CLEP College Algebra/Intro to Algebraic Proofs

To certain students, math feels like a memorized set of rules. However, the only things that students may ever need to memorize in algebra are the basics of what is true about the object being studied and some notation. Any and all properties that students may feel like they need to remember are actually something someone can derive from the foundations they actually should remember.

In this page, we will introduce laws of logic and properties of conditional statements. We will be applying the laws of real numbers and algebra to prove some often-used theorems. Finally, we will use algebra to demonstrate a mathematical statement as true.

Introduction to Sets

The basic understanding of anything related to mathematics begins with learning about sets. While this will not be a substitute to a discrete math class (or the CLEP College Mathematics exam), some concepts you usually learn in there will be introduced much earlier on. As such, we will attempt to make this introduction intuitive and easy-to-understand. Finally, learning about sets and logic can easily help you in real world situations (moreso logic), especially in the context of probability.

Most people will imagine a set as an aggregation of objects (called elements) that which is defined. However, there needs to be a restriction. Template:Calculus/Def Here is a question: can a set contain itself? If there is such a thing as the set that contains all sets, then the set that does not contain itself would be in the set. However, that is impossible; that set would not be in itself but can not be defined as such because this set contains all sets. Since it is impossible to define whether it is a set under this such definition, the set that contains all sets cannot be possible (this paradox is known as Russel's Paradox). The restriction that a set must be well-defined is a necessity for this reason.

"Well-defined" is vague for now, but the restrictions for what constitutes "well-defined" is beyond the scope of this text. For those that are curious, understanding Zermelo–Fraenkel set theory would be important (although a little high level for what we want to accomplish in College Algebra).

A set can contain any number of objects. For example, we could have a set that contains all of the positive, even, single-place numbers, $A$. We could also have a set that contains nothing (we call that the empty-set, $\mathrm{\varnothing }$). We could also have the set of common breakfast drinks, $D$.

• $A=\{0,2,4,6,8\}$
• $\mathrm{\varnothing }=\{\}$
• $D=\{\text{coffee},\text{milk},\text{water},\text{orange juice},\text{apple juice}\}$

Note how we define the set. The way we define it is with curly braces, $\{\}$ with the elements denoted and separated by commas. Sets are also unordered, meaning that the order of the elements in the set does not change whether one set is equal to another. Thus, if $A$ is defined as earlier, and $B=\{0,4,2,8,6\}$, then $A=B$. This tells us something very important about sets.

The Elements Define a Set

By what we defined, a set is a collection of elements. This means the objects define the set. Remember, this is true by definition, so this is sound logic. Template:Calculus/Def If there are two sets, $A$ and $B$, then one of the sets can contain the other as an element, such as $A\in B$. If otherwise, then $A\notin B$. Consequently, the question of whether or not a set is contained within another is a question of conditional logic: it is "true" that $A\in B$ or it is "not true" that $A\in B$. There is no in-between option.

A set can only ever equal another if they contain the same elements. The order of the elements do not matter, so long as the element of each corresponds through its definition, and the number of the elements within each set is also identical. This therefore means a set is uniquely determined by its elements.

Template:ExampleRobox Let $A=\{0,2,4,6,8\}$ and $B=\{0,1,\{2,4\},3,6,4\}$.

(a) For all elements found in the both sets, write them as elements in the set $S$.

(b) For all non-common elements, write all such "$e\in A$ but $e\notin B$" or vice-versa, where $e$ is an element found in one and only one set but not in the other.

Answers:

(a) $S=\{0,4,6\}$.

(b) $2,4,8\in A$ but $2,4,8\notin B$ AND $1,\{2,4\},3\in B$ but $1,\{2,4\},3\notin A$.

Explanation for (a) and (b):

These are mostly self-explanatory. Keep in mind that any set within another set is defined as the element of the set. However, if that element contains any objects, then it is not part of the "parent" set. For example, let $E=\{2,4\}$. Here $B=\{0,1,E,3,6,4\}$. We know $E\in B$ and $2,4\in E$. However, any element in $E$ does not define $B$. The elements only define its own set. Template:Robox/Close

How to define a set contains an element is done in two different ways, explicitly and implicitly. Template:Calculus/Def The explicit way of defining a set is what we have been using throughout the entire section. However, this can also tend to be very useless when there are an infinite number of elements in the set. The next definition should held us in that aspect. Template:Calculus/Def A more robust definition of a set is thus the following: Template:Calculus/Def Because elements define the set, it is often important to know the size of the set. This is known as the cardinality or size of the set. Because we want to keep this part of the subject intuitive, we will use size from here on. Template:Calculus/Def Template:ExampleRobox Let $D$ be the set of common breakfast drinks. A non-comprehensive list of the most common breakfast drinks in the U.S. are given: Coffee, Milk, Orange Juice, Apple Juice, Water.

(a) Write the set $D$ in set notation.

(b) Find the size of the set. Write it in official notation

Answers:

(a) $D=\{\text{coffee},\text{milk},\text{water},\text{orange juice},\text{apple juice}\}$.

(b) $|D|=5$.

Explanation for (a) and (b):

These are mostly self-explanatory. The number of the items in the list is five, so the size of the set is five. Template:Robox/Close

Check your Understanding

Directions: Some questions will require you to select from among five choices. For these questions, select the BEST of the choices given.
Some questions will require you to type a numerical answer in the box provided.
Some questions will require you to select one or more answer choices. <quiz shuffle="none" display="simple"> {Let $A=\{0,\{0\},1,\{0,1\},2,\{0,1,2\},3,\dots ,100\}$ and $B=\{0,1,\dots ,99,A\}$. Which of the following statements are true? Select all that apply. |type="[]"} + $0,1,2,3,\dots ,99\in A$ and $0,1,2,3,\dots ,99\in B$. || Because sets are unordered, and this set is implicit, any and all numbers below 100 and above 0 are in set $A$. Further, as listed in $B$, those are elements of the list. - $\{0,1,2,3,\dots ,99\}\in A$ and $\{0,1,2,3,\dots ,99\}\in B$. || While it is true $\{0,1,2,3,\dots ,99\}\in A$, it is not necessarily true that it is in $B$. It is true that $A\in B$. However, elements of $A$ do not automatically become objects of $B$. + $0,1,2,3,\dots ,99\in \{0,1,2,3,\dots ,99\}$ and $\{0,1,2,3,\dots ,99\}\in A$.

