A University-Level Understanding of Number with an eye towards elementary education

Subsection 1.3.1 Intro To Sets

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• We let \(\mathbb{N} = \{1,2,3,4, \ldots \}\) and call it the set of natural numbers, or counting numbers.

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• We let \(\mathbb{N}_0 = \{0,1,2,3,4, \ldots \}\) and call it the set of natural numbers with zero, or whole numbers.

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• We let \(\mathbb{Z} = \{ \ldots, -3, -2, -1, 0, 1, 2, 3, \ldots \} \) and call it the set of integers. Note that some other textbooks might use the symbol \(\mathbb{I}\) instead, but \(\mathbb{Z}\) is most common.

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• We let \(\mathbb{Q} = \{ \dfrac{a}{b} \ | \ a,b \in \mathbb{Z} \text{ and } b \neq 0 \text{ and the usual rules for equivalent fractions hold} \}\) and call it the set of rational numbers. Note that this is saying that the set contains all fractions with integer numerators and denominators, but we cannot have a denominator of 0.

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• We let \(\mathbb{R} = \{ \text{all numbers on the number line} \}\) be the set of real numbers. This set includes numbers like \(\pi, \sqrt{2},\) and all others.

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Checkpoint 1.3.7.

Let A,B,C be sets that aren’t the empty set (so \(A,B,C \neq \emptyset\)). Show that if \(A \subseteq B\) and \(B \subseteq C\) then \(A \subseteq C\text{.}\)

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Solution.

Note that this is an if-then statement. So we need to show that it’s never the case that the if-part is true and the then-part is false. But, in the definition of subsets we have additional if-then statements as well, so we have to be careful with these "nested" conditional statements.

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Let’s assume the if-part is true: that \(A \subseteq B\) and \(B \subseteq C\text{.}\) So this means that if \(x \in A\) then \(x \in B\) (since A is a subset of B) and if \(x \in B\) then \(x \in C\) (since B is a subset of C).

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We need to show that the then-part is true; that A is a subset of C, so if \(x \in A\) then \(x \in C\text{.}\)

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So now, using the if-part assumption, let’s show that the then-part is true. Let \(x \in A\text{.}\) By our assumption that \(A \subseteq B\) we know \(x \in B\text{.}\) Now, by the second part of our assumption, that \(B \subseteq C\) we know that \(x \in C\text{.}\)

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Since we started with an element of A and showed that it must be an element of C, we have shown the then-part; that \(A \subseteq C\text{,}\) and this means we’ve proved that the if-then statement is always true.

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Note that you can see this visually using the following Venn diagram:

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Subsection 1.3.2 Combining Sets

Example 1.3.14.

Suppose we were only interested in the counting numbers up to 10. Let’s define the universe \(U = \{ 1,2,3, \ldots , 9, 10 \}\) and the following sets:

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1. \(\displaystyle A = \{1,2,3,4,5,6,7\}\)

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2. \(\displaystyle B = \{2,4,6,8\}\)

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Then

\begin{equation*} A \cup B = \{ 1,2,3,4,5,6,7,8\}, \ A \cap B = \{2,4,6\}, \ A - B = \{1,3,5,7\}, \ B - A = \{8\}, \ \sim A = \{8,9,10\} \end{equation*}

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Here are some Venn diagram showing the sets above:

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Checkpoint 1.3.15.

Let the universe \(U = \{ \text{ all polygons } \}\) and define the following sets:

\begin{equation*} A = \{ \text{ all triangles } \}, \end{equation*}

\begin{equation*} B = \{ \text{ all polygons with at least one obtuse angle (angle larger than 90 degrees) }\}, \end{equation*}

\begin{equation*} C= \{ \text{ all polygons with at most 5 sides } \}. \end{equation*}

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For each set:

• Describe the set in words

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• Give an example of an element contained in the set

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• Give an example of an element not contained in the set (but still in the universe!)

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1. \(\displaystyle A \cap B\)

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2. \(\displaystyle A \cup C\)

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3. \(\displaystyle \sim B\)

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4. \(\displaystyle C - B\)

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5. \(\displaystyle A - A\)

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Solution 1.

Since elements in an intersection need to satisfy being a member of *both* A and B, we have that \(A \cap B = \{ \text{ obtuse triangles} \}\text{.}\)

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A possible example of an element in this set is a triangle with one angle of 100 degrees:

A possible example of an element not in this set is a square:

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Solution 2.

Since elements in a union need to satisfy being a member of *either* A or B (or both), we have that \(A \cup C = \{ \text{ all polygons that are triangles or have at most 5 sides.} \}\text{.}\) However, triangles are polygons of at most 5 sides (they have three). Thus we have that \(A \subseteq C\) and thus we can express \(A \cup C = \{ \text{ all polygons of at most 5 sides.} \}\) and note \(A \cup C = C\text{.}\)

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An example of an element of the set is the square:

and a non-example would be a hexagon since it has more than 5 sides:

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Solution 3.

Since elements in complement are all elements not in the original set, we have that \(\sim B = \{ \text{ all polygons with no obtuse angles } \}\text{.}\) In a future Geometry chapter we will show that this set will only contain certain triangles and quadrilaterals (polygons with 4 sides).

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An example of an element in the set is a square, since it has all angles of measure exactly 90 degrees (and not larger):

A non-example is a triangle with one angle with measure 100 degrees:

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Solution 4.

Since elements in the set difference need to satisfy being a member of C but not B, we have that \(C - B = \{ \text{ polygons with at most 5 sides but without any obtuse angles } \}\text{.}\) Again, in a future chapter we will show that \(\sim B = C - B\)

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Note that the examples and non-examples from the previous part work for this set as well.

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Solution 5.

Elements in the set difference need to satisfy being a member of A but not A. However there are, of course, *no* elements that are both a member and not a member of a set. Thus, we have that \(A - A =\emptyset\text{.}\)

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Note that since the set is empty, no polygon is an example of an element in the set, and every polygon is a non-example.

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