Great Ideas of Primary Mathematics

Strategy and Content

Exploding Dots

Exploding dots is a concept developed by James Tanton. ⁵ In its simplest form I think it is a great tool to introduce the concept of place value. Most students can tell you that in the number 345 there are three hundreds, four tens and five ones. By using exploding dots students will have an illustrated representation of the base ten number system or any base number system you choose.

Exploding dots consist of a series of boxes constructed together horizontally. I would recommend starting with no more than three boxes. Start with the box farthest to the right and draw in dots up to a maximum of nine. When the tenth dot is added the dots explode and a lone solitary dot will appear in the box to the left of the box. The boxes are called a machine. When this is carried out we will label it a 10 -> 1 machine.

When trying to add the tenth dot to the far right box the teacher should make a loud sound simulating an explosion, wipe the dots away and then place the dot in the middle box. This will be true for any box. The maximum number of dots that can appear in any box is nine. The tenth causes the explosion and a dot to appear in the immediate box to the left. Question the students at this point and ask them how many dots are represented in total?

Write this box and the dots. First ask students how many dots does the first box represent? The answer should be 1. Ask them how many dots are represented in the second box? The answer should be 20. Then ask them how many dots are represented in the third box (going from right to left)? The answer should be 300. If the students don't get this immediately, the process can be explicated in more detail.

Students will then be asked to write down what numbers different boxes and dots represent. The concept is that they will quickly understand that a 10 -> 1 machine represents our base ten number system. The importance of zero will become more evident as you proceed with these exercises.

Different base number systems can and should be represented with exploding dots as well. A machine that is 2 -> 1 represents a binary system.

If we wanted to convert this number back to base ten then we simply write it in expanded form 1(2 ²) + 1(2 ¹) + 1 (2 ⁰)= 4 +2+1= 7 in base 10. I think it is important that teachers show students at least one example of a number written in another base system. Exploding dots can help illustrate why 9 is the highest digit in base 10 and 1 is the highest digit in base 2.

A widely held belief in education is that students who learn a foreign language will in turn learn the English language better. That is because they are forced to see something familiar to them, the English language, through the lens of learning a foreign language. I believe that if students are exposed to other base number systems and understand how they work then they will understand the base ten number system on a deeper level. It is beyond the scope of this unit to investigate this idea in detail, but some teachers might find it interesting.

Expanded Form

Being able to write numbers out in expanded form is important. It may be seem like the long way of doing problems by students but it does force them to see the significance of place values by writing numbers in expanded form and then being able to regroup them and perform various operations.

Expanded form can be shown in different ways. If we take the number 4,082 it can be written 4,082 = 4,000 + 80 + 2 .

In exploding dots this number would be written as :

In this form it is not necessary to explicitly indicate that there are no separate 100s present, since each non-zero digit is represented in the correct place. The boxes themselves function as placeholders, and an empty box replaces a zero in the corresponding place. That is, we don't have to write

4,082 = 4(1000) + 0 (100) +8 (10) + 2 (1)

Using Exploding Dots:

In the above form a zero has been added to indicate that there is a zero amount of hundreds. This zero must appear in the compressed version of the number, since in the compressed form, the only way to indicate the unit that a digit represents is by its place in the number. We can also write our number as:

4,082 = 4(10 x 10 x10) + 0(10 x 10) + 8(10) +2(1)

In this more refined version of expanded form the powers of ten are being broken into the factors of 10 that they are made from, with the exception of the units digit. Finally, we can rewrite our number a fourth time:

4,082 = 4(10 ³) + 0(10 ²) + 8(10 ¹) + 2(10 ⁰).

In this form, the various base ten units are now being written as powers of 10, using exponential notation. The representation using exploding dots would be the same for each expanded from version.

I think this is the best expanded form as it clearly shows a relationship to our base ten number system and allows for the importance of zero to be discussed as a place holder. This concept is referred to as the order of magnitude. For example the order of magnitude of 4,000 is three which is not only equivalent to the number of zeros but also the power of ten involved in the expression 4,000 = 4(10 ³). The order of magnitude of 12, 456 is four as the highest power of ten is 1(10 ⁴) which is the ten thousands place. The order of magnitude for 5 is zero. This is written as 5 (10 ⁰) and five is the highest number and it is multiplied by ten to the power of zero.