{Let there be a set $B=\{A\}$. Find $|A|$ and $|B|$. If unknown, write unknown. |type="{}"} $|A|=$ { unknown | Unknown _10 }, and $|B|=$ { 1 _10 } || Although it is unknown how many elements are found in the set $A$, it is known that there is one set $A\in B$. Therefore, we know the number of elements within set $B$ is $1$.

{Use the following information to answer items 3-4 below.
It is known that $a\in A$, $a,b\in B$, and $c\in C$.}

{Which of the following is true of $D=\{A,B,C\}$? |type="()"} - $a\in D$. - $a,b\in D$. - $c\in D$. - $a,b,c\in D$. + None of the above. </quiz> More questions added later.

Properties Define an Element, Ergo, the Set

An object can have properties that would be important to write down. If the object itself changes because of it, then it is important to have that defined. These are referred to as conditionals (the more common one is propositional functions, but this is avoided in the interest of providing a definition of functions that is not treated as tautological when the definition of a function is non-tautological). Template:Calculus/Def Since a set is defined by its objects, the conditional also defines the set. The notation for a given set that has a conditional is

$S=\{x\mid C(x)\}$ or $S=\{x:C(x)\}$

where such a set is read as $S$ is the set of all $x$ such that $C(x)$ is true. The term "such as" originates from either the vertical bar ( $|$ ) or the colon ( $:$ ). In this textbook, we will use the vertical bar $|$ to refer to "such as" since this is standard in most College Algebra courses.

A special type of notation is used when a set can better represented through conditionals. Take the set of all Fibonacci numbers, for example, $F$:

$F=\{0,1,1,2,3,5,8,13,21,34,55,89\dots \}$.

This set can be listed implicitly. However, when there is a pattern to the set's elements, then a set can become much more useful. Let $f(k)$ represent the index of the Fibonacci sequence, where $k$ is an index starting at zero. The set can be equivalently written as the following:

$F=\{f(k)|f(k)=f(k-2)+f(k-1),f(k),k\in {\mathbb{Z}}^{+},f(0)=0,f(1)=1\}$.

While this notation seems to hurt rather than help (and seem more confusing), in actuality, this tells us a lot of information about the set, including the pattern used, what initial conditions are necessary for the set to exist, and what set $f(k)\text{ and }k$ need to belong to for this pattern to work. Template:Calculus/Def We can now easily read sets. Looking back at the set $F$ (the set of Fibonacci numbers):

$F=\{f(k)|f(k)=f(k-2)+f(k-1),f(k),k\in {\mathbb{Z}}^{+},f(0)=0,f(1)=1\}$

.

$F$

is the set of all

$f(k)$

such that

$f(k)=f(k-2)+f(k-1)$

,

$f(k)\text{ and }k$

belong to the set of positive numbers, and there are two initial conditions,

$f(0)=0\text{ and }f(1)=1$

.

Template:ExampleRobox Write each of the following using set-builder notation and implicit or explicit notation.

(a) The set of all even numbers.

(b) The set of all odd numbers.

(c) The set of all prime numbers.

(d) The set of all real solutions to the equation $\frac{12(x+2)(x\sqrt{2}-1)(x-1)(x-2)}{(x+2)-(x+4)}=0$.

(e) The set of all natural number solutions to the equation $\frac{12(x+2)(x\sqrt{2}-1)(x-1)(x-2)}{(x+2)-(x+4)}=0$.

(f) The set of all integer solutions to the equation $2x+5=8$.

Answers:

(a) $\{2n|n\in \mathbb{Z}\}=\{n\in \mathbb{Z}|n\text{ is even}\}=\{\dots ,-4,-2,0,2,4,\dots \}$.

(b) $\{2n+1|n\in \mathbb{Z}\}=\{n\in \mathbb{Z}|n\text{ is odd}\}=\{\dots ,-3,-1,1,3,\dots \}$.

(c) $\{n\in \mathbb{N}|n\text{ is a prime number}\}=\{2,3,5,7,11,13,\dots \}$.

(d) $\{x\in ℝ|\frac{12(x+2)(x\sqrt{2}-1)(x-1)(x-2)}{(x+2)-(x+4)}\}=\{-2,\frac{1}{\sqrt{2}},1,2\}$.

(e) $\{x\in \mathbb{N}|\frac{12(x+2)(x\sqrt{2}-1)(x-1)(x-2)}{(x+2)-(x+4)}=0\}=\{1,2\}$.

(f) $\{x\in \mathbb{N}|2x+5=8\}=\mathrm{\varnothing }$.

Explanations:

Items (a)-(c)

These are mostly self-explanatory. Note that there are two possible answers we could write. $n\in \mathbb{Z}$ and $2n$ are both expressions. However, to make sure both are equivalent, we need a rule so that the expressions can the elements in the set are equal to the other. Similar for (b).

Items (d)-(f)

The pair of solutions is given by the expression of some $x\in ℝ$ or $x\in \mathbb{N}$. The rule is the equation itself since it describes how one finds the solution set. This is all that is needed for the set-builder notation. To obtain the solutions in the set, simply solve:

$\begin{array}{ccc}\frac{12(x+2)(x\sqrt{2}-1)(x-1)(x-2)}{(x+2)-(x+4)}& =0& \text{(Original equation)}\\ \frac{12(x+2)(x\sqrt{2}-1)(x-1)(x-2)}{-2}& =0& \text{(Distributive property)}\\ (x+2)(x\sqrt{2}-1)(x-1)(x-2)& =0& \text{(Division property of equality)}\end{array}$

By the zero factor theorem,

$\Rightarrow x=-2,\frac{1}{\sqrt{2}},1,2$

This is where the explicit set notation comes. Because the natural numbers do not have negative numbers, fractions, or square roots, the natural number set of solutions is only $1,2$.

