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.

Definition: Set
A set is a "well-defined" collection of objects, known as elements. A set is usually denoted by a singular, capital letter.

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, $\emptyset$ ). We could also have the set of common breakfast drinks, $D$ .

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

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.

Definition: $s$ is (Not) In $S$
Let there be a set

S

containing (or not containing) an object

s

. The object $s$ is in $S$ is written as

s\in S

. Otherwise, if $s$ is not in $S$ , then

s\notin S

.

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.

Example 1.1.(a): Set Conundrum

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.

How to define a set contains an element is done in two different ways, explicitly and implicitly.

Definition: Explicit Set

Let there be a set $S$ such that $a,\,b,\,c\in S$ . The explicit definition of a set is written using curly brackets, known as set braces— $\left\{\right\}$ — as below:

S=\left\{a,\,b,\,c\right\}

, whereby all and only the elements found in

S

must be written within the set braces.

The way the above set is read is $S$ is the set that contains $a,\,b,\,c$ .

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.

Definition: Implicit Set

Let there be a set $S$ such that $a,\,b,\ldots ,\,z\in S$ . The implicit definition of a set is written using set braces and ellipses given a pattern is demonstrated with the elements of the set, as below:

S=\left\{a,\,b,\ldots ,\,z\right\}

.

The way the above set is read as follows: $S$ is the set that contains $a,b$ , and so on, up to $z$ .

A more robust definition of a set is thus the following:

Definition: Set (Final Version)
A set, usually denoted by a singular letter, is a "well-defined" collection of objects, known as elements. A set is also uniquely defined by its elements.

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.

Definition: Size of a Set

A set $S$ with $n$ elements has a size of $n$ , denoted as

|S|=n

Example 1.1.(b): Breakfast set

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.

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.

Let $A=\{0,\{0\},1,\{0,1\},2,\{0,1,2\},3,\ldots ,100\}$ and $B=\{0,1,\ldots ,99,A\}$ . Which of the following statements are true? Select all that apply.

	$0,1,2,3,\ldots ,99\in A$ and $0,1,2,3,\ldots ,99\in B$ .
	$\{0,1,2,3,\ldots ,99\}\in A$ and $\{0,1,2,3,\ldots ,99\}\in B$ .
	$0,1,2,3,\ldots ,99\in \{0,1,2,3,\ldots ,99\}$ and $\{0,1,2,3,\ldots ,99\}\in A$ .

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).

Definition: Conditionals
Let there be an operation

C\left(x_{1},\,x_{2},\,\ldots \right)

such that the object variables

x_{1},\,x_{2},\,\ldots

give a truth value depending on either the values of the variables or the operation of

C

(its "nature"). This is a conditional.

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=\left\{x\mid C(x)\right\}

or

S=\left\{x:C(x)\right\}

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 ( ${\displaystyle$ ). 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\ldots \}

.

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.

Definition: Set-Builder Notation

A special case of set notation in which a set of elements can be more usefully listed with or too large to list with normal set notation.

E=\{f(n)|C(n)\}

We read the notation, "

E

is the set of all

f(n)

such that

C(n)

is true." Here,

f(n)

is an expression and

C(n)

is a rule or condition.

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

Example 1.2.(a): Set-builder notation

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

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

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

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

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

(f) $\left\{x\in \mathbb {N} |2x+5=8\right\}=\emptyset$ .

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 $x\in \mathbb {R}$ 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{aligned}{\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{aligned}}

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, $\emptyset$ .

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.

Suppose $A$ and $B$ are well-defined sets. Let $A=\{a|1\leq a\leq n,\,a,n\in \mathbb {N} \}$ and $B=\{b|b\geq 1,b\in \mathbb {N} \}$ . If $|B|=|A|+1$ , which of the following must be true?

	If $a\in A$ , and $\|A\|<\|B\|$ , then $A\in B$ .
	If any $a\in A$ , then $a\in B$ .
	If any $a\in \mathbb {N}$ , then $a\in A$ .
	If any $b\in B$ , then $b-1\in A$ .
	If $b\in B$ , and $\|B\|<\|A\|$ , then $B\in A$ .

