Standard algorithms

I’m working on a couple of short videos comparing the standard algorithm for a multiplication and addition, and considering some ways of using other algorithms which are more likely to make sense for students.

To be clear, these presentations do not enough justice to student-created algorithms, which I would strongly recommend as a starting place for exploration of algorithmic reasoning. These two videos are merely an attempt to compare two different algorithms against each other, and to provide some support for teachers interested in learning more about why we might want to use something other than the standard algorithms in our classrooms.

I also want to make it clear that how I am describing these distinctions in the videos is not how I would introduce them to children (or adults for that matter, if I have the time). Instead, I would start with setting up realistic situations, whole group, and small group discussions around investigating these operations, using both manipulatives and symbols to represent numbers in the algorithms. A really useful activity, for example, is for children to be making these comparisons themselves, so that they can look for patterns in different operations, and abstract these patterns into general rules they can follow to make their use of any algorithm easier.

 I see these algorithms as a useful way to get started with abstract reasoning, provided they are framed in a certain way, as described below.

 

Multiplication

 

Addition:

 

 

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4 Comments

  • A. Stokke wrote:

    Hi David,

    I just watched your first video and have a couple of comments.

    The distributive law is as follows: a(b+c)=ab+ac or (b+c)a=ba+ca (where a,b,c are real numbers, but for the discussion here we’re talking about whole numbers, or counting numbers). This is what is used in the standard algorithm. This is easy to explain using rows of pennies. If you have 4 rows and 12 columns of pennies, you can figure out how many pennies you have by considering how many pennies there are in 4 rows and 10 columns (40) and in 4 rows and 2 columns (8) and then adding the two numbers.

    From the distributive law follows this corollary: (a+b)(c+d)=a(c+d) + b(c+d)=ac+ad+bc+bd, which involves two applications of the distributive law. This is what you used in the second algorithm.

    If you were to teach kids 2-digit by 2-digit multiplication without first teaching 2-digit by 1-digit this would all surely be very confusing so I’m assuming you would thoroughly cover the 2-digit by 1-digit case first (and probably 3-digit by 1-digit, etc). The 2-digit by 1-digit case is an immediate application of the distributive law:
    (23 x 7)=(3+20)x7=3×7+20×7=…=161.

    The standard algorithm gives us a shortcut for keeping track of our calculations: we have 3 ones times 7 ones is 21 ones – that’s 1 one and 2 tens, which is why we carry a 2 to the tens columns. (I realize you know all this but I think when we’re teaching this algorithm we need to say these things instead of just “carry the 2” – at least at first, when students are just learning.) Now we have 2 tens times 7 tens which is 14 tens + 2 tens, is 16 tens, which is 1 one hundred and 6 tens. Thus we get 161. You are correct that the standard algorithm is often taught by saying things like “put down your 1 and carry your 2”, but we do not have to teach it that way. Instead of arguing against standard algorithms, I prefer to argue that standard algorithms should (absolutely) be taught but they should be taught properly.

    Once students understand the 2-digit x 1- digit case, using the standard algorithm, we can move onto the 2-digit by 2-digit case, which contains within it an application of the 2-digit by 1-digit case.

    A natural conclusion to come to using the distributive law is as follows: 23+17=23(7+10)=23 x 7 + 23 x 10=161+230=391, which is exactly the form we use with the standard algorithm. To find 23 x 7, we build on our knowledge from the 2-digit by 1-digit case and use the standard algorithm for 23 x 7 within the 2-digit by 2-digit case.

    You are right that students often don’t understand the standard algorithm but that’s because it is often not taught properly, not because it’s easier to see the distributive law in the second algorithm. In fact, if both were taught properly, illustrating where the distributive law is used, I would argue that it’s easier to understand how the distributive law is used in the first (standard) algorithm.

    As for the picture, a similar (and much simpler) picture can be used to describe 23 x 17 as 23 x (7 + 10)= 23 x 7 + 23 x 10=161+230=391, in the same way I mentioned above. If you have 23 rows and 17 columns of pennies, you can break this up into two pieces: 23 rows and 7 columns and 23 rows and 10 columns – in other words 161 pennies plus 230 pennies = 391 pennies. Your picture will then be split into two boxes (one for 23 x 7 and one for 23 x 10) instead of 4 boxes.

  • David Wees wrote:

    I’m still on the fence as to the utility of the standard algorithms when the slight variation I described is only slightly slower, and allows for much easier error checking. I’m obviously familiar with the distributive rule, so I assume that you included it as an aside to the audience in our discussion. That being said, while I can obviously see how the distributive rule is applied in the standard algorithm, I’m really not clear on why you think it is more obviously present in the standard algorithm than in my counter example (where the result of each multiplication that occurs in the distributive rule is visible, and easily verified for accuracy).

    We both agree that how something is taught makes a difference as to how it is understood. My argument is that the variation I described is more intuitive, and that students may even start to simplify some of the steps when doing it, in much the same way as the standard algorithm does, but that they will do so with understanding.

  • A. Stokke wrote:

    I should have written “Now we have 2 tens times 7 ONES which is 14 tens” in my comment above. I accidentally wrote 2 tens times 7 tens.

  • David Wees wrote:

    I won’t hold it againt you. 🙂 We all make mistakes, the frustrating thing about most web commenting forms is we have very little ability to correct them.

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