mi i mm
2. The way the selectable collections are required to compare with the gauge collection can be to be:
• larger-in-size than the gauge collection, or
• smaller-in-size than the gauge collection, or
• same-in-size as the gauge collection. or
• different-in-size from the gauge collection, or
• no-larger-in-size than the gauge collection, or
7.2. COLLECTIONS MEETING A REQUIREMENT
77
• no-smaller-in-size than the gauge collection, EXAMPLE 10. Jack has the following collection of one-dollar bills
empty full
So the bids that he can at all make in an auction correspond to the collections of one-dollar bills that he can use (set of selectable collections):
EI3
If it is the starting bid for a particular object that is three dollars, then the bids that Jack could make (select subset) would be:
If it is the current bid for a particular object that is three dollars, then the bids that Jack could make would be:
3. Occasionally, the subset of selected collections can be empty meaning that none of the selectable collections meets the given requirement. EXAMPLE 11. Jack has the following collection of one-dollar bills
LfiZl
So the bids that he can at all make in an auction (set of selectable collections) are:
If the starting bid (gauge collection) for a particular object is seven dollars, then Jack cannot make any bid so that the select subset is empty.
4. Occasionally, the subset of selected collections can be full meaning that all of the selectable collections meet the given requirement. EXAMPLE 12. Jack has the following collection of one-dollar bills
'(Hi
So the bids that he can at all make in an auction (set of selectable collections) are:
^fifJ
If the starting bid (gauge collection) for a particular object is one dollars, then the select subset is full.
There is of course nothing difficult with the one-to-one matching process involved in checking whether selectable collections compare or do not compare in a given way with a given gauge collection, but, as with most real-world processes, all this one-to-one matching of items is certainly going to
CHAPTER 7. BASIC PROBLEMS 1 COUNTING NUMERATORS
unspecified-numerator
x
specifying-formula
formula
equation
inequation
get very quickly very tedious.
7.3 Basic Formulas
In order to represent on paper real-world the various situations involving the selection on the basis of a requirement of a subset of selected collection from among a set of selectable collections we will use:
• Number-phrases to represent the collections,
• The six verbs that were introduced in Chapter 2 to compare collections
> < = ^ < >
• A special kind of form to represent the requirement.
1. The main difficulty with forms as we discussed them in Section 7.1 above is with the blanks. So we begin by introducing a kind of form that will be appropriate for "computations".
a. Instead of blanks, we will use an unspecified-numerator such as, for instance, the letter x. EXAMPLE 13. Instead of writing
<5
we will write
x < 5 b. A specifying-formula —we will often say formula for short—is a kind of forms in which:
• the verb can be any one of:
> < = ^ < >
• the nouns are numerators
• the common denominator is factored out. EXAMPLE 14. The following are specifying-formulas
x^8
x + 3 ^ 8
3 x x < 12
+3® x ©-7= -12
We will distinguish between:
• Equations, that is specifying-formulas that involve the verb
• Inequations 1 , that is specifying-formulas that involve any one of the
1 Although supposedly exceedingly concerned with the relevance of mathematics to the "ordinary life" of their students—as opposed to their "school life" one can only suppose, but judging by the textbooks they produce in vast numbers, Educologists are strangely indifferent to the fact that, in the real world, inequations are vastly more prevalent than
other five verbs: instruction
-. , / < > replace
code c. Then, instead of giving the instruction ,.
vertical bar enter the given numerator in the blank.
we will give the instruction
replace the unspecified numerator x by the given numerator.
EXAMPLE 15. Instead of giving the instruction
Enter 7 in the blank of the form: <5.
we will give the instruction
Replace x by 7 in the formula: x < 5
d. While a formula is not a sentence because it does not say anything about the real world (how could it since all that x stands for is a blankl), once we have replaced in a formula the unspecified numerator x by a given numerator, we have of course a sentence. (That this sentence is going to be either TRUE or FALSE depending on the given numerator is beside the point here.)
EXAMPLE 16. The specifying-formula
x < 5 is not a sentence because it does not sat anything about the real world since x does not stand for a given numerator. The instruction to replace a: by 7 in the specifying-formula
x < 5 results in
7<5 which is a sentence. (That it happens to be false is beside the point here.)
e. What will complicate matters a bit is that we will often code the instruction to replace the unspecified numerator x by some given numerator into the specifying-formula itself. For that, we will
i. draw, to the right of the specifying formula a vertical bar extending a bit below the line, which we read as "where" ii. write to the bottom right of the vertical bar:
- the unspecified numerator x followed by
- the symbol :=, to be read as "is to be replaced by", followed by
- the given numerator
equations.
CHAPTER 7. BASIC PROBLEMS 1 COUNTING NUMERATORS
EXAMPLE 17. Instead of using the instruction
Replace £ by 7 in the specifying-formula: x < 5 we shall write the instruction right into the specifying formula as follows:
x < 5 L=7
and the result is to be read as:
x < 5 where x is to be replaced by 7. The reason this complicates matters is that while
x < 5 is a specifying-formula,
x < 5 where x is to be replaced by 7. is a sentence since it is the same as the sentence
7<5 f. In particular, we have that: • replacing the unspecified numerator by a given numerator in an inequation results in an inequality. Example 18.