This type of analysis allows us to determine that one of the sets, see item (f), is empty, and is therefore equivalent to the empty set, $\mathrm{\varnothing }$. Template:Robox/Close

Check your Understanding

Directions: Some questions will require you to select from among five choices. For these questions, select the BEST of the choices given.
Some questions will require you to type a numerical answer in the box provided.
Some questions will require you to select one or more answer choices. <quiz shuffle="none" display="simple"> {Suppose $A$ and $B$ are well-defined sets. Let $A=\{a|1\le a\le n,a,n\in \mathbb{N}\}$ and $B=\{b|b\ge 1,b\in \mathbb{N}\}$. If $|B|=|A|+1$, which of the following must be true? |type="()"} - If $a\in A$, and $|A|<|B|$, then $A\in B$. || This is only true if the set $A$ was an element of $B$. Based on all the information given, the set-builder notation gives all $b\in [1,n+1]$. + If any $a\in A$, then $a\in B$. || The set-builder notation implies that $a$ is between $1$ and $n$ (inclusive), while $b$ is between $1$ and $n+1$ (inclusive). Because any element in $a$ is necessarily contained within that of $b$, $a\in B$. - If any $a\in \mathbb{N}$, then $a\in A$. || If $a=n+1$, then $a\notin A$. One counterexample can quickly disprove this. - If any $b\in B$, then $b-1\in A$. || If $b=1$, then $1-1=0\notin A$. Pay close attention to the set-builder notation. - If $b\in B$, and $|B|<|A|$, then $B\in A$. || The size of $B$ cannot be less than $A$ because the size is necessarily positive. Even if true, however, $B\notin A$ because the set-builder notation does not give us the directions to place a set anywhere in $A$. </quiz> More questions added later.

Comparing Sets

Of course, many sets are similar to another. As such, many mathematicians find it very helpful to compare sets. Some vocabulary terms will be listed here. Template:Calculus/Def The idea of the subset is very simple; however, it can be very powerful when comparing two sets. Nevertheless, many students may neglect to take care in their notation while trying to compare sets. The examples presented below are correctly stated. The reasoning is given as well. This type of reasoning will be required for you to identify subsets and non-subsets.

• $\{1,3,2\}\subseteq \{1,2,3,4,5\}$.
  • Because $1,3,2\in \{1,2,3,4,5\}$, $\{1,2,3\}$ (the set) is a subset of $\{1,2,3,4,5\}$.
• $\{1,3,2\}\subseteq \{1,2,3\}$.
  • Because $1,3,2\in \{1,2,3\}$, $\{1,2,3\}$ is a subset of $\{1,2,3\}$.
• $\{1,2\}⊈\{4,3,\{2,1\}\}$.
  • Although $\{1,2\}\in \{4,3,\{2,1\}\}$, the elements of set $\{1,2\}$ cannot be found in the set $\{4,3,\{2,1\}\}$. In other words, $1,2\notin \{4,3,\{2,1\}\}$.
• $\{\{1,2\}\}\subseteq \{4,3,\{2,1\}\}$.
  • Because $\{1,2\}\in \{4,3,\{2,1\}\}$, $\{\{1,2\}\}\subseteq \{4,3,\{2,1\}\}$. If one cannot see this, let $A=\{1,2\}$. Notice that $\{4,3,\{2,1\}\}=\{4,3,A\}$. Because $\{2,1\}=A\in \{4,3,\{2,1\}\}=\{4,3,A\}$, $\{\{1,2\}\}\subseteq \{4,3,\{2,1\}\}$.

One other definition that is useful to know is the superset. The definition is complex yet the intuition is rather simple. Template:Calculus/Def Some examples below will be listed below, along with an explanation. Notice they are the same as above.

• $\{1,3,2\}⊉\{1,2,3,4,5\}$.
  • Because $\{1,2,3,4,5\}⊈\{1,3,2\}$, $\{1,3,2\}⊉\{1,2,3,4,5\}$.
• $\{1,2,3,4,5\}\supseteq \{1,3,2\}$.
  • Because $\{1,3,2\}\subseteq \{1,2,3,4,5\}$, $\{1,2,3,4,5\}\supseteq \{1,3,2\}$.
• $\{1,3,2\}\supseteq \{1,2,3\}$.
  • Because $\{1,2,3\}\subseteq \{1,3,2\}$, $\{1,3,2\}\supseteq \{1,2,3\}$.
• $\{4,3,\{2,1\}\}⊉\{1,2\}$.
  • Because $\{1,2\}⊈\{4,3,\{2,1\}\}$, $\{4,3,\{2,1\}\}⊉\{1,2\}$.
• $\{4,3,\{2,1\}\}\supseteq \{\{1,2\}\}$.
  • Because $\{\{1,2\}\}\subseteq \{4,3,\{2,1\}\}$, $\{4,3,\{2,1\}\}\supseteq \{\{1,2\}\}$.

Check your Understanding

Directions: For each of the following conditions stated below, four to five lines will be provided. Identify the correct definition that would apply to the given set for the one provided in the problem. Some problems require multiple correct answers and as such multiple selections. <quiz shuffle="none" display="simple"> {Suppose $A=\{4,5,10,20\}$. |type="[]"} |Subset|Superset|Neither -+- $\mathbb{N}$. || The set of natural numbers is a set that contains only numbers. Applying the definition of superset here, for all $a\in A$ that can be found in $\mathbb{N}$. As such, $\mathbb{N}$ is a superset +-- $\{4,5\}$. || $4\in A$ and $5\in A$, so this is a subset. +-- $\{5,10,20\}$. || $5,10,20\in A$, so this is a subset. --+ $\{3,4,5\}$. || While $4,5\in A$, $3\notin A$, so this is not a valid subset of $A$. From this, it cannot be a valid superset either since $10\notin \{3,4,5\}$, for example. --+ $4$. || Although $4\in A$, $4$ is not a set. Because this is not a set, this could never be a valid subset nor superset.

{Suppose $B=\{4,5,\{10,20\}\}$. |type="[]"} |Subset|Superset|Neither ++- $B$ || Because all elements of $B$ can also be found in $B$, $B\subseteq B$. In a sense, it could also be a superset. +-- $\{4,5\}$ || Since $4,5\in B$, $\{4,5\}\subseteq B$. --+ $\{5,10,20\}$ || Although $5\in B$, $10,20\notin B$ (notice $\{10,20\}\in B$), so $\{5,10,20\}⊈B$. From this, $\{5,10,20\}⊉B$. +-- $\{\{10,20\}\}$ || Notice that the set $\{10,20\}\in B$, so $\{\{10,20\}\}\subseteq B$.

