3.1: Definition and Properties

You will need: Coins (Material Card 1), A-Blocks (Material Cards 2A-2E), Centimeter Strips (Material Cards 17A-17L)

Whole numbers are made up of zero and the counting numbers 0, 1, 2, 3, 4, 5, . . . We begin this exercise set by exploring how to add two whole numbers together. To add 3 + 4, we find a set with three elements and a different set with four elements (having no elements in common with the first set). We combine the two sets and then count how many elements are in the union. This method is illustrated in the following exercise.

Exercise 1

Take out the set of 25 coins from Material Card 1. Put three coins in one hand and call this set of coins set A. Put four different coins in your other hand and call this set of coins set B. Use appropriate set notation and coin abbreviations when completing the following:

a. A =	b. n(A) =
d. n(B) =	d. n(B) =

Put the coins from both hands into one pile to form the union of A and B.

e. \(A \cup B\)

f. n\(A \cup B\)

So to add two numbers, 3 and 4, we must find a set A such that n(A) = 3 and a set B such that n(B) = 4. Then, 3 + 4 is how many elements are in \(A \cup B\). In Exercise 1, since n(\(A \cup B\)) = 7, then 3 + 4 = 7. Of course, we already knew that from years of drill and practice, flashcards, etc. To the right is a picture which shows three coins in one hand and four in another. Combining the coins into one big pile, they can be counted to find the sum.

There is one restriction on the two sets chosen which you will discover by doing the next exercise.

Exercise 2

We are going to try and find the answer to 3 + 6 by using the same method just employed. But this time, I want you to work with sets A and B as I define them here: A = {t, u, v} and B = {u, v, w, x, y, z}. Note that n(A) = 3 and n(B) = 6.

a. List the elements in \(A \cup B\):

b. How many elements are in \(A \cup B\)? ____

c. Is 3 + 6 = n\(A \cup B\)? ____

d. Is n(A) + n(B) = n\(A \cup B\)? ____

e. How might the sets A and B be chosen so that the answers to both c and d would be yes? If you need a hint, look at what is in parentheses in the beginning paragraph of this exercise set!

By the way, since n(A) = 3 and n(B) = 6, part c and d are asking the same question!

Use your A-Blocks for the following exercises. As a reminder, the letter that is in BOLD type in each word below is the abbreviation used for the nine main subsets of A-Blocks. RED, BLUE, GREEN, YELLOW, SMALL, LARGE, CIRCLE, TRIANGLE, SQUARE (To denote a particular A-Block, use three letters in the format Size, Color, Shape — e.g., Large Blue Triangle is abbreviated LBT, Small Red Square is abbreviated SRQ, etc.)

Exercise 3

a. Name two subsets of A-Blocks (each containing 6 elements) that could be used to find the sum of 6 and 6: ____ and _____

b. Illustrate how the two sets you chose will give the correct sum by listing both the elements in each set you named and the elements in the union of the two sets (using abbreviations). The number of elements in the union should be 12!

Exercise 4

Since n(L) = 12 and n(T) = 8, let's find out if the subsets of L and T can be used to find the sum of 12 and 8.

a. List the elements in (\(L \cup T\)):

b. n\(L \cup T\) = ____

c. Is n (\(L \cup T\)) equal to 12 + 8? ____

d. Why didn't it work to use the sets of Large A-Blocks and Triangular A-Blocks to find the sum?

A formal, more general definition of addition of whole numbers now follows.

Set Theory Definition of Addition: If A and B are two disjoint sets, that is \(A \cap B\) = Ø, then n(A) + n(B) = n(\(A \cup B\)).

Following are some examples of how to use the set theory definition of addition to add.

Example 1

Use the set theory definition of addition to show that 3 + 2 = 5.

Solution

Let A = {x, y, z} and B = {m, r} Since n(A) = 3, n(B) =2, and \(A \cap B\) = Ø then

3+2 = n(A)+n(B)	by substituting n(A) for 3 and n(B) for 2
= n(\(A \cup B\))	by the set theory definition of addition
= n({x, y, z, m, r})	by computing A
= 5	by counting the elements in A

Therefore, 3 +2 = 5.

Example 2

Use the set theory definition of addition to show that 1 + 3 = 4

Solution

Let A = {x} and B {a, b, c}. Since n(A) = 1, n(B) = 3 and \(A \cup B\) = Ø, then

1 + 3 = n(A) + n(B)	by substituting n(A) for 1 and n(B) for 3
= n(\(A \cup B\))	by the set theory definition of addition
= n({x, a, b, c})	by computing \(A \cup B\)
= 4	by counting the elements in \(A \cup B\)

Therefore, 1 + 3 = 4

Example 3

Use the set theory definition of addition to show that 0 + 4 = 4.