Adding with Expanded Form and Exploding Dots

Adding numbers in expanded form can be done in a variety of ways. Let's look at adding 4,082 + 988. In expanded form and moving numbers around because of the commutative law of addition I it would look like 4(10 ³) +0(10 ²) + 8(10 ¹) + 2(10 ⁰) + 9(10 ²) + 8 (10 ¹)+ 8(10 ⁰). When we add the numbers together and combine like terms we get 4(10 ³) + 9(10 ²) + 16 (10 ¹) + 10(10 ⁰). Since our number system is based on the number ten then we do not have a number higher than nine as any single digit. Starting with our last number in the 0 magnitude place we have to reduce the digit to a single digit below nine that is in the single digits spot of the number appearing. The digit is being reduced by removing the next highest magnitude. This might be a better demonstration to students of what is occurring during the addition process. This will allow students to see that the excess dots equal the number times a particular magnitude in addition with expanded form.

Note: In representing a number in standard base ten form, we cannot have more than 9 copies of ten in any place.

Note how whenever when the addition process creates more than 9 copies of ten we have to convert 10 of them to the next larger unit, and then add this to the place value to the left. This is a big concept and should not be overlooked. For instance when there is 16 (10 ¹) in the above problem the question is should I just subtract 7 to get to 9? How many do I subtract? Simply write the number out. 16(10 ¹) = 160.Then break it into its base ten components: 160 = 100 + 60. The 100 belongs in the next larger place. So we know we have to convert 10 tens into 1 hundred or 1 (10 ²). This of course is how the number is represented when added to the hundreds column. Remember the value never changes, only the location of the values. I always use exploding dots to guarantee a clear illustration.

Algebra Relationship (Addition)

In Algebra numbers are represented by variables. Students should be familiar with variables. While doing this unit you may need to familiarize students with the concepts of variables if they have not been exposed to them previously. I would use a variable to represent the powers of ten and the digit indicating how many copies of a given power of ten are present to be the coefficient of the corresponding power of the variable. The above problem would look like this algebraically:

The number 4,082 could be shown as 4x ³ +0x ²+8x+ 2 or simplified it would be 4x ³+8x+2. The number 988 would be shown as 9x ²+ 8x+8 and it is simplified as much as possible. Here again we can use an array to add our numbers. The important concept here is that the x's could represent any number so we do not have to get our coefficients below nine, as we do when working in base ten.

If we now substitute a ten back in for the variable x then we get the following

4(10 ³) + 9(10 ²) + 16(10 ¹) + 10(10 ⁰) is the same number as subtotal A in the addition table. This illustrates the fact that addition in base ten can be thought of as having two stages. In stage 1, digits are added as if they were coefficients of polynomials, and we were doing polynomial addition. In stage 2, we remember that x = 10, and do regrouping of coefficients that ended up being larger than 10. In the standard algorithm for addition, these two stages are mixed together.

Subtracting in Expanded Form and Exploding Dots

Subtracting numbers in expanded form can also be done with the same concepts in mind. Here the idea of borrowing is revealed. We must start from the right to make sure that we have a high enough number to subtract so the top number is greater than the number being subtracted. The example below shows 4,082-988.

This is of course not an answer since standard base ten notation does not use negative multiples of the base ten units. This is the point where students should understand how the concept of "borrowing" is used. Showing this with exploding dots will help illustrate what is being done.

Subtraction with exploding dots is a little different than addition. I introduce the concept of anti-dots. The great thing with subtraction and exploding dots is how clearly it illustrates the concept of borrowing. If we take the same two numbers we were using in expanded form 4,082- 988 we can illustrate what happens:

This can get real messy in such a confined space but as you can see I am matching up an anti-dot with a dot and creating a zero pair. This is using the additive inverse property (see Appendix B ). When we are completed matching up all the possible zero pairs we end up with this:

If we look at the exploding dots above we clearly have: 4(1000) -9(100) + 0(10) -6(1)

This is exactly what we had when we finished subtracting with expanded form. The problem is I cannot write this number out in condensed form as it has negatives in it. What I can do is spread the dots down from the thousands place and then I can eliminate the anti-dots by having enough dots to make zero pairs. This is where the concept of borrowing occurs.