{Suppose $C=\mathbb{N}\cup \{0\}$ (i.e., the set that contains the set of natural numbers including zero). |type="[]"} |Subset|Superset|Neither -+- $\{\sqrt{x}|x>0,x\in \mathbb{Z}\}$ || Notice how for every natural number and zero, it can be found $\{\sqrt{x}|x>0,x\in \mathbb{Z}\}$. Because of this, $C\subseteq \{\sqrt{x}|x>0,x\in \mathbb{Z}\}$, so $\{\sqrt{x}|x>0,x\in \mathbb{Z}\}\supseteq C$. +-- $\mathbb{N}$ || Because all elements of $\mathbb{N}$ can also be found in $C$, $\mathbb{N}\subseteq C$. -+- $\mathbb{Z}$ || Notice how for every natural number and zero, it can be found $\mathbb{Z}$. Because of this, $\mathbb{Z}\supseteq C$. +-- $\{x^2|x\in \mathbb{Z}\}$ || Since $\{x^2|x\in \mathbb{Z}\}=\{\dots ,9,4,1,0,1,4,9,\dots \}$, we can show that for all elements found in $\{\dots ,9,4,1,0,1,4,9,\dots \}$, it can also be found in $C$. As such, $\{x^2|x\in \mathbb{Z}\}\subseteq C$.

{Suppose $D=\{\mathbb{N},\{\mathbb{Z}\}\}$. |type="[]"} |Subset|Superset|Neither +-- $\mathbb{N}$ --+ $\{\mathbb{N}\}$ --+ $\mathbb{Z}$ +-- $\{\mathbb{Z}\}$ --+ $ℝ$ </quiz>

Combining Sets

The process of combining sets can be very useful, especially when working in contexts of probability.

If you want to explain this a five-year-old: the "union set" is the set that contains all things in $P$ and $Q$.

A few examples are presented below, where $A=\{\text{a},\text{b},\text{h}\}$, $B=\{\text{b},\text{d},\text{e},\text{f}\}$, and $C=\{\text{c},\text{e},\text{g}\}$. These are all the possible combinations of the sets (excluding combinations with its own set).

• $A\cup B=\{\text{a},\text{b},\text{d},\text{e},\text{f},\text{h}\}$
• $A\cup C=\{\text{a},\text{b},\text{c},\text{e},\text{g},\text{h}\}$
• $B\cup C=\{\text{b},\text{c},\text{e},\text{g},\text{d},\text{f}\}$
• $A\cup B\cup C=\{\text{a},\text{b},\text{c},\text{d},\text{e},\text{f},\text{g},\text{h}\}$

If you want to explain this a five-year-old: the "intersection set" is the set that contains all things that are found in both $P$ and $Q$.

A few examples are presented below, where $A=\{\text{a},\text{b},\text{h}\}$, $B=\{\text{b},\text{d},\text{e},\text{f}\}$, and $C=\{\text{c},\text{e},\text{g}\}$. These are all the possible combinations of the sets (excluding combinations with its own set).

• $A\cap B=\{\text{b}\}$
• $A\cap C=\mathrm{\varnothing }$
• $B\cap C=\{\text{e}\}$
• $A\cap B\cap C=\mathrm{\varnothing }$

If you want to explain this a five-year-old: the "difference set" is the set that contains elements of $P$ but excluding what are also found in or are in $Q$.

Several examples are presented below, where $A=\{\text{a},\text{b},\text{h}\}$, $B=\{\text{b},\text{d},\text{e},\text{f}\}$, and $C=\{\text{c},\text{e},\text{g}\}$. These are all the possible combinations of the sets (excluding combinations with its own set and any combinations with all three sets).

• $A-B=\{\text{a},\text{h}\}$
• $A-C=\{\text{a},\text{b},\text{h}\}$
• $B-A=\{\text{d},\text{e},\text{f}\}$
• $B-C=\{\text{b},\text{d},\text{f}\}$
• $C-A=\{\text{c},\text{e},\text{g}\}$
• $C-B=\{\text{c},\text{g}\}$

Introduction to Logic

Logic is important in many fields. However, it is especially in mathematics since without this tool, we could not be able to come up with theorems nor algorithms. That is why it is especially important for us to teach this tool. While this may not be a philosophy class, we will be teaching as much as needed for us to be able to present proofs to you without too much trouble.

The process of deduction is simply the ability to "connect" statements through some related way.

For example, we know that Aristotle is a "grefunkle," and "grefunkles" are "prostireoni." Therefore, Aristotle is a "prostireonis."

Now, "grefunkles" and "prostireoni" are made up words; these things do not exist. However, the sentence we made is logically correct. Remember, logic does not require the correct information, so this sentence is logically correct while not essentially correct.

Statements

Logic must begin with "statements." After all, the example we presented could not exist without "statements."

Statements need a "verb" of some kind, mostly "declaratory verbs," like "is" or "are." For example, "Apples are blue" is a statement since it is either true or false (here, it is false for many apples). Questions are not statements. More examples are provided.

1. Every even number is divisible by itself and two. True.
2. All prime numbers are odd. False.
3. Twelve more than twenty-two less than the initial means that the initial is twenty-two more than twelve less than the result. True.
4. If a side of a parallelogram is $p$, then the area of the parallelogram is $\frac{1}{2}{p}^2$. False.
5. The side length of any one side of an isosceles triangle is equal to the square root of the two remaining sides. False.
6. If one and only one side of a quadrilateral is not parallel to its opposing side, then it is a trapezoid. True.

Statements #4 and #6 are a type of statement we will be discussing very closely in this section. Keep in mind the form of the sentence for future reference.

Notice how statement #4 contains a variable $p$. A statement may use variables so long as no matter what the variable stands for or what the variable will equal to, the statement will always either be true or false by simply the statement alone. We will discuss this more in this section.

Also the statements do not have to be sentences, as has been stated in the definition:

• $\pi =3.141519265358979...$ True
• $\pi =2.718281828459045...$ False.
• $2\in \mathbb{N}$ True
• $2\in {\mathbb{Z}}^{-}$ False.
• $\{\}\subseteq \mathrm{\varnothing }$ True
• $\{\}\supseteq \mathrm{\varnothing }$ True

Keep in mind that the mathematical expressions you use need to state something in the "declaratory verb" form we describe. For example, $2\in \mathbb{N}$ states in a more compact form $2$ is an element of the set of natural numbers, so this is a valid statement. What would not be a valid statement is simply $\mathbb{N}$ (the set of natural numbers). This tells us nothing.