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.

Definition: Subset

Suppose $P$ and $Q$ are sets. If each element $p\in P$ is also in $Q$ , then $P$ is a subset of $Q$ . This is written as

P\subseteq Q

If there is at least one element $p\in P$ such that $p\notin Q$ , then $P$ is not a subset of $Q$ , or

P\nsubseteq Q

.

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.

$\{1,3,2\}\subseteq \{1,2,3,4,5\}$ $\{1,3,2\}\subseteq \{1,2,3,4,5\}$ .
- Because $1,3,2\in \{1,2,3,4,5\}$ , $\{1,2,3\}$ (the set) is a subset of $\{1,2,3,4,5\}$ .
$\{1,3,2\}\subseteq \{1,2,3\}$ $\{1,3,2\}\subseteq \{1,2,3\}$ .
- Because $1,3,2\in \{1,2,3\}$ , $\{1,2,3\}$ is a subset of $\{1,2,3\}$ .
$\{1,2\}\nsubseteq \{4,3,\{2,1\}\}$ $\{1,2\}\nsubseteq \{4,3,\{2,1\}\}$ .
- 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, $1,2\notin \{4,3,\{2,1\}\}$ .
$\{\{1,2\}\}\subseteq \{4,3,\{2,1\}\}$ $\{\{1,2\}\}\subseteq \{4,3,\{2,1\}\}$ .
- 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.

Definition: Superset

Suppose $P$ and $Q$ are sets. $Q$ is the superset of $P$ if $P\subseteq Q$ , which is denoted as

Q\supseteq P

If $P\nsubseteq Q$ , then $Q$ is not a superset of $P$ , or

Q\nsupseteq P

.

Some examples below will be listed below, along with an explanation. Notice they are the same as above.

$\{1,3,2\}\nsupseteq \{1,2,3,4,5\}$ $\{1,3,2\}\nsupseteq \{1,2,3,4,5\}$ .
- Because $\{1,2,3,4,5\}\nsubseteq \{1,3,2\}$ , $\{1,3,2\}\nsupseteq \{1,2,3,4,5\}$ .
$\{1,2,3,4,5\}\supseteq \{1,3,2\}$ $\{1,2,3,4,5\}\supseteq \{1,3,2\}$ .
- Because $\{1,3,2\}\subseteq \{1,2,3,4,5\}$ , $\{1,2,3,4,5\}\supseteq \{1,3,2\}$ .
$\{1,3,2\}\supseteq \{1,2,3\}$ $\{1,3,2\}\supseteq \{1,2,3\}$ .
- Because $\{1,2,3\}\subseteq \{1,3,2\}$ , $\{1,3,2\}\supseteq \{1,2,3\}$ .
$\{4,3,\{2,1\}\}\nsupseteq \{1,2\}$ $\{4,3,\{2,1\}\}\nsupseteq \{1,2\}$ .
- Because $\{1,2\}\nsubseteq \{4,3,\{2,1\}\}$ , $\{4,3,\{2,1\}\}\nsupseteq \{1,2\}$ .
$\{4,3,\{2,1\}\}\supseteq \{\{1,2\}\}$ $\{4,3,\{2,1\}\}\supseteq \{\{1,2\}\}$ .
- 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 choices will be provided. Identify the valid subsets of the one provided in the problem. Almost all of the questions below require multiple selection to be answered correctly.

	$\mathbb {N}$ .
	$\{4,5\}$ .
	$\{5,10,20\}$ .
	$\{3,4,5\}$ .
	$4$ .

	$B$
	$\{4,5\}$
	$\{5,10,20\}$
	$\{\{10,20\}\}$

	$\{{\sqrt {x}}\|x>0,\,x\in \mathbb {Z} \}$
	$\mathbb {N}$
	$\mathbb {Z}$
	$\{x^{2}\|x\in \mathbb {Z} \}$

	$\mathbb {N}$
	$\{\mathbb {N} \}$
	$\mathbb {Z}$
	$\{\mathbb {Z} \}$

Combining Sets

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

Definition: Union
Suppose

P

and

Q

are sets. The union of sets

P

and

Q

is the set

P\cup Q=\{x|x\in P{\text{ or }}x\in Q\}

.

