Adding Fractions With Pictures! (The Crisscross Method) December 3, 2012

Fraction Addition (And Subtraction): We’re not in kindergarten anymore

Addition and subtraction are only easy in elementary school.  Once middle school starts, continuing throughout any Math class taken that point forward, addition and subtraction are much harder than multiplication and division.  Why?  The Common Denominator.  To a kid who is not fluent in his multiplication facts, finding The Common Denominator is an exercise in torture.

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What is a common denominator?  A common denominator is a multiple of both denominators in a fraction addition (or subtraction) problem.  For example:

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In the above example, 6 is a common denominator of 2 and 3.  But is it the only one?  No.  How many common denominators are there between two fractions?  Infinite.  For example:

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Why would we want to use 7830 as a common denominator?  Why not?  The point is that any number that both denominators divide into evenly can act as a common denominator.  We are far less restricted than we thought.

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So if we’re virtually unrestricted in choosing a common denominator, why not pick the one that is the product (multiply) of the two denominators?  For example:

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Just multiply the denominators to find a common denominator.  This is easy.

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At this point in the traditional method of adding fractions, we’d begin to ask our questions: “How many 8’s go into 16?”  Ok, 2.  “2 times 3 is …?”  Ok 6.  So 3/8  =  6/16 .  Though this process is easy to a person who is fluent in their multiplication and division, it will give reason for a non-fluent Math student to seize up.

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A great alternative way of adding fractions is the Crisscross Method of adding (and subtracting) fractions.  In this method, we use the common denominator just once (this method will not create two equivalent fractions to the original two) and multiply “crisscross” to find two new numerators.

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In  3/8  +  5/2, we’ll first multiply the denominators to find our new, common denominator:

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Next, we’ll multiply 3 • 2 (always starting our crisscross in the top left corner) to find the first missing numerator:

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And then 8 • 5 to find the second missing numerator:

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But why are we allowed to do this?  Let’s back up to see what really happened.-

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First, we found the common denominator 16 by multiplying the denominators (8 and 2) of both fractions.  We’re guaranteed that our denominator is common if we created it by multiplying the two original denominators to get it.  To get the first numerator 6, we multiplied the numerator of the first fraction (3) by the denominator of the second fraction (2).

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In the process, we multiplied both numerator and denominator by 2.  In other words, we multiplied  3/8 by  2/2.  Any number divided by itself is just a fancy 1, and multiplying any number by 1 does not change the number’s value.  As a check to see if this process worked,  3/8  =  6/16 .  The old and new fractions are equivalent.

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The same is true to get the second numerator 40:

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Both numerator and denominator were multiplied by 8.  In other words, we multiplied  5/2  by 8/8, which is just a fancy 1.  Multiplying by 1 does not change a number’s value.  As a check,  5/2   =  40/16.  The old and new fractions are equivalent.

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Now we simply add the numerators:

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The Crisscross method also works for fraction subtraction – we’d have a subtraction in the numerator.  Why was this method not taught in school?

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Hurray for Fraction Addition (and Subtraction)!

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You can download a PDF ebook that uses pictures to explain fraction division, multiplication and addition on CurrClick at Fractions: A Picture Book!

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New (free) ZeroSum ruler – for teaching addition with negative numbers September 30, 2012

Below is a new version of the ZeroSum ruler.  This one needs no hardware to construct, just scissors and glue.  You can download, print and use this proven tool right now by clicking on the picture, which will bring you to the PDF file that contains 2 ZeroSum rulers. 

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The Language of Math Poster August 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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Link to Google doc:

https://spreadsheets.google.com/spreadsheet/ccc?key=0Asra4GjkRBNidGhoZlZYcjk4dmhISDlSNHJDbjBPTXc&hl=en_US

Multiplying Fractions with Pictures! June 15, 2011

Fraction Multiplication: Of what?

Fractions are probably the most troublesome topic in Math.  As soon as a problem involves a fraction, kids freeze up.  In Math, of tells us to multiply.  How many shrimp are in five pounds of shrimp?  We multiply the number of shrimp in a pound by five.  Once we know this, fraction multiplication becomes a bit easier to understand.

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The How of fraction multiplication is easy – multiply the numerators and multiply the denominators.  When we show fraction multiplication with pictures, we need to remember of.

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Now to the Why.  To start, we’ll look at a relatively easy problem so that we can develop a pattern to follow with more difficult fraction multiplication problems:

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(1/2)(1/2)

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Translated into English, this problem reads “one-half of one-half”.  Here’s a picture of  1/2:

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And the area below in red is “one-half of one-half”:

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It’s easy to see that one-half of one-half is (1/4).  And in fact:

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(1/2)(1/2)   =   1/4

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Before moving on, let’s look more closely at one aspect of the problem above: the denominator 4.  Where did this come from?  To get that denominator, we needed to keep the entire circle (whole) in mind.  In other words, we needed to say that the red piece was “1 out of something”.  (Confusingly, out of means to divide in Math!)  The denominator is 4 because the red pie piece is 1 out of 4 total pie pieces in the circle.  Always remembering the entire original area is key in fraction multiplication.  Later, we’ll see the same is true with fraction division.

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To make the denominator easier to see, we can divide the circle twice: first vertically for the first fraction, then horizontally for the second fraction:

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It’s then easy to see that the overlapped area (numerator) and the entire number of pie pieces in the circle (denominator) create our answer.  This will always be the case.  It wasn’t a coincidence that the denominator was naturally created as we divided the circle twice.  Let’s use this pattern to solve a more complicated fraction multiplication problem:

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(2/7)(3/5)

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Because these two fractions do not have a common denominator, it would be hard to divide a circle into 7 (and take 2), then into 5 (and take 3), and analyze the overlapped area.  So instead, we’ll use nice, easy rectangles.  First,  2/7:

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In the rectangle above, two of the rectangle’s 7 horizontal bars are colored green to represent 2/7.  Now, keeping the whole rectangle in mind, let’s take 3/5:

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In the rectangle above, three of the 5 vertical columns are colored blue to represent 3/5.  Where to the two colored areas overlap?

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In the above picture, we can see that the overlapped area consists of 6 purple boxes.  But 6 of what?  Remembering our easy example (1/2)(1/2), where our denominator was the total number of pie pieces in the circle after our two rounds of dividing, let’s count the total number of boxes in the above rectangle.  The total number of boxes is 35.

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And in fact:   (2/7)(3/5)   =   6/35

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OLD FASHIONED CHECK: We know that (2/7)(3/5) = 6/35 from the algorithm “multiplying across”.

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To see this more clearly, we can look at the below picture and see that the area the fractions share, or the overlapped region, is 6 boxes, and the area of the entire region is 35 boxes.  “6 out of 35 boxes are double shaded”.

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In this last picture we can see that the area of the double shaded region is 6, or (2×3), and the area of the entire region is 35, or (7×5), which is why we multiply the numerators (2×3) and the denominators (7×5) when we find the product of two fractions.

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Also see Dividing Fractions With Pictures!

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If this was helpful, here is a free poster download for your classroom:

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Also see Dividing Fractions With Pictures! and Differences of Squares with Pictures!

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You can download a PDF ebook that uses pictures to explain fraction division, multiplication and addition on CurrClick at Fractions: A Picture Book!

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contact blog author Shana Donohue: shanadonohue@gmail.com

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The long division algorithm, explained at last March 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. 

 

Comments welcome!

 

 

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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