Algebra & Trigonometry Super Review

Geometry

CHAPTER 1

Sets and Set Operations

1.1 Sets

A set is defined as a collection of items. Each individual item belonging to a set is called an element or member of that set. Sets are usually represented by capital letters, elements by lowercase letters. If an item k belongs to a set A, we write k ∈ A ("k is an element of A"). If k is not in A, we write k ∉ A ("k is not an element of A"). The order of the elements in a set does not matter:

{1, 2, 3} = {3, 2, 1} = {1, 3, 2}, etc.

A set can be described in two ways: 1) it can be listed element by element, or 2) a rule characterizing the elements in a set can be formulated. For example, given the set A of the whole numbers starting with 1 and ending with 9, we can describe it either as A = {1, 2, 3, 4, 5, 6, 7, 8, 9} or as {the set of whole numbers greater than 0 and less than 10}. In both methods, the description is enclosed in brackets. A kind of shorthand is often used for the second method of set description; instead of writing out a complete sentence in between the brackets, we write instead

A = {k | 0 < k < 10, k is a whole number}

This is read as "the set of all elements k such that k is greater than 0 and less than 10, where k is a whole number."

A set not containing any members is called the empty or null set. It is written either as φ or { }.

Problem Solving Examples:

List the elements of the set:

{n | 27 ≤ n ≤ 216, n is a perfect cube}

This expression may be read as: the set of all elements (numbers) n, such that n is a perfect cube between 27 and 216 inclusive. One way to list the elements is:

C = {27, 64, 125, 216}

Note that we could have chosen any other letter for the set, but "C suggests the word cube."

Given that N = {9, 15, 21, ..., 99}, describe N in words.

First, we observe that 9, 15, 21, and 99 are all odd numbers. However, two odd numbers are missing between 9 and 15 and also between 15 and 21. If we continue skipping two odd numbers, we should finally arrive at 99. So we can describe the set as: all third odd numbers between 9 and 99, inclusive. (Though technically correct, this is an awkward construction.)

We notice also that 9, 15, 21, and 99 are all odd multiples of 3. So an alternative description would be: the set of odd multiples of 3 between 9 and 99 inclusive. (This is a more elegant description.)

List the elements of the set:

{p | p is a quadrilateral whose diagonals bisect each other at right angles}

In words, this may be read as the set of all elements, p, such that p is a quadrilateral whose diagonals bisect each other at right angles. There are only two such quadrilaterals, namely, the square and the rhombus.

Note that rectangle and parallelogram are not acceptable responses, since the diagonals of all rectangles do not necessarily bisect each other at right angles and the diagonals of all parallelograms do not necessarily bisect each other at right angles.

1.2 Subsets

Given two sets A and B, A is said to be a subset of B if every member of set A is also a member of set B. A is a proper subset of B if B contains at least one element not in A. We write A ⊆ B if A is a subset of B, and A ⊂ B if A is a proper subset of B.

Two sets are equal if they have exactly the same elements; in addition, if A = B then A ⊆ B and B ⊆ A.

Example:

Then 1) A equals C, and A and C are subsets of each other, but not proper subsets, and 2) B ⊆ A, B ⊆ C, B ⊂ A, B ⊂ C. (B is a subset of both A and C. In particular, B is a proper subset of A and C.)

A universal set U is a set from which other sets draw their members. If A is a subset of U, then the complement of A, denoted A′, is the set of all elements in the universal set that are not elements of A.

Example: If U = {1, 2, 3, 4, 5, 6,...} and A = {1, 2, 3}, then A′ = {4, 5, 6,...}.

Figure 1.1 illustrates this concept through the use of a Venn diagram.

Figure 1.1

Problem Solving Examples:

List all the subsets of C = {1, 2}.

{1}, {2}, {1,2}, and φ where φ is the empty set. Each set listed in the solution contains at least one element of the set C. The set {2, 1} is identical to {1, 2} and therefore is not listed. φ is included in the solution because φ is a subset of every set.

Find four proper subsets of P = {n|n ∈ I, – 5 < n ≤ 5}. (I stands for integer.)

P = { – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, 5}. All these elements are integers that are less than or equal to 5 and greater than – 5. A set A is a proper subset of P if every element of A is an element of P and in addition there is an element of P that is not in A.

A) B = { – 4, – 2, 0, 2, 4} is a subset because each element of B is an integer greater than – 5 but less than or equal to 5. B is a proper subset because 3 is an element of P but not an element of B. We can write 3 ∈ P but 3 ∉ B.