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).

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

Definition: Intersection
Suppose

P

and

Q

are sets. The intersection of sets

P

and

Q

is the set

P\cap Q=\{x|x\in P{\text{ and }}x\in Q\}

.

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).

$A\cap B=\{{\text{b}}\}$
$A\cap C=\emptyset$
$B\cap C=\{{\text{e}}\}$
$A\cap B\cap C=\emptyset$

Definition: Difference
Suppose

P

and

Q

are sets. The difference of sets

P

and

Q

is the set

P-Q=\{x|x\in P{\text{ and }}x\notin Q\}

.

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).

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

Check your Understanding

Introduction to Laws of Logic

Symbolic Logic

Properties of Conditional Statements

De Morgan's Laws

Proving Mathematical Properties

Applying Laws of Real Numbers to Verify and Prove Mathematical Properties

Example 3.1(a): Verify

{\frac {a+b}{c}}={\frac {a}{c}}+{\frac {b}{c}}

where

c\neq 0

.

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.

(3.1.1.1) ${\frac {a+b}{c}}={\frac {a}{c}}+{\frac {b}{c}}$

According to the definition of division, ${\frac {a}{c}}=a(c)^{-1}$ . Since this is true per each division found in the right expression, (3.1.1.1) can be rewritten as the following:

(3.1.1.2) ${\frac {a}{c}}+{\frac {b}{c}}=a(c)^{-1}+b(c)^{-1}$

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.

(3.1.1.3) $a(c)^{-1}+b(c)^{-1}=c^{-1}(a+b)$

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

(3.1.1.4) $c^{-1}(a+b)={\frac {a+b}{c}}$

Notice how the chain of equations herein can connect to every equality. Since (3.1.1.3) and (3.1.1.4) is true, we know the following equation below is true:

(3.1.1.5) $a(c)^{-1}+b(c)^{-1}={\frac {a+b}{c}}$

Because (3.1.1.2) and (3.1.1.5) is true, we know this below equation is true, and so on:

(3.1.1.6) ${\frac {a}{c}}+{\frac {b}{c}}={\frac {a+b}{c}}$

(3.1.1.7) ${\frac {a+b}{c}}={\frac {a+b}{c}}$

As shown in (3.1.1.7), 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{aligned}{\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{aligned}}

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 (3.1.1.1). 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.

Example 3.1(b): Verify

x=12

in

12x-12=11x

.

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.

One more example will be added later.

Example 3.1(d): Prove

{\frac {a}{b}}\cdot {\frac {c}{d}}={\frac {ac}{bd}}

, where

ac\nmid bd

,

bd\neq 0

,

a,b,c,d\in \mathbb {Z}

, and

\gcd(b,d)=1

.

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:

(3.1.1.8) ${\frac {a}{b}}\cdot {\frac {c}{d}}=a(b)^{-1}\cdot c(d)^{-1}$

Notice how we can apply a property of multiplication, that multiplying real numbers (and by extension integers) is associative. Therefore, we may rewrite (3.1.1.8) as the following:

(3.1.1.9) $a(b)^{-1}\cdot c(d)^{-1}=a\cdot c\cdot b^{-1}\cdot d^{-1}$

(3.1.1.9) 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

(3.1.1.10) $a\cdot c\cdot b^{-1}\cdot d^{-1}=ac\cdot (bd)^{-1}={\frac {ac}{bd}}\blacksquare$

With this, we are done with the proof.

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.

Example 3.1(e): Prove

{\frac {a}{b}}+{\frac {c}{d}}={\frac {ad+cb}{bd}}

, where

(ab+cd)\nmid bd

,

bd\neq 0

,

(a,b,c,d)\in \mathbb {Z}

, and

\gcd(b,d)=1

.

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):

(3.1.1.11) ${\frac {a}{b}}+{\frac {c}{d}}={\frac {a}{b}}+{\frac {c}{d}}$

From there, we multiply ${\frac {bd}{bd}}$ to the right side. After all, this will simplify to (3.1.1.11), so nothing changes.