Statements can be used many times in a proof or even a math problem. This is why mathematicians have developed tools to be able to declare statements using only one letter, usually $P,Q,R,S$. Subscripts are used when more than four statements are necessary, although one may be allowed to use more than the letters provided (if somewhat unconventional).

• $P:$ There are more than two different letters in the word "cranberry." True.
• $Q:$ The sum of the digits of any three digit number is between $1$ and $27$. True.
• $R:$ Any two consecutive primes have a distance of six between them. False.
• $S(x,y):\{x,\{y\}\}\cap \{\{x\},y\}=\mathrm{\varnothing }$. True.

With $S(x,y)$, the parenthetical statement tells us that there are two variables within a statement. This is a notation because at times, the variables become very important to determine its truth value. This will be explained shortly.

As another rule of thumb, if a sentence does not describe anything about a subject, then it is not a statement. Here are a few examples of non-statements.

• The set $S$ and $P$.
• What is $S\cup P$?
• $2+2$
• $\frac{1}{2}=2x-1$.

The final bullet point is a tricky one for students unfamiliar with identifying statements. Some will say this is a statement because it is either true or false that $2x-1$ is equal to $\frac{1}{2}$. However, if you think about it, this statement is entirely dependent on the value of $x$. This is called an open sentence.

Look back at the open sentence. Let $P(x):\frac{1}{2}=2x-1$. If $x=\frac{3}{4}$, then $P\left(\frac{3}{4}\right)$ is true. Otherwise, it is false. This is why $P(x)$ is not a statement. It depends on the value of $x$. A statement needs to be either true or false but not due to a changing artifact.

One important topic to discuss when working with statements is the negation.

When we take a statement and we negate it, we make a true statement false or a false statement true. Take the following as a couple or examples:

• Every even number is not divisible by itself and two. False.
• Some prime numbers are not odd. True.

The second example required us to make the "all" into "some." More information about how and when this is done will be shown later.

The examples above are from the beginning. Notice how we made these statements opposite their truth value from the beginning. Mathematicians use these so often that we have a notation for it.

Suppose $R:\mathrm{\neg }P$. For $R$ to have a truth value of $T$ (true), it is required for $P$ to be false. Otherwise, $R$ is $F$ (false). This can be further shown using this simple truth table (a table that shows the truth values of each individual statement and the resulting column):

$\begin{array}{cc}P& \mathrm{\neg }P\\ T& F\\ F& T\end{array}$

Properties of Conditional and Biconditional Statements

Conditional Statements

If you want to have any hope of understanding mathematical theorems at a deeper level, then you will need to understand conditional statements. Template:Calculus/Def Other ways in which this sentence can be written is presented below:

• "$Q$ if $P$"
• "$P$ only if $Q$"
• "Whenever $P$, then also $Q$"
• "$P$ is a sufficient condition of $Q$"
• "$Q$ is a necessary condition for $P$"

Here are some examples:

• If you pass this final exam, you pass the course.
• You can travel to Spain if you learn the Spanish language and culture.
• Garfunkles are prostireoni only if at least 100 years have passed between garfunkles and prostireoni.
• Whenever you lose power, then you also lose heat.
• Composing good music is a sufficient condition of getting paid $100,000.
• Getting famous is a necessary condition for writing the best calculus textbook.

Conditional statements have a sufficient condition (after the "if") – or a hypothesis – and a necessary condition (after the "then") – or a conclusion. This is why if our true hypothesis leads to a false conclusion (the result is different from our stated conclusion), then the conditional statement is false.

Mathematics is a big friend of the conditional statement. Here are some familiar statements using the conditional statement.

• If a right triangle has leg lengths $a,b$ and hypotenuse $c$, then $a^2+b^2=c^2$.
• $A\cap \mathrm{\varnothing }=\mathrm{\varnothing }$ if $A$ is a finite set.
• $f(x)$ are one-to-one only if $f(a)=f(b)$ for some real numbers $a\ne b$.

Because these type of statements are so important, it is necessary to understand how they work. There is a short-hand notation for these statements, which is presented below.

The way we determine the truth value for $P\Rightarrow Q$ is to write a truth table. We can determine each row of the truth table by simply looking back at the definition of conditional statements.

$\begin{array}{ccc}P& Q& P\Rightarrow Q\\ T& T& T\\ T& F& F\\ F& T& T\\ F& F& T\end{array}$

A conditional statement is true if the sufficient condition leads to the stated conclusion. That is why the second row gives us the result of a false truth value. This still explains the two final statements: if we have a false hypothesis but the conclusion is true, then either way, something true did happen but not as a result of the hypothesis. The hypothesis is not necessarily required for a conditional statement to be true so long as the conclusion is true. Finally, whenever a false hypothesis leads to a false conclusion, then that makes the conditional statement true because something false leading to something false is itself a confirmation of the conditional statement.

Template:ExampleRobox There are six rules to be able to enter or stay in a non-specific kid's music club. If at least one of these rules does not apply to the student, they will be placed in the teen's music club (covered by rule #5) or will be given a one month grace period if not over twelve years of age. The one month grace period will not be invoked if one is in the kid's or teen's music club.

1. To enter, it is necessary for one to be twelve years or younger.
2. If one is older than twelve years, then one can only teach the young musicians.
3. Grade 3 music or higher gives you automatic acceptance only if you perform grade 3 or higher music thrice per month.
4. Passing a mandatory music theory test at the end of the year is sufficient to maintain being in the kid's music club.
5. Being in the teen's music club is necessary for one that is older than twelve years.
6. Performing with "superior" rating on one's determined grade level is necessary to stay in the kid's music club.

Items (a)-(d) below have a few situations. Determine if there is enough information to kick these people out (whether temporarily or permanently) of the club based on their descriptions. Provide a reason for each of the young people below. Otherwise, state that individual as allowed to stay.

(a) George is thirteen and plays grade 3 music or higher three times a month.

(b) Miranda is ten and failed the music theory test at the end of the year. Plays grade 2 music with "excellent" rating.

(c) Carlos passed the music theory test and is nine. Plays grade 1 music with "superior" rating.

