DrDelMathIntermediate Algebra

Important:The Law of Trichotomy is a property of the Real numbers. It is true for any subset of the Real numbers. It is not true for any other sets of numbers. It is not true for sets of any other mathematics objects.
This Law of Trichotomy states that when comparing two real numbers, or expressions representing real numbers, there are ALWAYS three possibilities. These possibilities are represented by the normal symbols <, =, > for the three binary relations.
It is beneficial to consider all three any time you want to consider any one of them. Think of them as inseparable siblings  triplets.
Any time you are working with an equation think of the two siblings obtained by replacing the = symbol with < and >.
Any time you are considering an inequality think of the two siblings obtained by replacing the inequality with the other inequality and the equal symbol =.
Examples: When considering 3x + 5 = 9 you should also consider 3x + 5 < 9 and 3x + 5 > 9
Examples: When considering 2x^{3} + 5x < 9 you should also consider 2x^{3} + 5x > 9 and 2x^{3} + 5x = 9
Examples: When considering 2x^{2} + 5x  4 > 0 you should also consider 2x^{2} + 5x  4 < 0 and 2x^{2} + 5x  4 = 0
Linear equations in one variable are equations of the form ax + b = 0.
Linear inequalities in one variable are inequalities of the form ax + b < 0 and ax + b > 0.
Note that in each of these three statements, an algebraic expression ax + b is
compared to the number 0.
Clearly these three statements resemble the Law of Trichotomy.
In fact The Law of Trichotomy tells us that for each real number x:
As we examine linear equations in one variable and linear inequalities in one variable side by side we will uncover even stronger ties between the three expressions ax + b < 0, ax + b = 0 and ax + b > 0
When solving ax + b = 0, the other two siblings
ax + b < 0 and ax + b > 0 cry out for attention.
When solving ax + b < 0, the other two siblings
ax + b = 0 and ax + b > 0 cry out for attention.
When solving ax + b > 0, the other two siblings
ax + b = 0 and ax + b < 0 cry out for attention.
You should recognize that when you solve any one of the three, solutions to the other two are close at hand and may be determined without another "solution process" or another bunch of calculations.
The relationship between solution sets for linear equations in one variable and linear
inequalities in one variable is an example of relationships between equations and inequalites
of any degree involving any number of variables.
It is an important principle to observe and learn!
About Solutions of Equations and Inequalities. Graphs of Equations and Inequalities. Equivalent Equations and Inequalities
The following properties of equations and inequalities are extremely important. These properties are the first things that should come to mind when you encounter an equation or inequality in any context.
These properties are the basic tools for:
Exercises for Section B1:a
Solve Exercises 3, 4, 9, 15,and 17. Be sure you can state the mathematics principle which permits you to perform each step in the process.
Sketch the graph of each of the equations in Exercises 3, 4, 9, 15, and 17.
Write the corresponding inequalities for the equations in Exercises 3, 4, 9, 15, and 17.
Use the first three properties of inequalities to solve the inequalities you associated with the equations in Exercises 3 and 4.
Graph each inequality on its own line, then sketch the graph of the equation and both inequalities on the same line.
What observations can you make about this last graph?
Use the graph of the equation, results of testing one point, and deductive reasoning to solve the two inequalities you associated with the equations in Exercises 9, 15, and 17.
Graph each inequality on its own line, then sketch the graph of the equation and both inequalities on the same line.
What observations can you make about this last graph?
Answer Question 47. Be sure to write your conclusion/answer and all reasoning which produced that conclusion.
Answer Questions 51 and 52. Be sure you can state the mathematics principle which permits you to perform the steps in your work.
Because some may not yet own a textbook, I will reproduce exercises here.
Solve each of the following:
3) 3x  4  5x = x + 4 + x
4) 13x  15x + 8 = 4x + 2  24
9) 5(y + 4) = 4 (y + 5)
15) 6x  2(x  3) = 4 (x + 1) +4
17) 3/8 + b/3 = 5/12
47) Which solution strategies are incorrect and why?
a) Solve (y  2)(y + 2) = 4 by setting each factor equal to 4.
b) Solve (x + 1)(x + 3) = 0 by setting each factor equal to 0.
c) Solve z^{2} + 5z + 6 = 0 by factoring z^{2} + 5z + 6 and setting each factor equal to 0.
d) Solve x^{2} + 6x + 8 = 10 by factoring x^{2} + 6x + 8 and setting each factor equal to 10.
