Conjectures in the Math Class



We have been practicing mental math in grade 5/6.  We are working on ways to divide by one digit divisors in our heads.  A great place to start was dividing by two.  The kids realized, very early on, that when you divide a number with an odd number in the ones column, you will get a remainder (57, 31, 77).  Most of them were using a chunking or splitting strategy.

Forgot to double the remainder...


84 ÷ 2

is the same as

80 ÷ 2 = 40
÷ 2 = 2

40+2 = 42

Today, we moved onto dividing by four.  After an initial discussion on the differences, we were all the same page about how to do it.  Dividing by 4 was dividing by 2 and then dividing by 2 again.  I wrote a couple of problems on the board and asked them to try.

Something wasn't fitting with their previous knowledge this time.  One chlid noticed that when he did 54 ÷ 4 he got a remainder.  Why?  I thought that when we divide by an even number we get no remainder?

Note: My role during what proceeds was to capture what they were thinking on the whiteboard, and to ask for clarification.  The ideas presented came out of the group, though I had a very powerful position in that group. I am also doing my best to remember what was said, I am writing this a couple of hours after it happened.  None of this was planned.  I had originally thought that we would do games to practice mental math, dividing by 4.  However, when a student presents a conjecture, I cannot let it go.

Conjecture #1

Another child put forth a conjecture.  If the dividend has an odd number in the tens place and an even number in the ones place, there would be a remainder (when dividing by four).

They set to work to try and disprove this theory.  It wasn't long until somebody found a number that goes against this.

36 ÷ 4 = 9

3 in the tens place is odd, 6 in the ones place is even.  No remainder.  Conjecture busted.

(At this point I wanted to push them in the direction of why this disproved the conjecture.  I was hoping for them to see that because this was a factor of four, it had no remainder, and all numbers that were factors of four would follow this pattern.  However, another conjecture was presented, so we went in that direction instead)


Feb 15 dividing by 4



Conjecture #2

Once the previous conjecture was disproven, another child immediately put forth a new one.  If the dividend has an odd number in the tens place, and either 4 or 8 in the ones place, there will be a remainder (when dividing by 4).

Again we set out to disprove it.  Though this time it was a little more difficult.  Everybody was doing their own thing, selecting random numbers that followed this pattern and trying to disprove this conjecture.



Somebody took charge and made a list of all the numbers that fit in this set (due to the nature of the conjecture, it was confined to numbers between 1-100).


Now they had a frame of reference and they quickly went through all of them and found the conjecture to be true.  Each one of these numbers (when dividing by 4) had a remainder.

Now, I took over and played the teacher role asked them to think of why this was true.  How can we explain this?

Proving it

We started to think of why this was the case.  There was, literally, three minutes of silence while everybody stared at the white board, deep in thought (or not, how am I to know?).  Finally, I heard a little squeak from a rather quiet child.

S: "Because of the 5 in the ones column."
T: "Can you tell us more about that?"
S: If you have a number in the tens column that is odd, when you divide it by 2, you will get a number with a 5 in the ones column.


This kicked off some thinking.  

S:  "if you divide all the numbers in the ones in half, the biggest number you can get is 4 (from 8), or 4 and a remainder from 9."
S:  "these are added to the five we already have in the ones column (from the division of the odd tens column)."
S:  "So, if you add an even number from the ones column to the 5, you will get an odd number, but if you add an odd number you will get an even number. 

for example

5 + 1 = 6 
5 + 3 = 8
even numbers

5 + 2 = 7
5 + 4 = 9
odd numbers

S: "and when you divide 4 by 2, you will get 2, and when you divide 8 by 2 you will get 4.  They are both even numbers, which means you will get an odd number when added to 5, which means you will have a remainder."
S: "what about 2 and 6?  They are even numbers as well."
S: "No, because 2 divided by 2 is 1, which is odd.  And 6 divided by 2 is 3, which is also odd.  And anything divided by 1, 3, 5, 7, or 9 can't be divided evenly.  
S: "So, the only way we could get an odd number (because that will give us a remainder) is by adding 5 to an even number, and the only way you can get an even number is by dividing 4 or 8 by 2.




At this point they were pretty happy with what they had done.  They weren't quite sure what they had done, or why they had done it, but they knew they accomplished something.  I asked them to reflect on how they were feeling:

- I think I learned something a new about numbers
- I don't know if this will help me do division, but it was fun
- We had to think really hard
- It was interesting to listen and understand, even though I didn't say anything

Another Way with Patterns

During recess another student presented a different way to look at.  She said she didn't want to share it with the class because she was afraid of being wrong.  I respected her decision and let her explain her way to me.  She showed me a list of all the factors of four.  She noticed a pattern.  When the tens digit was even, the ones digit were either 0 or 4 or 8.  But, when the tens digit was odd (as per the conjecture) the ones digits were 2 and 6.  Therefore, all numbers with an odd numbers in the tens and 4 and 8 in the ones, could not be factors of 4.  That is why it works.











Comments

  1. I asked the child (afterwards) about why she presented conjecture #2. She could not explain why those numbers came to her. She said it was a feeling....

    Interesting.

    ReplyDelete

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