(d) Paul is thirteen and is the head teacher of music theory for the young kids. Plays grade 4 music.

Items (e)-(f) ask you a couple of questions regarding the situation provided. Answer items (e)-(f).

(e) Is there a situation that would make it impossible to determine whether a student may stay in the kid's music club? What is that situation if possible?

(f) Rewrite rule #2 into an equivalent statement not in "if...then" form.

Answers:

(a) George is rejected from the kid's music club and hereby placed in the teen's music club for failure to be a teacher.

(b) Miranda is hereby rejected from the kid's music club for a duration of one month because of her failure to pass the music theory test and her failure to gain a superior rating for her grade 1 music.

(c) Carlos may stay in the kid's music club.

(d) Paul may stay in the kid's music club.

Explanation

More will be added later. Template:Robox/Close

Biconditional Statements

Conditional statements also have its different, non-equivalent forms. These statements are not equivalent because their truth values can be different. Notice how the end result is different simply by switching the conclusion and the hypothesis.

$\begin{array}{cccc}P& Q& P\Rightarrow Q& Q\Rightarrow P\\ T& T& T& T\\ T& F& F& T\\ F& T& T& F\\ F& F& T& T\end{array}$

By this brute force method, we were able to show that these statements are not equivalent. These statements are equivalent if both $P$ and $Q$ are true. However these two statements can be true depending on the statements. Real life examples are rare.

Here is a conditional statement that has a true converse: "If a phrase is chromatic, then it contains notes separated by half-steps." The converse of the statement is "If a phrase contains notes separated by half-steps, then the phrase is chromatic." The reason why is because a chromatic phrase, by definition, uses notes separated by half-step, so both $P\Rightarrow Q$ (the conditional statement) and $Q\Rightarrow P$ (the converse) are true.

In situations where both the conditional and converse is true, you obtain a biconditional statement.

Biconditional statements require that the conditional statement is true. Therefore, $P$ only if $Q$. However, the converse is also true, so $P$ if $Q$. As such, we say a statement is biconditional by stating that "$P$ if and only if $Q$."

Because it requires that $P\Rightarrow Q$ and $Q\Rightarrow P$ be true, it will mean that the middle rows will be false. Thus, the following truth table results:

$\begin{array}{ccc}P& Q& P\iff Q\\ T& T& T\\ T& F& F\\ F& T& F\\ F& F& T\end{array}$

More detail will be provided as to why this is true by looking at the next section.

More will be provided later.

Related Forms of Conditional Statements

The related forms of conditional statements are of great importance. However, some proofs classes ignore the different forms of the conditionals statements. However, at times, proving a conditional statement involve proving all of the three related forms.

A mathematician's job is to test the limits of their statements and find what can be true or false.

There are some properties of the related statements that are not often taught. However, we feel it may be at least somewhat important to understand that these statements exist and that there may be some logical equivalences.

A good way to see this property is true is with a truth table:

$\begin{array}{cccc}P& Q& P\Rightarrow Q& \mathrm{\neg }Q\Rightarrow \mathrm{\neg }P\\ T& T& T& T\\ T& F& F& F\\ F& T& T& T\\ F& F& T& T\end{array}$

Hopefully, it is clear why these are related statements. Because the negation of $Q$ and the negation of $P$, and the order of the hypothesis and conclusion require that the truth values consequently change. Thus, these statements are logically equivalent.

The same can be shown with the inverse and the converse.

$\begin{array}{cccc}P& Q& Q\Rightarrow P& \mathrm{\neg }P\Rightarrow \mathrm{\neg }Q\\ T& T& T& T\\ T& F& T& T\\ F& T& F& F\\ F& F& T& T\end{array}$

More will be added later.

Combining Statements

There are many times in life we have to think about the truth of a statement which contains two statements. For example, "The light is on and the light is off." This statement is definitely false. How do we know? Because of the word and.

Let $A$: "The light is on and the light is off." Notice how we can break the one large statement into two small statements thanks to the conjunction and. As such, let $P$: "The light is on" and $Q$: "The light is off." According to the English language, "on" is opposite in meaning to "off." However, a lightbulb can only even be on or off. Thus, if $P$ is true, then $Q$ is false by default. Therefore, $R$ is false.

The process of combining statements will be very useful in proofs. Since these combinations are defined by themselves, we do not need to prove them. Finally, because this combination is based entirely on the english language, the truth of a statement can be easily determined.

Based on what we did with the introductory example, let us generalize what we did.

Suppose $R:P\wedge Q$. For $R$ to have a truth value of $T$ (true), it is required for both $P$ to be true and $Q$ to be true. If either one is false, or both is false, then $R$ is $F$ (false). This can be further shown using a truth table.

$\begin{array}{ccc}P& Q& P\wedge Q\\ T& T& T\\ T& F& F\\ F& T& F\\ F& F& F\end{array}$

Suppose $R:P\vee Q$. For $R$ to have a truth value of $F$ (false), it is required for both $P$ to be false and $Q$ to be false. If either one is true, or both is true, then $R$ is $T$ (true). This can be further shown using a truth table:

$\begin{array}{ccc}P& Q& P\vee Q\\ T& T& T\\ T& F& T\\ F& T& T\\ F& F& F\end{array}$

By using these conjunctions and truth tables, we can easily determine when a open sentence will be true and whether a statement will be true or false.

Template:ExampleRobox Read the open sentence below that someone claims to always be false for any $(a,b)$.

$a$ is negative and $b$ is positive but $a-b\ne 0$.

Answer items (a)-(c) below.

(a) Write this in symbolic notation.

(b) Find a counter example. If possible, how many possible counter-examples exist. If not possible, explain why.

(c) For this combination, how many possible false truth values are there?

Answers:

(a) $(a<0)\wedge (b>0)\wedge \mathrm{\neg }(a-b=0)$

(b) Let $b=2$ and $a=-1$. $-1-(2)=-3\ne 0$, which is true. The open sentence is false in infinitely many instances so long as $a<0$ and $b>0$.

(c) 7.

Explanation

(a): Because $a$ is negative, $a<0$. Because $b$ is positive, $b>0$. Finally, because $a-b\ne 0$, it means that it is not the case that $a-b=0$, so $\mathrm{\neg }(a-b=0)$. Knowing that these statements are in tandem with each other according to the English language, all statements are combined with an and. All a "but" means is an "and," with emphasis.