Solution

Let A = { } and B = {a, b, c, d}. Since n(A) = 0, n(B) = 4 and \(A \cup B\) = Ø, then

0 + 4 = n(A) + n(B)	by substituting n(A) for 0 and n(B) for 4
= n(\(A \cup B\))	by the set theory definition of addition
= n({a, b, c, d})	by computing \(A \cup B\)
= 4	by counting the elements in \(A \cup B\)

Therefore, 0 + 4 = 4

The definition of addition allows us to verify some properties about addition. Example 3 verifies the property from Set Theory that states that for any set A, \(A \cup Ø\) = A.

Let n(A)= a. Since the number of elements in the null set is zero, then a + 0 = a and 0 + a = a.

Exercise 5

Use the set theory definition of addition to do the following additions. Write it out in detail! Follow the examples above.

a. Show that 2 + 4 = 6

b. Show that 2 + 2 = 4

c. Show that 3 + 0 = 3

Take out your set of Centimeter Strips (C-Strips). There are twelve different kinds of strips, each having a different length. Below is a listing of the different kinds, with their dimensions, colors and the abbreviations we'll be using to refer to each. Let's agree to use the abbreviations in more than one context — a letter may refer to the actual physical strip itself or it might refer to the length of the strip. This isn't exactly perfect notation wise, but it will make the work that follows less cumbersome.

1 cm by 1 cm.................................... WHITE................... W

1 cm by 2 cm......................................... RED.................... R

1 cm by 3 cm....................... LIGHT GREEN.................... L

1 cm by 4 cm.................................. PURPLE..................... P

1 cm by 5 cm................................ YELLOW.................... Y

1 cm by 6 cm........................ DARK GREEN.................... D

1 cm by 7 cm.................................... BLACK.................... K

1 cm by 8 cm.................................. BROWN.................... N

1 cm by 9 cm....................................... BLUE.................... B

1 cm by 10 cm.............................. ORANGE.................... O

1 cm by 11 cm................................. SILVER..................... S

1 cm by 12 cm............................. HOT PINK.................... H

We will be using the C-Strips to discover and reinforce concepts and properties about addition. For many of the problems, we will be working with "trains", which are made by putting one or more strips end to end in a straight line. So the width will remain 1 cm but the length of the train will vary.

To designate a train consisting of more than one strip, we write an addition problem so that the C-Strip on the left is written first and the rest, if any, are written from left to right. Below, note the way trains are formed and how we denote each one.

Exercise 6

Use the C-Strips to make the train indicated. Then, find a C-Strip of equal length and fill in each blank with the abbreviation of that C-Strip

a. P + L =	b. S + W =	c. R + K =
d. B + W =	e. N + L =	f. L + W =
g. P + Y =	h. D + R =	i. B + L =
j. Y + P =	k. R + D	l. L + B =

m. What do you notice about j & g, k & h and l & i?

Parts j & g, k & h and l & i of Exercise 6 illustrate an important principle about whole numbers called the Commutative Property of Addition.

The Commutative Property of Addition states that if a and b are any two whole numbers, then a + b = b + a.

Using the definition of addition and this important fact from Set Theory,

\(A \cup B = B \cup A\), we can show why the Commutative Property of Addition is true:

a + b = n(A) + n(B) = n(\(A \cup B\)) = n(\(B \cup A\)) = n(B) + n(A) = b + a

Most of us have been adding numbers together for years and years and so this property may seem obvious — it is second nature that the order in which we add two numbers is irrelevant. But if you take the opportunity to ask a child who is just learning to add the following two questions, even one right after the other, you may notice the child can do the first one quickly and easily but then struggles a little longer at the second one. That is usually the case if the child hasn't yet discovered the commutative property of addition. As adults, we take this property for granted.

First question: What is 7 + 2? Second question: What is 2 + 7?

Exercise 7

How do you think a child might figure out the answer to the first question and how might he or she figure out how to do the second question? Be specific and assume the answers haven't been memorized yet! Why might the second question be perceived as a harder problem?

When adding two numbers together, some people start with the first number and count on the second number. So if you think about 7 + 2, start with 7 and count two more in your head — eight, nine. For 2 + 7, start with 2 and count seven more in your head — three, four, five, six, seven, eight, nine. Although the answer is the same, 7 + 2 was quicker and easier to keep track of. Just think about the difference between 1000 + 1 and 1 + 1000 using this counting on method! Thank goodness for the commutative property of addition!