B) C = {3} is a subset of P, since 3 ∈ P. However, 5 ∈ P but 5 ∉ C. Hence, C ⊂ P.

C) D = { – 4, – 3, – 2, – 1, 1, 2, 3, 4, 5} is a proper subset of P, since each element of D is an element of P, but 0 ∈ P and 0 ∉ D.

D) φ ⊂ P, since φ has no elements. Note that φ is the empty set. φ is a proper subset of every set except itself. Note that there are many other possible sets which comprise the proper subset of P.

1.3 Union and Intersection of Sets

The union of two sets A and B, denoted A ∪ B, is the set of all elements that are either in A or B or both.

The intersection of two sets A and B, denoted A ∩ B is the set of all elements that belong to both A and B.

If A = {1, 2, 3, 4, 5 } and B = { 2, 3, 4, 5, 6 }, then A ∪ B = {1, 2, 3, 4, 5, 6} and A ∩ B = {2, 3, 4, 5}.

If A ∩ B = φ, then A and B have no elements in common and are said to be disjointed. Figures 1.2 and 1.3 are Venn diagrams for union and intersection. The shaded areas represent the given operation.

A ∪ B

Figure 1.2

A ∩ B

Figure 1.3

Problem Solving Examples:

If A = {1, 2, 3, 4, 5} and B = {2, 3, 4, 5, 6}, find A ∪ B.

The symbol ∪ is used to denote the union of sets. Thus, A ∪ B (which is read "the union of A and B) is the set of all elements that are in either A or B or both. In this problem, if

A = {1, 2, 3, 4, 5} and B = {2, 3, 4, 5, 6},

then

A ∪ B = {1, 2, 3, 4, 5, 6}.

If A = {1, 2, 3, 4, 5} and B = {2, 3,4, 5, 6), find A ∩ B.

The intersection of two sets A and B is the set of all elements that belong to both A and B; that is, all elements common to A and B. In this problem, if

A = {1, 2, 3, 4, 5} and B = {2, 3, 4, 5, 6},

then

A ∩ B = {2, 3, 4, 5}.

1.4 Laws of Set Operations

If U is the universal set and A, B, and C are any subsets of U, then the following hold for union, intersection, and complement:

Identity Laws

Idempotent Laws

Complement Laws

Commutative Laws

Associative Laws

Distributive Laws

De Morgan’s Laws

CHAPTER 2

Number Systems and Fundamental Algebraic Laws and Operations

2.1 Decimals

If we divide the denominator of a fraction into its numerator, we obtain a decimal form for it. This form attaches significance to the placement of an integer relative to a decimal point. The first place to the left of the decimal point is the units place; the second to the left is the tens; third, the hundreds, etc. The first place to the right of the decimal point is the tenths, the second the hundredths, etc. The integer in each place tells how many of the values of that place the given number has.

Example:

721 has seven hundreds, two tens, and one unit.

.584 has five tenths, eight hundredths, and four thousandths.

, b ≠ 0, then all rational numbers can be expressed as decimals by dividing b into a. The resulting decimal is either a terminating decimal, meaning that b divides a with remainder 0 after a certain point; or repeating, meaning that b continues to divide a so that the decimal has a repeating pattern of integers.

Example:

A) and C) are terminating decimals; B) and D) are repeating decimals. This explanation allows us to define irrational numbers as numbers whose decimal form is nonterminating and nonrepeating.

Example: √3 = 1.732...

′).

Problem Solving Examples:

as a repeating decimal.

To write a fraction as a repeating decimal, divide the numerator by the denominator, until a pattern of repeated digits appears:

2 ÷ 7 = .285714285714...

Identify the entire portion of the decimal which is repeated. The repeating decimal can then be written in the shortened form:

Find the common fraction form of the repeating decimal 0.4242....

Let x represent the repeating decimal.

Divide both sides of the difference by 99, and reduce the fraction.

The repeating decimal of this example had two digits that repeated. The first step in the solution was to multiply both sides of the original equation by the 2nd power of 10, or 10² or 100. If there were three digits that repeated, the first step in the solution would be to multiply both sides of the original equation by the 3rd power of 10, or 10³ or 1,000.

as a quotient of integers.

. Multiply both sides of this equation by 100:

(1)

Multiplying by 100 is equivalent to moving the decimal point two places to the right, and since the digits 25 are repeated, we have:

(2)

Now subtract Equation (1) from Equation (2):

(3)

(Note that this operation eliminates the decimal point.)

Dividing both sides of Equation (3) by 99:

Therefore,

, was multiplied by 100 or 10², where the power of 10 (which is