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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.

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  • $\begingroup$ Thanks for the answers. I have learnt the Austrian method when I was little, Bavaria had switched to the American method something like 20 years ago. This was discussed here in Germany quite a bit. I didn't have an opinion until I made this App, but now I must say, the Austrian method is "better". This recursive borrowing thing, regardless if it is a hidden repeat until, is quite hard to understand for kids. Btw. my App got approved in Google Play for Education, so chances are good that some kids are learning subtraction with your input, next year. I will roll out an update soon. $\endgroup$ – user1704369 Jul 16 '14 at 22:28
  • $\begingroup$ The result of subtracting one from X is the number that comes just before X. The number that comes just before 1000 is 999. Any explanation requiring a "method" (rather than simply relying on understanding of countring) seems to me perverse as a manner of solving the particular problem $1000 - 1$. $\endgroup$ – Dan Fox Jul 18 '14 at 11:01
  • $\begingroup$ Dan, I am not sure about your point. Subtraction using the American method is what kids learn at shool. So to me it seems that the American method is not negotiable. I have used 1000-1 as an example for the odd behaviour of that algorithmus, when it comes to borrowing over multiple digigits. $\endgroup$ – user1704369 Jul 19 '14 at 12:46
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    $\begingroup$ Teaching a student to subtract 1 from 1000 by using a formal algorithm is potentially even damaging. It is substituting mindless computation for a tiny bit of thought. If the goal is to teach a method for subtraction, then subtraction of 1 is not a good example, because no method should be needed. One needs to learn to recognize when to use complicated general machinery and when it is not necessary to do so. Making a kid subtract 1 from 1000 by carries is taking a simple thing and making it difficult. $\endgroup$ – Dan Fox Jul 21 '14 at 8:27
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    $\begingroup$ Just because a method is what is typically taught does not mean that it is reasonable. I'm with Dan. Students should not just know algorithms, they should know when and why they are useful. A much better solution to 1000 - 1 is to ask yourself, what number comes before 1000? $\endgroup$ – David Wees Jul 22 '14 at 10:53
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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. $999$.


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. enter image description here

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: enter image description here

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    $\begingroup$ Hm, doesn't this cause the question, how to explain 100-1? $\endgroup$ – user1704369 Jul 16 '14 at 21:23
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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.

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  • $\begingroup$ Thanks to your wife, this is much better then what I had. However, we replace the recursion with a kind of a repeat until. $\endgroup$ – user1704369 Jul 16 '14 at 22:17
  • $\begingroup$ My App actually supports up to five digits. May I ask if you could provide a few more of this nice explanations? I like the idea of not saying the same over and over again. An other problem of mine is the limited english on my side. $\endgroup$ – user1704369 Jul 16 '14 at 22:59
  • $\begingroup$ I think @mweiss has nicely shown the way to think about it now. If you still need me to write more then let me know. $\endgroup$ – James S. Cook Jul 17 '14 at 5:12

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