51) Find the value of K which will make the two equations equivalent x/6 + 4 = x/3 and x + K = 2x
52) Find the value of K which will make the two equations equivalent 5x/4 + 1/2 = x/2 and 5x + K = 2x
If the reader needs a quick review of basic factoring techniques it is available HERE
These factoring techniques are absolutely essential. If the reader is rusty  study the review.
The Zero Factor Property may be generalized to the following very useful statement.
If each of two expressions represent real numbers and their product is zero then one (or both) of the expressions is zero.
If the quadratic in a quadratic equation can be factored, then the Zero Factor Property (generalized form) may be used to conclude that the original quadratic equation is equivalent to the compound equality (obtained by setting each factor equal to zero) with the conjunction OR.
Example: The quadratic polynomial in the equation x^{2}  4x  21 = 0 may be factored so that x^{2}  4x  21 = 0 is equivalent to (x + 3)(x  7) = 0.
As a result of The Zero Factor Property (x + 3)(x  7) = 0 is equivalent to the compound equality x + 3 = 0
OR x  7 = 0.
The solution set for x + 3 is {3} and the solution set for x  7 = 0 is {7}.
The solution set for the compound equality is the union {3} ∪ {7} = {3, 7}.
Therefore the solution set for the original quadratic equation is {3, 7}.
The role of equivalance and The Zero Factor Property in the above process for solving a quadratic equation was probably neglected in previous courses. Because we want to generalize this process to many other situations, it is important now to pay attention to the importance of these two properties.
Exercises for Section B1b:
Solve Exercises 5, 7, 14, 16, 26, 29, 30, and 38. Be sure you can state the mathematics principle which permits you to perform each step in the process.
Sketch the graph of each of the equations in Exercises 5, 7, 14, 16, 26, 29, 30, and 38.
Write the corresponding inequalities for the equations in Exercises 5, 7, 14, 16, 26, 29, 30, and 38.
Use the first three properties of inequalities to solve the inequalities you associated with the equations in Exercises 5, 7, 26 and 30.
Graph each inequality on its own line, then sketch the graph of the equation and both inequalities on the same line.
What observations can you make about this last graph?
Use the graph of the equation, results of testing one point, and deductive reasoning to solve the two inequalities you associated with the equations in Exercises 14, 16, 29, and 38.
Graph each inequality on its own line, then sketch the graph of the equation and both inequalities on the same line.
What observations can you make about this last graph?
Answer Question 47. Be sure to write your conclusion/answer and all reasoning which produced that conclusion.
Was the answer to this question more obvious after studying Section B1b?
Answer Question 48. Be sure you can state the mathematics principle which permits you to perform the steps in your work.
NOTE: In Question 48 the reference is to equations IN ONE VARIABLE.
Because some may not yet own a textbook, I will reproduce exercises here.
Solve each of the following:
5) 12x^{2} + 5x  2 = 0
7) z^{2} + 9 = 10z
14) n(2n  3) = 2
16) 10x 2(x + 4) = 8(x  2) + 6
26) y^{2}/30 = y/15 + 1/2
29)(3t + 1)/8 = (5 + 2t)/7 + 2
30)4  (2z + 7)/9 = (7  z)/12
38)7c  2(3c + 1) = 5(4  2c)
47) Which solution strategies are incorrect and why?
a) Solve (y  2)(y + 2) = 4 by setting each factor equal to 4.
b) Solve (x + 1)(x + 3) = 0 by setting each factor equal to 0.
c) Solve z^{2} + 5z + 6 = 0 by factoring z^{2} + 5z + 6 and setting each factor equal to 0.
d) Solve x^{2} + 6x + 8 = 10 by factoring x^{2} + 6x + 8 and setting each factor equal to 10.
48)Describe two ways a linear equation in one variable differs from a quadratic equation in one variable.
Solving equations or inequalities involving absolute values is based on the definition of absolute value.
Notice the precise definition of absolute value has two cases:
Case 1: The expression inside the absolute value symbol is positive or zero.
Case 2: The expression inside the absolute value symbol is negative.
Procedure: To solve an equation or an inequality involving absolute values of variable expressions, it is necessary to consider the two equations which naturally result from the definition of absolute value.
Every equation or inequality involving absolute value is solved by considering the two cases as in this general example:
Generic Example: To solve an equation or inequality involving GLOB, two cases must be considered. The two cases arise from the definition
of absolute value. Therefore to solve any equation or inequality involving GLOB, we consider:
Case 1: The equation that results from relacing GLOB with GLOB
Case 2: The equation that results from relacing GLOB with the opposite of GLOB.
The following two properties of equations and three properties of inequalities are also important tools when solving equations involving absolute value.
generating equivalent inequalities generating equivalent equations These five properties might be used to:
Simplify the equation or inequality before considering the two cases.