(b): Explanation already given in Answers.

(c): Because there are three variables, and each of either a $T$ or $F$ truth value, there are $2\cdot 2\cdot 2=2^3=8$ possible truth values for the final open sentence. The most organized way to check how many possible open sentences are false is through the truth table.

$\begin{array}{cccccc}P& Q& R& P\wedge Q& \mathrm{\neg }R& P\wedge Q\wedge \mathrm{\neg }R\\ T& T& T& T& F& F\\ T& T& F& T& T& T\\ T& F& T& F& F& F\\ T& F& F& F& T& F\\ F& T& T& F& F& F\\ F& T& F& F& T& F\\ F& F& T& F& F& F\\ F& F& F& F& T& F\end{array}$

More will be added later.

Logical Equivalences

More will be added later.

Symbolic Logic

More will be added later.

Quantifiers

More will be added later.

Proving Mathematical Properties

Applying Laws of Real Numbers to Verify and Prove Mathematical Properties

Template:ExampleRobox Verification is an easy task since all you have to do is rewrite the equation we see above and change only one side. Because only one side changes, a chain of transitive properties may be applied to that side, until one reaches the final conclusion: a simple statement as a property of equality to itself. This may be easier to show than explain, so simply follow along with us.

According to the definition of division, $\frac{a}{c}=a(c{)}^{-1}$. Since this is true per each division found in the right expression, Template:Eqnref can be rewritten as the following:

Notice how each term in the right-hand side of the equation has a factor of $c^{-1}$ therein. Because that is true, the following is true of the right-hand side of the equation.

By the definition of division, the right-hand side of the equation is equivalent to $\frac{a+b}{c}$, so

Notice how the chain of equations herein can connect to every equality. Since Template:Eqnref and Template:Eqnref is true, we know the following equation below is true:

Because Template:Eqnref and Template:Eqnref is true, we know this below equation is true, and so on:

As shown in Template:Eqnref, we have verified the truth, and are therefore done with the problem. The "chain of equations" below shall perhaps show this chain of "change of one side only."

$\begin{array}{cc}\frac{a+b}{c}& =\frac{a}{c}+\frac{b}{c}\\ & =a(c{)}^{-1}+b(c{)}^{-1}& \text{Definition of fractions.}\\ & =c^{-1}(a+b)& \text{Factor }{c}^{-1}\text{ from }a(c{)}^{-1}+b(c{)}^{-1}\text{.}\\ & =\frac{a+b}{c}& \text{Definition of fractions.}\end{array}$

It is important to take note the language of the directive verbs found within the prompt in the example above. By asking the examinee to verify, it, in effect, tests the understanding of the writer to communicate how a certain statement can be made true by only validly changing one side.

In the example above, we decided to change the right-hand side of Equation Template:Eqnref. Here is a question for the reader: "Why did we decide to change that side only?" If you can answer this question with precision before the next two examples, then it is safe to infer that this student understands how to verify literal equations.

Note: despite being an important skill, it would be impossible to show this type of understanding on a multiple choice exam. Nevertheless, the journey of the mathematician is of the skeptic of statements made. You cannot fully regard a statement as true if you do not know the proof of it or the verification of it. Template:ExampleRobox This equation is not literal, so it is not necessary to change only one side to become the other. Nevertheless, it is not necessary one needs to solve the single-variable equation either. Instead, it may be easiest to simply substitute $x=12$ into the above equation. We shall do exactly that.

$12(12)-12=11(12)$

$144-12=132$

$132=132$

Because both sides of the equation are equal, we have verified $x=12$ is true. Template:Robox/Close One more example will be added later. Template:ExampleRobox This problem is a little different from the previous ones because the examinee needs to show something is true by what they are given, as well as demonstrate something is true through the derivation of different formulae and equations from fundamental properties. The importance of mathematical communication is what is tested here. The mathematician needs to describe what is given, or else, the proof will not follow logically as a form of deductive reasoning.

We cannot follow the strategy we had for the previous two problems. This is because we are trying to communicate that a statement is beyond a shadow of a doubt true! With verification, we could assume the final statement is true, and thereby work backwards from where we started. However, when proving something, we have to show that when only given one statement, we can fully derive the other side without going backwards from where we started. Understanding this distinction is crucial!

Let us first start with rewriting the expression $\frac{a}{b}\cdot \frac{c}{d}$ into its equivalent: Template:Equation Notice how we can apply a property of multiplication, that multiplying real numbers (and by extension integers) is associative. Therefore, we may rewrite Template:Eqnref as the following: Template:Equation Template:Eqnref has the term $b^{-1}{d}^{-1}$. We can apply a property (proven in the exponents chapter) that $b^{-1}\cdot d^{-1}=(bd{)}^{-1}$. As such, we learn Template:Equation With this, we are done with the proof. Template:Robox/Close There is a very subtle distinction here between a proof and a verification. If we were asked to verify the statement above, we would assume what we are given here is true and simply only change one side. However, notice how we started the proof in Example 3.1.d: "Let us first start with rewriting the expression..."

In a proof, starting with the conclusion is absolutely wrong and should not be allowed because such an argument is circular! Whenever working with deductive proofs, we start with what is given and try to derive a true statement from what we started. This is one of the fundamental properties of deductive proofs.

Usually, in proofs, especially in deductive ones, there are two principles in mind that are necessary in the development of further understanding: A "universal proposition," such as a theorem or definition, will imply a "singular proposition," such as a premise, conclusion, or intermediate conclusion. Either that or a singular proposition implies another singular proposition.

For this proof, the former is true (a universal proposition implied a singular proposition). By such an action, we can verify the validity of this proof. Although one may not call this a formal proof, per se, such proof did indeed follow from verified axioms, and since those axioms are true, and it led to a truth statement, this proof is valid.

As a student further explores mathematics, principles of proof and the formality of such proofs will be very important in the life of the mathematician. In exploration of such concepts, a deeper understanding will be essential, where further examination of logic will be in order, and different types of proofs will be categorized, explained, and used, sometimes with little delay in thought.