(3.1.1.12) ${\frac {a}{b}}+{\frac {c}{d}}=\left({\frac {a}{b}}+{\frac {c}{d}}\right){\frac {bd}{bd}}$

Next, apply the distributive property to the right hand side of (3.1.1.12).

(3.1.1.12) $\left({\frac {a}{b}}+{\frac {c}{d}}\right){\frac {bd}{bd}}={\frac {a}{b}}\cdot {\frac {bd}{bd}}+{\frac {c}{d}}\cdot {\frac {bd}{bd}}$

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.

(3.1.1.13) ${\frac {a}{b}}\cdot {\frac {bd}{bd}}+{\frac {c}{d}}{\frac {bd}{bd}}={\frac {a\cdot {\color {Red}b}d}{{\color {Red}b}\cdot bd}}+{\frac {c\cdot b{\color {Red}d}}{{\color {Red}d}\cdot bd}}$

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:

(3.1.1.14) ${\frac {a\cdot bd}{b\cdot bd}}+{\frac {c\cdot bd}{d\cdot bd}}={\frac {ad}{bd}}+{\frac {bc}{bd}}$

We can rewrite the rightmost side of (3.1.1.14) as $ad{\color {Blue}(bd)^{-1}}+bc{\color {Blue}(bd)^{-1}}$ . The factorable term is in blue. Therefore, we know the following can be rewritten as the final equation.

(3.1.1.15) ${\frac {ad}{bd}}+{\frac {bc}{bd}}=(bd)^{-1}(ac+bc)={\frac {ad+bc}{bd}}\blacksquare$

The entire argument above can be condensed into the following compact argument:

{\begin{aligned}{\frac {a}{b}}+{\frac {c}{d}}&={\frac {a}{b}}+{\frac {c}{d}}&({\text{Equality postulate.}})\\&=\left({\frac {a}{b}}+{\frac {c}{d}}\right){\frac {bd}{bd}}&\left({\frac {bd}{bd}}=1\right)\\&={\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)&\left({\text{Factor }}(bd)^{-1}{\text{.}}\right)\\&={\frac {ad+bc}{bd}}\blacksquare \end{aligned}}

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.

Example 3.2(a): Examine the two below properties listed below when dividing. Try to find a possible error in the proofs.

(a) Proof: A rational number with numerator $0$ is equal to zero when the rational denominator $n\neq 0$ .

Imagine a number

{\frac {0}{n}}

that equals some number, say

f

. If

{\frac {0}{n}}=f

, then we know

0=fn

. Given

n\neq 0

, the only way

0=fn

is if

f=0

. Because the result can only ever equal zero, and

n\neq 0

, by the transitive property,

{\frac {0}{n}}=0

.

(b) Proof: For a rational number, the denominator cannot be $0$ .

Let the following be true:

{\frac {m}{0}}=f

. This implies

m=f\cdot 0

. Is there any number that can give a non-zero

m

when multiplying by zero? No, because any number times zero equals zero. Therefore, because

m

cannot be defined,

{\frac {m}{0}}

cannot be defined to be some number

f

. Before moving on, we need to talk about a special case,

\left(m=0\right)\in {\frac {m}{0}}

. Let

{\frac {0}{0}}=c

. We know that

0=0c

. However, if that is true, then any number

c

could equal

{\frac {0}{0}}

. Therefore, we say

{\frac {0}{0}}

is undefined.

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.

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.

Example 3.2(b): Prove any real number multiplied by zero gives zero.

Let $a\in \mathbb {R}$ be multiplied by zero.

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

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\left(0-0\right)=0

By the $b-b=0$ property,

\Rightarrow a\cdot 0=0\blacksquare

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.

Example 3.2(c): Does adding an even number to another even number equal an even number?

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{aligned}a+b&=2m+2k\\a+b&=2(m+k)\end{aligned}}

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

\Leftrightarrow a+b=2p

From this, if $a,b$ are even, then $a+b$ is even. $\blacksquare$

Example 3.2(d): Prove any two odd numbers give an even number.

This proof will be finished later.

More will be added later.

This article is issued from Wikibooks. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.