What if you had to add three numbers together? Actually, we only add two numbers together at a time and then add the third one to the sum of the first two numbers. We will use C-Strips to illustrate ways of adding three numbers together. Consider putting the C-Strips L, R and P together to form a train. Let's compute the sum of these three in two different ways, working inside parentheses first.

Exercise 8

Fill in each blank with the correct abbreviation of a C-Strip by first computing the sum in the parentheses and putting that answer in the first blank. Then add once more to get the answer. An example with numbers: (5 + 3) + 9 = \(\underline{8}\) + 9 = \(\underline{17}\).

Exercise 8 illustrates another important principle about whole numbers called the Associative Property of Addition.

The Associative Property of Addition states that if a , b and c are any three numbers, then (a + b) + c = a + ( b + c )

Using the definition of addition and this important fact from Set Theory, \((A \cup B) \cup C = A \cup (B \cup C)\), we show why the Associative Property of Addition is true: (a + b) + c = (n(A) + n(B)) + n(C) =n(\(A \cup B\)) + n(C) = n(\((A \cup B) \cup C\)) = n(\((A \cup B) \cup C\) = n(A) + n(\((B \cup C)\)) = n(A) + (n(B) + n(C)) = a + (b + c)

Exercise 9

Using the model of Exercise 8 and the C-Strips, illustrate another application of the associative property of addition. Then write down an equivalent statement using numbers.

The root in the word "associative" is associate. Think about whether the middle number will associate with the first number or the last number for the first computation. As for adding numbers in your head, it may be a lot easier to do it one way instead of the other. For instance, to add 58 + 39 + 41, you could think (58 + 39) + 41 = 97 + 41 = 138 or you could think 58 + (39 + 41) = 58 + 80 = 138. 97 + 41 and 58 + 80 both equal 138, but that isn't obvious until after you add. I prefer to add the 39 and 41 together first. The associative property allows that. But again, the associative property is something that most adults take for granted and use but they don't really think about it.

Here is another example. Although the sum of the three numbers is the same for each problem, the way you do the computations are different.

(84 + 56) + 73 140 + 73 213

84 + (56 + 73) 84 + 129 213

Exercise 10

Make up an example of your own showing an application of the associative property

It might be helpful to note that the commutative property implies there has been an order change. An order change means that the numbers written are in a different order. The associative property implies there has been a parentheses change. A parentheses change means there are different numbers in the parentheses and that the parentheses are now around different numbers.

Exercise 11

State which property (Commutative or Associative Property of Addition) is being used in each equation. Ask yourself: Is the difference between the left and right side due to order (commutative property) or parentheses (associative property)?

a. _____	(99 + 76) + 38 = (76 + 99) + 38
b. _____	(65 + 22) + 56 = 56 + (65 + 22)
c. _____	(57 + 88) + 43 = 57 + (88 + 43)
d. _____	(a + b) + (c + d) = ((a + b) + c) + d

We use the commutative and associative properties of addition all the time to do simple arithmetic — of course, the arithmetic might not be quite so easy if we weren't allowed to use these properties. Imagine neither property existed and the only way to get the sum of three numbers was to add them in order from left to right. Consider this arithmetic problem: 17 + 59 + 83. Going from left to right, we first add 17 and 59 to obtain 76 and then add 83 to 76 to get the answer of 159. Okay, so that's not so difficult. But it becomes a breeze if we are allowed to use the commutative and associative properties of addition as shown below (three ways are shown using different steps):

(17 + 59) + 83 = 83 + (17 + 59) Commutative property

= (83 + 17) + 59 Associative property

= 100 + 59

= 159

(17 + 59) + 83 = (59 + 17) + 83 Commutative property

= 59 + (17 + 83) Associative property

= 59 + 100

= 159

(17 + 59) + 83 = 17 + (59 + 83) Associative property

= 17 + (83 + 59) Commutative property

= (17 + 83) + 59 Associative property

= 100 + 59

= 159

Of course, most of us probably automatically add the 83 and 17 in our head first without bothering to think about the properties. But it is because of these fundamental properties that we arrive at the correct answer by adding the numbers in any order or combination we find convenient.

Exercise 12

Use the commutative and associative properties to add (135 + 384) + 165. Assume you prefer to add 135 and 165 in your head. Show each step and state which property is being used at each step. Then do it again another way using different steps.

We say two trains are equal if they have the same length. For instance, look at the six trains shown below. All of these have the same length of 6 cm and so they are all equal.