The first thing I could do is implode a one thousand to create ten hundreds.

Then I could implode a dot from the hundreds place to create ten dots in the tens place.

This actually still represents 4,000 in an unexploded sense. Check it to see.

3(1000) + 9(100) + 10(10) = 4,000

We have one more step and that is to implode a dot in the tens place and create ten ones.

We can again check our number in expanded form as it is represented in the unexploded dot form to make sure we still have 4,000.

3(1000) + 9(100) + 9(10) + 10(1) = 4,000

Now we have an equal or greater number of dots then anti-dots and we can eliminate all the anti-dots through the additive inverse property and get our results.

The clear dots are called anti-dots and represent the number being subtracted. Some students might want to call them negative numbers. Based on prior knowledge students should realize that adding a negative number is the same as subtracting a positive number.. When more anti-dots appear in a box than dots then obviously there is a problem. The key here is to implode a dot in the next box to the left. When a dot is unexploded then ten dots will appear in the box to the right. This illustrates the concept of borrowing that students should be familiar with but it focuses them on what is really happening when they borrow.

Algebra Relationship (Subtraction)

Subtraction is done in a very similar way as addition. I have constructed an array to show exactly what would occur. I have already commented on how to make this as parallel as possible to the straight numerical calculation. I suggest that you also now substitute x = 10 and show that the regrouping process leads to the same answer.

If we go back and substitute 10 in for the variable x we get the same answer that we arrived at in expanded form.

4(10 ³) -9(10 ²) +0(10 ¹) – 6(10 ⁰) = 4(1000) -9(100) + 0(10) -6(1)

Multiplying in Expanded Form

Multiplication can be done using the same principles. I think it is actually easier to see.

Simply carry out the multiplication using an extended version of the distributive property. This means that each term in one factor must be multiplied every term in the other factor. This could be called the Each With Each rule since the total product is the sum of the products of each component of the first number multiplied by each component of the second number. Demonstrating the Each With Each rule can be done using the area model. It is beyond the scope of this essay but it could be demonstrated to students.

I have set up a multiplication array to show this more clearly. What I am demonstrating in the array below is 4,082(988). I am however breaking the condensed numbers up and rewriting them in expanded form:

{4(10 ³) x 9(10 ²) + 4(10 ³) x 8(10 ¹) + 4(10 ³) x 8(10 ⁰) + 8(10 ¹) x 9(10 ²) + 8(10 ¹) x 8(10 ¹) + 8(10 ¹) x 8(10 ⁰) + 2(10 ⁰) x 9(10 ²) + 2(10 ⁰) x 8(10 ¹) + 2(10 ⁰) x 8(10 ⁰)

Multiplication Array

Once the multiplication has been carried out then addition has to occur. I have written in arrows to show how we add the like terms which means we are adding together the multiples of the same power of ten In the above problem, for 10 ³, this sum would be 32 (10 ³) + 0(10 ³) + 72 (10 ³) = 104(10 ³). I don't add different powers of ten together here. This relates directly to the algebra example later in the unit. The same problem arises as with addition before. We will have to subtract from the place value to the right and add to the place value to the left in order to get the correct numbers. The numbers must be placed in an array and lined up by their respective magnitudes. Notice how the powers arrange themselves diagonally. When we add up our like terms we get the following answer:

Algebra Relationship (Multiplication)

The use of the distributive property is evident when representing this situation.

Multiplication is carried out by each term from one number with each term from the other number. In this problem a polynomial is being multiplied by another polynomial. Students should quickly learn how to multiply polynomials by using the array method. This will help avoiding having to teach the FOIL method with no reasoning. The reason that FOIL works is through the each with each concept. Here I again I am substituting the variable x for the powers of ten.

Compare the above array to the array completed with the powers of ten. Once the multiplication is completed then simply adding by using like terms is done and the problem is complete.

If I substitute a ten back in for the variable x then we will see the relationship to the expanded work previously demonstrated. The answer below is the exact same answer we obtained in subtotal A from our array example.

