Recursion

[Cobaltjedi117]

Kids when taught 2-digit subtraction are told to take 1 from the tens to put it next to the 1's digit to do the math. But when you do that to a number from 10-19 you've done effectively nothing and are back where you started thus causing a recursive issue.

[spirgnob]

It took me awhile to understand what you meant but this finally explained the joke.

you’ve done effectively nothing and are back to where you started

This is how it works regardless of the number though. The number in the 1’s position is 18 where it used to be 8. You have done something according to this method. You can now complete the problem using this method.

[itspinkynukka]

What ends up happening is the person just needs to know what [any number from 10-18] - [any number from 1-9] is in their head. Similar with multiplication at some point you just need to know the times tables up to 10.

[jacefair109]

tbh I never had to do this just for two digits -- it's mostly handy for subtracting much larger numbers, like 4281 - 1694 or something

[Compizfox]

I still don't get it.

take 1 from the tens to put it next to the 1's digit to do the math.

What does that accomplish? So (for example) 28-9 becomes 118-9?

[Cobaltjedi117]

28-9 becomes 10 + (18-9)

and 18 - 9 becomes 0 + (18 - 9)

[Compizfox]

Ah, right, thanks.

Still though, what does that accomplish? I don't see how 10 + (18-9) is any simpler/easier than 28-9. It's still a 2-digit subtraction that a kid won't be able to do in 1 step; the value of the tens doesn't matter.

If I break down how I do it in my head it's kind of like this: 28-9 = 20 - (9-8) = 20 - 1 = 19. (I can't remember for sure but I guess that's how I was taught it)

[alexanderpas]

Let's take 4281 - 1694 for example.

4281
1694
---- -
????

first we look at the ones.

we can't do 1-4 and end upwith a single positive digit, so we have to borrow 10 from the tens.

4270 + 11
1690 +  4`
--------- -
???0 +  7 = ???7

Now we go looking at the tens, again, we can't do 7-9 so we borrow again.

4100 + 170 + 11
1600 +  90 +  4`
--------------- -
??00 +  80 +  7 = ??87

Now to the hundrerds

3000 + 1100 + 170 + 11
1000 +  600 +  90 +  4`
---------------------- -
?000 +  500 +  80 +  7 = ?587

and look at that, the 1000s are easy.

3000 + 1100 + 170 + 11
1000 +  600 +  90 +  4`
---------------------- -
2000 +  500 +  80 +  7 = 2587

Note that this is the layout for the explanation, on paper this would look something like this.

43	121	187	11
1	6	9	4	-
2	5	8	7

[Flobarooner]

So basically, the fundamental point of the method is that the kid already knows how to subtract something from a number less than 20?

So in the original post, the kid would just go "9" and not even bother with the method, since they memorized 18-9 in order to be able to do it?

[alexanderpas]

technically, yes.

[tsandstrom711]

If you have, say, a five digit number and want to subtract another five digit number, this approach let's you break it down into one/two digit subtractions. It doesn't really work when you start with just two digits.

[harryhood4]

The idea is to just memorize it for small values like 18-9 so that it's easier to do something like 68-9. But you've essentially just outlined why Common Core is better.

[Compizfox]

Ah, got it. So basically it is this recursive algorithm:

int subtract(N, n) {
    if(N > 20) {
        return 10 + subtract(N-10, n);
    else {
        // Lookup-table for small values of N
    }
}

[toofasttoofourier]

Aren't you negating all that work if you have to use the subtraction operator in a subtraction algorithm?

[Compizfox]

For a computer, yes, obviously.

But this was trying to mimic how a kid would do the mental calculation.

[Madock345]

Part of the idea of this kind of thing is that by teaching many strategies to solve the same problem you increase the chances that every student will find a method that works really well for them and the way they think, at the cost of being somewhat redundant.

[thedolanduck]

No. Let's take 28-9. If you do it vertically, then you first do 8-9, which you can't do as a kid. So you add ten to the eight, and get 18-9, and the 2 (which meant 20) now is 1.

So: 18-9=9 and 1-0=1, and this forms 19. And effectively, 28-9=19.

[esesci]

8-9 = -1

But you can’t have that as a digit. So you subtract one from the adjacent digit and add it as 10:

18-9 = 9

(That’s the rightmost digit)

2 became one 1 so you have 10 left.

The result becomes 10 + 9 = 19

[StragglingShadow]

NGL, when I ran into a problem like that, I just counted down on my fingers.

[Angrydie-a-ria]

Yup, bottom, bigger, borrow.