Although the six trains shown above are equal, they are considered different because none of them have the exact same strips in exactly the same order. Here is a list of the six trains above: D, W + Y, R + P, L + L, Y + W, P + R. There are more trains having this same length but different from the ones listed so far. For instance, seven more are: W + R + R + W, W + P + W, P + W + W, W + W + P, W + L + W + W, R + L + W, L + W + W + W. In fact, there are several more trains that are equal in length to the Dark Green C-Strip than the 13 different ones listed so far . All of the trains listed translate into addition facts about combinations of numbers that add up to 6: W + Y = D translates to 1 + 5 = 6, W + P + W = D translates to 1 + 4 + 1 = 6, R + P = D translates to 2 + 4 =6, P + R = D translates to 4 + 2 = 6 and so on.

Exercise 13

Use the C-strips to list all the different trains that are equal in length to the Light Green C-strip. Different trains have different strips or and/or a different order of strips.

Exercise 14

Use the trains listed in Exercise 13 to write addition facts obtained about the number 3.

Exercise 15

Use the C-strips to list all the different trains that are equal in length to the Purple C-strip. Different trains have different strips or and/or a different order of strips.

Exercise 16

Use the trains listed in Exercise 15 to write addition facts obtained about the number 4

Listing all the different trains equal in length to the Dark Green C-Strip is quite a chore. We will define equivalent trains as equal trains that have the same make-up (same exact strips), but the strips are listed in a different order. For instance, W + R + L, W + L + R, L + R + W, L + W + R, R + L + W and R + W + L are all equivalent trains because each is made up of a red, light green and white C-strip. W+L and L+W are equivalent trains because they are all made up of exactly the same C-strips — a white C-Strip and a red C-Strip. W+L, R+R, P, W+W+R and W+W+W+W are equal but nonequivalent trains because they are composed of different strips. Nonequivalent trains are equal trains made up of different C-strips. If they are made up of the same C-strips but are in a different order, they are equivalent!

Exercise 17

List all nonequivalent trains (note the last sentence in bold in the above paragraph) equal in length to the Dark Green C-Strip. That means only one of the six ones listed in the middle of the paragraph above (and it doesn't matter which one) should be written down in this list. Remember to include the train of one single strip. How many nonequivalent trains equal in length to the Dark Green C-Strip are there?

Exercise 18

From your work in Exercise 17, list all the facts about what combinations of numbers add up to 6.

If you look at the C-Strips, it is clear that some are longer than others. We say that one C-Strip is longer than another C-Strip if there is a C-Strip that can be added to the second C-Strip to make a train equal in length to the longer C-Strip. For example, the Black C-Strip is longer than the Purple C-Strip because we can add the Light Green C-Strip to the Purple C-Strip to make a train equal in length to the Black C-Strip. We write K is longer than P because K = P + \(\bf \underline{\text{L}}\).

Exercise 19

Verify that each of the named C-Strips are longer than the other C-Strip by filling in the blanks. Abbreviations are used.

a. H is longer than O because H = O + ___

b. N is longer than L because N = L + ___

c. B is longer than W because B = W + ___

d. P is longer than R because P = R + ___

e. S is longer than K because S = K + ___

f. D is longer than Y because D = Y + ___

The same principle just used with C-Strips applies when comparing two whole numbers.

Definition: The whole number m is greater than the whole number n if there is a whole number k such that m = n + k.

In mathematics, the symbol used to denote "is greater than" is: " > ". Therefore, " 7 > 5 " is read "Seven is greater than five".

Exercise 19 can be modified to express information about numbers. For instance, part a corresponds to the fact "12 is greater than 10 because 12 = 10 + 2" or if we want to write it symbolically, we write "12 > 10 because 12 = 10 + 2".

Exercise 20

Verify each of the following (I've done part a for you):

a. 9 > 3 because \(\underline{9 = 3 + 6}\)	b. 30 > 22 because ____
c. 156 > 96 because ____	d. 80 > 0 because ____
e. 231 > 195 because ____	f. 987> 967 because ____

Now that you've worked a little with greater than, we may as well define less than.

Definition: The whole number r is less than the whole number s if there is a whole number k such that r + k = s.

In mathematics, the symbol used to denote "is less than" is: " < ". Therefore, " 4 < 9 " is read "Four is less than nine".

Exercise 21

Verify each of the following (I've done part a for you):

a. 9 < 22 because \(\underline{9 + 13 = 22}\)	b. 30 < 99 because
c. 19 < 70 because	d. 0 < 32 because
e. 489 < 500 because	f. 65 < 201 because

Exercise 22

"Less than" and "greater than" can be used to order numbers on a number line. Fill in each blank with right or left.

a. If m < n, then m is to the _____________ of n on the number line

b. If m > n, then m is to the _____________ of n on the number line.

