# ZeroSum Ruler (home)

## Blogging on math education and other related things

### The Language of Math PosterAugust 19, 2011

Below is a poster I hang in my classroom every fall.  Each year it grows longer as more and more terms come up for the different operations of math.  When I was a kid, no one told me to look out for these words, or that math was even a language at all, which made word problems pretty tough.  By clicking on the poster you will be sent to the original Excel file on Google Docs.  Do you have any words to add?

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### The Coolest Math Class Ever. Seriously!June 6, 2011

I was lucky to get into Oliver Knill’s class Teaching Math With a Historical Perspective, one of the choices within the Harvard Extension School’s ALM in Mathematics for Teaching program.  It changed the way I think and the way I teach.  He explained complex topics, such as code breaking and non-verbal proofs, with such ease.  He was inspiring and made me look at math in a way I had never before – from a historical perspective!  His site is worth checking out, especially if you can not get into his class!  (Click on the red circle to go to Oliver’s site)

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### The long division algorithm, explained at lastMarch 29, 2011

We’ll stop taking long division for granted in a minute, but first, let’s take a general look at division.  What is division?  Division is the shortcut for subtracting one value from another value repeatedly until we reach zero.  We say that “90 divided by 10 is equal to 9″ because 90-10-10-10-10-10-10-10-10-10 = 0.  Using the distributive property, we can rewrite this as 90 – 9(10) = 0.

If we then made steps to solve this equation, we’d get:

90 – 10(9) = 0

+10(9)    + 10(9)

90            =  10(9)      90 is “ten, nine times”.  Division is the inverse of multiplication.

But division also works in situations where we would not eventually subtract to end with zero.  For example, we may end with the number 5.  “95 divided by 10 is equal to 9.5″ because 90-10-10-10-10-10-10-10-10-10-5, and that last 5 is one-half (.5) of the 10 we had been subtracting from 90.  This 5 is also known as a “remainder”.  The result of “95 divided by 10” can then be written as “nine with a remainder of 5”.  This remainder is a smaller number than the number being repeatedly subtracted, so does not quite fit in (“remains” outside).

95 – 10(9) = 5

+10(9)     + 10(9)

95            = 5 + 10(9)

95            = 10(9) + 5     [commutative property]

95 is “ten nine times with five left over”

However, subtracting repeatedly can be tedious for large numbers, such as “950 divided by 10”, so the long division algorithm was developed.  Formatting “950 divided by 10” to the algorithm, we would write:

The sideways “L” is referred to as the “division bar” in this paper.

Here, the decimal can be expanded to 950.00 because of the definition of our number system.  This also makes room for any possible “remainders”, or numbers less than 10 that are left over once we subtract all the 10’s from 950 that we can.

To work this algorithm, we want to first ask ourselves, “how many 10’s can be subtracted from 9?”  We want to remember both: that our definition of division is “repeated subtraction” and that “multiplication is division’s inverse”.  10×9 is 90, not “9”.  We’d answer with “zero” and imagine a “0” above the 9 in 950.  Here, we’ll put in a zero, but usually we wouldn’t.

Usually, when the number above the division bar is not zero, we would move to the second step of the algorithm.  But since our “0” is imaginary, we won’t move to step two quite yet.

Starting the algorithm over, we’d then move in from the “9” and look at “95” and ask ourselves, “How many 10s cam be subtracted from 95?”  We’d answer with “9” and place it above the tens spot in 950.

Our placement of the “9” actually represents a “90” as it is vertically aligned with the 950’s tens placement.  Had we placed this “9” over the 9 in 950, it would have represented a 900.  By placing our “9” over the 5 in 950, we’re saying that “ninety 10s can be subtracted from 950”, which we know to be true.  If you notice the placement of this “90”, you will see that its 9 is vertically aligned with the 950’s hundreds placement.  Does this make sense? (remember the definition of division as multiplication’s inverse!)