Solve the equations or inequalities in Case 1 and Case 2.
In this section of the textbook the expression inside the absolute value symbol is always a linear expresion in one variable. Each of these equations or inequalities can be solved by directly using the two cases that arise from the definition of absolute value together with the two properties for generating equivalent equations and the three properties for generating equivalent inequalities. That is what is done in most elementary textbooks. HOWEVER, there is an easier method. We will explain and demonstrate the easier method. This easier method depends on deductive reasoning as opposed to computations.
The Law of Trichotomy informs us that the constant k on the right side is negative, zero, or positive.
If k is negative, we easily observe:
The only remaining case is when k is positive. Therefore in the remaining discussion we will assume that the constant k on the right is positive.
Fact:When k is positive:
The word equivalent as used in the above fact means  X  < k has the same solution set as k < X < k.
Fact: Neither the corresponding "greater than" inequality nor the corresponding equality can be written in such a compact form.
The equation is called the boundary equation because its graph forms a boundary between the graphs of the two inequalities.
The fact that the "less than" inequality is equivalent to a compact compound inequality makes it the easiest to solve because the compact form incorporates both cases into one computational structure.
The Law of Trichotomy informs us that when we consider any one of ax + b = k, ax + b < k, or ax + b > k we should, in fact, consider all three of them.
The Law of Trichotomy informs us that each real number is a solution to one of
ax + b = k, ax + b < k, or ax + b > k.
Therefore the union of the three solution sets
is R.
Therefore when all three are graphed on the same number line, the entire
number line is used.
The Law of Trichotomy informs us that each real number is a solution to only one of the three.
Therefore the intersection of any two of the solution sets is the empty set.
Therefore when all three are graphed on the same number line, none of the graphs overlap.
Page 563: Exercises: 1, 2, 9, 10. These should be solved by inspection using nothing but the definition of absolute value.
Page 563: Exercises: For each of Exercises 5, 15, 39, and 59 follow each of the following instructions in the order presented. Think about what you are doing. How does The Law of Trichotomy enter these considerations?
Write the two sibling/corresponding inequalities.
Observe that the "less than" inequality is "the easy one".
Write the less than inequality as a compact compound inequality which does not involve abasolute value.
How is this compact compound inequality related to the original less than inequality?
Solve the compact compound inequality.
Write that solution set in interval notation. Write that solution set with set builder notation.
Sketch the graph of the less than inequality.
Without computation determine the solution set of the original absolute value equation.
Write the solution set of the original equation using the roster method.
Graph the solution set of the equation on the graph of the less than inequality.
Without computation determine the solution set of the greater than inequality.
Write the solution set of the greater than inequality using interval notation.
Write the solution set of the greater than inequality using set builder notation.
Graph the solution set of the greater than inequality on the graph of the less than inequality and the equation.
What is the answer to the original question?
Look at the graph and determine the union of the three solution sets.
Look at the graph and determine the intersection of each pair of solution sets.
Page 568: Exercises: 1, 11, 22, 26. These should be solved by inspection using nothing but the definition of absolute value.
Page 568: Exercises: For each of Exercises 6, 15, 24, 34, and 63 follow each of the following instructions in the order presented. Think about what you are doing. How does The Law of Trichotomy enter these considerations?
Write the two sibling/corresponding equation and inequality.
Observe that the "less than" inequality is "the easy one".
Write the less than inequality as a compact compound inequality which does not involve abasolute value.
How is this compact compound inequality related to the original less than inequality?
Solve the compact compound inequality.
Write that solution set in interval notation. Write that solution set with set builder notation.
Sketch the graph of the less than inequality.
Without computation determine the solution set of the original absolute value equation.
Write the solution set of the original equation using the roster method.
Graph the solution set of the equation on the graph of the less than inequality.
Without computation determine the solution set of the greater than inequality.
Write the solution set of the greater than inequality using interval notation.
Write the solution set of the greater than inequality using set builder notation.
Graph the solution set of the greater than inequality on the graph of the less than inequality and the equation.
What is the answer to the original question?
Look at the graph and determine the union of the three solution sets.
Look at the graph and determine the intersection of each pair of solution sets.
Write the first two properties of equations.
Write the first three properties of inequalities.
Compute the exact area of a circle whose radius is $7$.
Write the standard formula for the circumference of a circle.
What is the definition of $\pi $.
How many irrational numbers are in $\left\{x3<x<5\right\}$?
How many irrational numbers are in $\left\{xx\in Q\text{}and\text{}3x5\right\}$?