Exercise 23

Form two trains of C-Strips, m and n, that are equal. If p is a C-Strip or train of C-Strips of any length, is it always true than m + p = n + p?

Exercise 24

Take two different C-Strips or form a train of C-Strips, \(m\) and \(n\), such that \(m > n\). If p is a C-Strip or train of C-Strips of any length, is it always true than \(m + p > n + p\)?

Exercise 23 illustrates that the additive property of equality . If the same number is added to both sides of an equality, the equation remains true. Exercise 24 illustrates the additive property of inequality. If the same number is added to both sides of an inequality, the inequality remains true.

If you take any two trains, s and t, and measure them up by placing one next to the other, you will notice that either they are of equal length or one of them is longer that the other. This observation leads us to the trichotomy law.

Trichotomy law

If a and b are whole numbers, then exactly one of the following is true: a = b or a < b or a > b.

Exercise 25

Place the correct symbol ( <, =, or >) between each pair of numbers. Verify if you are correct if you use the less than or greater than symbol (as in exercises 20 and 21).

a. 8 ____ 19 ____

b. 24 ____ 13 ____

Exercise 26

We began this exercise set by defining the whole numbers. Using set notation, express the whole numbers using the listing method: ____

Exercise 27

If two whole numbers are added together, will the sum always be a whole number? ____

Exercise 28

If two whole numbers are added together, is it possible to get more than one answer? ____

Your answers to Exercises 24 and 25 should confirm that the sum of two whole numbers is always a unique whole number. This is called the closure property of addition for whole numbers. We say the set of whole numbers is closed under addition because when you add any two whole numbers together, you get another unique whole number.

A set is closed under addition if the sum of any two elements in the set (these could be the same elements or two different elements) produces a unique element in the same set.

Addition is actually just an operation we perform on two members of a set. Other operations you are no doubt familiar with are subtraction, multiplication and division. There are other operations and there are new ones you can make up, but we'll leave that discussion for another time. If the operation is defined to be performed on exactly two elements, then it is called a binary operation.

In general, a set is closed under an operation if when you perform the operation on any two elements in the set, the answer is a unique element of the same set.

To show that a set is not closed under addition, you must give a counterexample showing that when you add two numbers in the set, you get a sum that is not in the set. The two numbers you choose can be the same number or different numbers.

Example 1

Is the set {1, 2, 3, 4, 5} closed under addition?

Solution

Imagine you had two hats, each containing these five numbers in the set written on pieces of paper. So each hat contains the set {1, 2, 3, 4, 5}. If you pull a number out of each hat, you might get the same number twice or you might get different numbers. Some of the sums are 1 + 1 = 2, 2 + 3 = 5, 1 + 3 = 4, 3 + 3 = 6, etc. Note that the first three sums 2, 5 and 4 are in the set {1, 2, 3, 4, 5}. But for a set to be closed, the sum of every possible two numbers in the set must yield a number in the original set. Since 3 + 3 = 6, (where the two addends, 3 and 3, are in the original set, but the sum 6 is NOT in the original set), then the set {1, 2, 3, 4, 5} is not closed. There are any number of counterexamples you can use to show it is not closed. 3 + 3 = 6 will do. Or someone else might write 4 + 5 = 9. What are two other counterexamples you could use to show {1, 2, 3, 4, 5} is not closed under addition?

Example 2

Is {4} closed under addition?

Solution

The only possibility to try is 4 + 4, which is 8, and 8 is not in the set. So the set is not closed. The counterexample to show it is not closed is 4 + 4 = 8.

Example 3

Is {5, 10, 15, 20, . . .} closed under addition?

Solution

Start by trying a few examples: 5 + 5 = 10 which is in the set; 10 + 15 = 25 which is in the set (note the three dots which means the next few numbers are 25, 30 and 35. It looks like it is closed. But how can you convince someone that the sum of any two numbers in the set will always be in the original set. Note that the set is composed of multiples of 5. Note any two multiples of 5 can be written as 5 times something, like 5x and 5y. Let 5x and 5y denote two arbitrary elements in the original set. Adding them together, we get 5x + 5y = 5(x+y) which is another multiple of 5 (since it is 5 times something). So, you should be convinced that the sum of any two elements in the set is a multiple of 5, which is what the original set consisted of. Therefore, {5, 10, 15, 20, . . .} is closed under addition.

Exercise 29

State whether or not each set below is closed under addition. If it is not closed, provide an counterexample showing two elements in the set whose sum is not in the set. In general, providing an example showing that something is not always true is called providing a counterexample. If a set is closed, prove and explain why.

a. {0}

b. {1}

c. {0,2,4,6,. . .}

d. {1,3,5,7,. . .}