While this is not meant to be a further exploration nor explanation of proofs in general, this book does seek to further a student's foundation of such concepts before one gets into a discrete math class. Template:ExampleRobox Let us first start with what we are given: $\frac{a}{b}+\frac{c}{d}$. One common trick done in proofs is to find a way to change the expression to make things easier on us without changing the meaning of the expression itself. For example, if one wanted to find the solution to $2x+5=10$, one may want to isolate the variable $x$. However, that cannot be done if the five is in the way, so one could subtract the five. This would change the meaning of the equation unless it is also done to the other side.

This same principle could apply to proofs. Here are some common ways in which a mathematician could make their lives easier. Let $xyz$ be an expression of interest:

• One could square $xyz$, then take the square root: $xyz=\sqrt{(xyz{)}^2}$.
• One could take the natural logarithm of $xyz$, then have a base $e$ in the exponent: $xyz=e^{\ln (xyz)}$.
• One could add a constant to $xyz$, then subtract that constant again: $xyz=xyz+c-c$.
• One could multiply a constant $xyz$, then divide the constant: $xyz=xyz\cdot \frac{c}{c}$.

The last bullet point shall be the trick we employ for this proof. After all, we are working with fractions. However, this constant cannot simply be any number. We shall be clever and select a number that will eliminate certain terms within. This will make our life easier in the long run.

First, we begin by stating the Equality Postulate (an item must equal itself): Template:Equation From there, we multiply $\frac{bd}{bd}$ to the right side. After all, this will simplify to Template:Eqnref, so nothing changes. Template:Equation Next, apply the distributive property to the right hand side of Template:Eqnref. Template:Equation Here is the special part of the proof. Notice how that there are two terms in which two fractions are multiplied against another. We can use the property we proved in Example 3.1.d. Template:Equation As before, we colored all terms that are red as terms that can be canceled. We will leave this as a trivial exercise for the reader to prove this is true. This gives us our subsequent equation: Template:Equation We can rewrite the rightmost side of Template:Eqnref as $ad(bd{)}^{-1}+bc(bd{)}^{-1}$. The factorable term is in blue. Therefore, we know the following can be rewritten as the final equation. Template:Equation The entire argument above can be condensed into the following compact argument:

$\begin{array}{ccc}\frac{a}{b}+\frac{c}{d}& =\frac{a}{b}+\frac{c}{d}& (\text{Equality postulate.})\\ & =(\frac{a}{b}+\frac{c}{d})\frac{bd}{bd}& (\frac{bd}{bd}=1)\\ & =\frac{a}{b}\cdot \frac{bd}{bd}+\frac{c}{d}\cdot \frac{bd}{bd}& (\text{Distributive property.})\\ & =\frac{abd}{b\cdot bd}+\frac{cbd}{d\cdot bd}\\ & =\frac{ad}{bd}+\frac{bc}{bd}\\ & =ad(bd{)}^{-1}+bc(bd{)}^{-1}\\ & =(bd{)}^{-1}(ad+bc)& (\text{Factor }(bd{)}^{-1}\text{.})\\ & =\frac{ad+bc}{bd}◼\end{array}$

Applying Laws of Real Numbers to Derive New Mathematical Properties

Proofs are not only useful when it comes to algebraic statements but also mathematical properties in general. It may be the case that these such items are much more useful towards the latter than the former because the laws of real numbers are by themselves always true. Although empirical evidence satisfy scientists, mathematicians differ in that regard.

If at some point, we found a location in which an mathematical "property" were to be false, then we would need to modify that statement. From thereon, we need to do the same to other items that also use this property, which may include absolute statements, such as theorems. If a specific location fails and that specific location is integral to the proof, then the proof of that theorem is automatically invalid. We do not wish for this "domino effect" to occur, so we must prove statements are true through deduction (or induction, which we will not discuss).

Let us take a look at division properties again. Template:ExampleRobox In both proofs (a) and (b), it relied on the property that $a\cdot 0=0$. While there is a zero factor theorem, this does not necessarily mean that the products of two numbers, one being zero, equals zero. Therefore, this proof may be false. One needs to prove that any number times zero equals zero. Template:Robox/Close

This may seem a little nit-picky. However, this can be very important when it comes to mathematical properties. If we assumed this were true, we may fundamentally be wrong about some solutions to rational equations, may fundamentally break some models of the universe, and may fundamentally conclude absurd ideas due to a wrong assumption. Therefore, mathematicians need to be picky!

Let us therefore start this section with a proof of this statement: any number times zero is equal to zero. Template:ExampleRobox Let $a\in ℝ$ be multiplied by zero.

$\begin{array}{cc}a\cdot 0& =a\cdot 0\\ & =a\cdot (0+1-1)& \text{(Proven later that }b-b=0\text{ in Exponents chapter.)}\\ & =a\cdot 0+a\cdot 1-a\cdot 1& \text{(Distributive Property.)}\\ \end{array}$

Subtract $a\cdot 0$ to both sides of equation.

$\Rightarrow a\cdot 0-a\cdot 0=a\cdot 1-a\cdot 1$

Suppose $a\cdot 0=y$

$\Rightarrow y-y=a-a$

Already shown that $b-b=0$. Ergo,

$\Rightarrow a\cdot 0-a\cdot 0=y-y=a-a=0$

By transitive property,

$\Rightarrow a\cdot 0-a\cdot 0=0$

Factor $a$:

$\Rightarrow a(0-0)=0$

By the $b-b=0$ property,

$\Rightarrow a\cdot 0=0◼$

Template:Robox/Close In the mean time, the property we used for this proof will not be proven until the Exponents chapter. For now, this proof should be sufficient grounds to accept the properties of rational numbers as shown in Example 3.2.a.

The next property we will prove has usually been proven in Geometry classes due to its simplicity. However, will do these two proofs next because we want to show a semi-formal proof. Template:ExampleRobox Let $n\in \mathbb{Z}$, $m\in \mathbb{Z}$, and $k\in \mathbb{Z}$. If $N=2n$, then $N$ is even. If $N=2n+1$, then $N$ is odd. Let $a=2m$ and $b=2k$.

$\begin{array}{cc}a+b& =2m+2k\\ a+b& =2(m+k)\end{array}$

Let $m+k=p\in \mathbb{Z}$.

$\iff a+b=2p$

From this, if $a,b$ are even, then $a+b$ is even. $◼$ Template:Robox/Close

Template:ExampleRobox This proof will be finished later. Template:Robox/Close

More will be added later. Template:BookCat