Continuing to step two of the algorithm, we write this “90” below the “95” to find the next remainder, remembering that this “90” actually represents a “900” because of its placement.

We are left with “5”, which is actually a “50” because of its placement below the tens spot of the 950.  The “5” in the original 950 is in the tens spot, therefore our remainder of 5 is actually a remainder of 5×10, or 50.

The algorithm then tells us to “bring down” the next digit in the original 950 and repeat the algorithm.  But what is the purpose of this “bringing down”?  We always need to keep place value intact and our “5” is really a “50”.  By bringing down the “0”, we accomplish this.  Also, remembering that our “9” is a 90 because of place value and that division is repeated subtraction, we have:

950 – 10(90), which yields 50 (not 5)

We bring down the “0” from 950 and place it next to our 5 remainder to make a “50”, creating the 50 we already knew to be true.

If that remainder of “5” had been a “0”, we’d be moving into the ones place.  Because it is a “5” and aligned in the tens spot of “950”, we’re still working within the tens.  We then ask ourselves, “How many 10s can be subtracted from 50?”, and continue on with the algorithm.

We end here by saying that “950 divided by 10 is 95”.  In the division algorithm, when we end with a “0”, we stop the algorithm because it means that the divisor (10) can be subtracted from the divisee “950” an even amount of times with no remainders.  But we can use this same algorithm when dividing any number by any other number.

### My Harvard Math for Teaching Thesis: Complete! And ready to share…March 20, 2011

After many many years of jumping through many many hoops, I am finally graduating with my MA in Mathematics for Teaching in May.  My thesis, Negative Number Misconceptions in High School: An Intervention Using the ZeroSum Ruler is right now at the printers being printed and bound.  I don’t know about you, but that instantaneous feeling of relief after taking a final exam or passing in a final paper stopped hitting me sometime in college.  So now, I’m just feeling a bit burnt out.  OK, completely burnt out.  But I’m sure it will hit me soon since it kind of needs to; I need to now get in a post-Bach program to get my Initial teaching license.  I like to do things backwards.

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So here it is for download!  For all to read!  Or maybe to just glance.  In my study, the ZeroSum ruler proved effective in reducing eleventh grade error on integer addition and subtraction problems (especially with negative integers).  If I wasn’t so burnt out, I’d want to test it with younger kids.  Imagine how our world would be if my eleventh graders actually mastered integers when they learned them in, and only in, 7th grade.  But that’s in my thesis.]

### ZeroSum ruler’s 62% success rate!March 9, 2011

The ZeroSum ruler improved my student’s understanding

of integers

by 62%

in a very short 2 weeks

Surpassed even my high expectations!

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HaPpY Calculating!

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### Why we add fractions the way we do… a visual tourJanuary 15, 2011

Why do we add fractions the way we do- by getting the common denominator?  A legitimate question!  The following is a visual explanation of why we need to do so…

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When we add two fractions with different denominators, we have been taught to “find the common denominator”, then add the numerators only.  But why not add the denominators too?

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Let’s try to add 2/7 + 3/5.  We know from our standard algorithm that we would change both denominators to 35, then change the numerators by making sure to keep the ratio between the numerator and the denominator in tact.  We’d end up with 10/35 + 21/35 = 31/35.

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But how can we see why this works?  Let’s first look at both as pictures:

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The picture on the left represents “2 out of 7”, and the picture on the right, “3 out of 5”.

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To add means to combine, and fractions- with the exception of improper fractions- represent amounts less than one.  So, we want to combine these two shaded regions into one, or, if we can make more than one, we want to see how many “ones” we can make.  But the shaded bars aren’t the same size.  And how big is “one”?

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In the algorithm we’d “find the common denominator”, but what does this mean and look like?  It means we have to change the look of these two fractions so that their numerators represent portions of whole broken up into the same amount of pieces.  To do this, we break the picture on the left up into fifths and the picture on the right up in to sevenths.

