhaps we could think of this as a series of banks from which each level can only borrow from the level immediately above.