36(10 ⁵) + 32(10 ⁴) + 104 (10 ³) + 82 (10 ²) + 80(10 ¹) + 16 (10 ⁰)

Use of Integers and Base Ten

Most students in an honors level class should be able to solve basic integer problems. In this method students simply ignore carrying and borrowing when adding or subtracting. They rely on their knowledge of base ten and integers. In the below problem students are subtracting 988 from 4,082. Students have been taught to borrow to carry out this subtraction. I am showing that they do not need to borrow in order to carry out the operation. They can treat the subtrahend as a negative number and add the two numbers together. The only difference is they will be writing the problem and answers out in expanded form and then arrive at a solution.

Take the resulting numbers and add them together. 4000+ (-900) + (00) + (-6) = 3,100+ (-6) which if difficult can be expanded to 3,000 +90+10+ (-6) which is 3,000 + 90 +4 or 3,094. I would have the class work with various problems of this nature using addition, subtraction, and multiplication. The students should be familiar with expanded form and comfortable with demonstrating different ways of arriving at solutions.

Algebra Connection

Replacing the 10 with the variable x can help here also.

This answer is the same as our total in the array showing subtraction in expanded form.

Divisibility Rules

Place value is also a powerful tool to use when determining divisibility rules. Rote memorization is a tool that teachers often rely on to teach students. They want students to learn by memorizing a fact or formula. I often comment in my class that this memorization method is like giving my students an orange rather than giving them some seeds and having them plant, tend and harvest their own oranges and in turn planting more.

I myself cannot remember all the divisibility rules but I can use place value to help determine them. What about four? Let's take the number 334 and see if it is divisible by 4.First write it in expanded form.

300 + 30 + 4 = 3(25 × 4) + 3(2 × 4 +2) +4 = 4(3 × 25 + 3 × 2) + 3 × 2 + 4

The final expression shows that divisibility of 334 by 4 comes down to the divisibility of 3x2+ 4 by 4. Since 3x2+4 = 10, which is not divisible by 4, neither is 334.

If students do several more examples, some of them will probably see that 4 will always go into any hundred value as 100 is divisible by 4. Likewise, 1000, and 10,000, and all higher powers of 10 are divisible by 4. We can therefore focus on the tens and ones place. If we divide 30 by four we can rewrite it as 34 = 30 + 4 = 3x10+4 = 3x(2x4+2) + 4 = (3x2)x4 + 3x2 + 4 and we know that 6 is not divisible by 4 so 334 is not divisible by 4. What we did learn is that twice the 10s digit plus the 1s digit should be divisible by 4. Exploring how 8 would work might not be so hard if we understand how 2 and 4 are determined. The place value that becomes significant with 8 is the hundreds place. For 8: 4 times the 100s digit plus 2 times the 10s digit plus the ones digit should be divisible by 8. The number one thousand, ten thousand, one hundred thousand…. are all divisible by 8. If this is the case then the focus needs to be on the hundreds, tens and ones place. If a number represented in the hundreds place and lower is divisible by 8 then the whole number is.

When looking at whether a number is divisible by three the same approach can be taken. Let's take an Algebraic approach to this. If we use a three digit number then it could be written as 100a + 10b+ c where a, b, and c are the digits of the number. Here arithmetic really is being dealt with as algebra! Obviously 100 and ten are not divisible by 3 but 99 and 9 are so we could rewrite this number to be 99a+ a+ 9b+b+c and we have the same number. If we regroup the number we get 99a+9b +a +b +c. We know that 99a +9b is divisible by 3 so the focus would then be on whether a + b+ c is divisible by 3. This makes the rule that all we have to do is take the sum of digits of a number and if they are divisible by 3 then the number is divisible by 3. Here again determining the rule for 9 should be fairly obvious.

In order to investigate this take a number that we know is divisible by 3 such as 300. If we assume that because it is divisible by three it is divisible by 9 we might be mistaken.

3(100) equals 3(99+1)

Since 99 is clearly divisible by 9 a three is left over and is not divisible by 9. How does that help with developing a rule?

Go back to the aforementioned divisibility by 3 rule. With a three digit number 100a + 10b+ c is the same as 99a+1a+9b+1b+c and we know that 99a+9b is divisible by 9 so a+b+c must be divisible by 9. In simple words the digits have to add up to a multiple of 9.