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Now our two fraction pictures are broken up into the same amount of pieces, and each piece is the same size.

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To justify this, let’s look at the dimensions of each area…

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The fraction picture on the left has an area of (7×5).  The fraction picture on the right has an area of (5×7).  Because of the commutative property, we know that 7×5 = 5×7, so both fraction pictures have an equal total area (denominator), and that area is 35 spaces.  For this same reason, the shaded spaces in both pictures are also all the same size.

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But now we have 21 (out of 35) shaded pieces in the fraction picture on the left and 10 (out of 35) shaded pieces in the fraction picture on the right.  Can we do this?  Is 21/35 the same as 3/5?  Is 10/35 the same as 2/7?

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Because we can break a “whole” up into as many pieces as we want, these fractions are equal.  For example, if two people both had a liter of soda each, one could give small cups of soda to 21 friends, and the other friend could give larger cups of soda to 10 friends.  If both empty their bottles, both gave out the same amount of soda despite giving it out to different numbers of friends.

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It’s the same with these two fractions.  Once we have set shaded regions (3/5 and 2/7), we can break these regions up into an infinite number of pieces and still have 3/5 and 2/7.

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Now we can begin adding one to the other.

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A total of 31 spaces are filled in when we take the shaded spaces from the fraction picture on the left and add them to the picture on the right.  So, 31 out of 35 spaces are now filled in, or “31/35”.

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If we had used the standard algorithm, we would have added the numerators of the fractions (after we found the common denominators) to get this 31.

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2/7 + 3/5 = 10/35 + 21/35 = 31/35, or 4/35 less than a whole.

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For why we need to first find the common denominator, see two or three posts down…

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### Negative Numbers. OH NO!October 6, 2010

In our BPS high school, there’s a big focus on the “broken window theory”, made famous recently in The Tipping Point.  One broken window we’ve identified in the school as far as discipline goes is hats and ipods.  So, there’s been a big push to get rid of them.

I’d like to mention to you a “broken window” that has somehow gotten lost in the mess of school closings, going charter, union fighting, pension plans, longer days, MCAS scores.  As a high school math teacher, the biggest broken window I face – in fact, it’s a gaping hole not even bothered to be temporarily covered with plastic- is… negative numbers.

What do I mean by negative numbers?  I’ve done my research as they’re the topic of my Harvard thesis.  Students using the TERC Investigations curriculum in Boston elementary schools do not do problems like “-22 + 5″.  One TERC representative told me they “leave that topic to middle school”.  So, I looked at the middle school Connected mathematics Project 2 (CMP2) curriculum, and negative integer problems, like “-22 + 7″ are taught for 20 days total in the 7th grade.  20 days.  From then on, students are assumed to know how positives and negatives interact and to be able to evaluate “-22 + 5″.

Then students get to me, their 11th grade Algebra 2 teacher, and they can’t solve for y in “y + 22x = 5x – 7″ because they don’t know what “5 – 22″ is.  The kids think -22 + 5 = -27.  Why?  Maybe the rules of multiplication get mixed in.  I don’t know.  Or maybe it’s because these problems were taught to them for a total of 20 days four years earlier and were never touched n again except in the context of other problems.  Understanding why and how kids think is beyond the scope of my thesis and my means for data collection.  What I can tell you is that because my students don’t know what “5 – 22″ is, they can’t solve y + 22x = 5x – 7 for y.  Because they can’t solve the equation for y, they can’t graph the equation.  I assume you know where I’m going with this.

Please, as someone on the front lines of math education in Boston, I’m telling you that the biggest difficulty our students have in math is adding and subtracting positive and negative integers.  It seems ridiculous and that there are bigger fish to fry, some of which I have listed, but if you want more competency in math, please, heighten the focus on negative numbers.  It will lead to better test scores, more understanding, but most of all, to students who feel good about themselves when they’re not still making silly 7th grade mistakes in high school.