Using a form, we would write: Using a formula, we will write:
"Before": Inequation
"Action":
"After": Inequality
> 3.14
(neither true nor false) 7.82
> 3.14
Enter 7.82 in the blank 7^82 I >3.14
IS TRUE
x > 3.14 (neither true nor false)
x > 3.14| jr:=7-8a Replace x by 7.82
7.82 > 3.14
is TRUE
• replacing the unspecified numerator by a given numerator in an equation results in an equality.
Example 19.
Using a form, we would write: Using a formula, we will write:
2. Given a formula, the associated formulas for that formula are the formula, associated formulas that differ from the given formula only by the verb. associated equation
o ccnPl i^Tf^tfi ^iTT*lf~ k T
Crucial for the general procedure that we will develop in the next chapter inequation and given an inequation regardless of whether this given inequation is strict or lenient, are:
• The associated equation, that is the equation we obtain by replacing the verb in the given inequation by the verb =.
EXAMPLE 20. The equation associated with the lenient inequation
-3®x Z +90.43 is the equation
-3 <g> x = +90.43
EXAMPLE 21. The equation associated with the strict inequation
x © -14.08 < +53.71 is the equation
a; ©-14.08 = +53.71
• The associated strict inequation, that is the inequation we obtain by replacing the verb in the given inequation by the corresponding strict verb.
EXAMPLE 22. Given the lenient inequation
x + 6.08^ 17.82 the associated strict inequation is
a;+ 6.08 > 17.82 So, the strict inequation associated to a strict inequation is the strict inequation itself.
EXAMPLE 23. Given the strict inequation
x 0 -6.08 < -44.78 the associated strict inequation is
x 0 -6.08 < -44.78 While certainly surprising, this will help us developing a general procedure in the next chapter. In particular, we can say that a lenient inequation gives the choice between the associated strict inequation and the associated equation.
EXAMPLE 24. The lenient inequation
x <. +53.71 gives the choice between the associated strict inequation:
x < +53.71 and the associated equation
x = +53.71 For instance,
—61.05 is a solution of x ^ +53.71 because —61.05 is a solution of x < +53.71 and
+53.71 is a solution of x < +53.71 because +53.71 is a solution of x = +53.71
CHAPTER 7. BASIC PROBLEMS 1 COUNTING NUMERATORS
basic formulas unspecified numerator gauge numerator
equation, basic inequation, basic simple inequation, basic strict inequation, basic lenient
3. The simplest kind of specifying formula, which we will call basic formulas, are formulas involving two nouns related by a verb in the following manner:
i. The first noun is the unspecified numerator x,
ii. The verb is any of the verbs introduced in Chapter 2 to compare collections: iii. The second noun is a given gauge numerator
EXAMPLE 25. The following specifying-phrases are basic formulas:
x < 5 x^ -3 x=£ -52.19 but the following specifying phrases are not basic formulas:
<r + 3^ 8
3xi<12
3<g>a:©-7 = -12
In order to talk in general about basic formulas, we will use the symbol xn
to stand for the gauge numerator.
4. We will sort basic formulas according to the kind of verb that is involved and we will distinguish four types of basic formula corresponding to the four types of comparison sentences that we encountered in Chapter 2.
• Basic equations are basic formulas of the type:
X = Xq
EXAMPLE 26. The formula
a; = 31.19 is a basic equation
• Basic simple inequations are basic formulas of type:
x / x 0 EXAMPLE 27. The formula
x / 742.05 is a basic simple inequation
• Basic strict inequations are basic formulas of type:
x > xo or x < xo
EXAMPLE 28. The formulas
x > 132.17 and
x < -283.41
are both basic strict inequations
• Basic lenient inequations are basic formulas of type:
x S x 0 or x ^ x 0 basic problem
graph
Example 29. The formulas dot solid
x ^ 132.17 dot, hollow
and name
x ^ +283.41 are both basic lenient inequations
5. A basic problem with thus be a problem in which
• the data set consists of number-phrases
• the formula is a basic formula
• the common denominator has been factored out and declared up-front.
EXAMPLE 30. Given the basic problem in Dollars where
■ The data set is:
{2,3,4,5,6,7,8}
■ The formula is:
x > 5 (where x is an unspecified numerator and 5 is the gauge numerator) the solution subset is
{6,7,8}
7.4 Basic Problems
Given a basic problem involving counting number-phrases,
i. We determine the solution subset by replacing the unspecified numerator successively by each and every numerator in the data set. We then have comparison sentences that are true or false depending on
• which one of the six verbs is the verb in the formula.
• which way, up or down or not at all, we have to count from the numerator replacing the unspecified numerator to the given gauge numerator
(See Chapter 2.)
ii. We represent the solution subset:
• To graph the solution subset, we will use:
— a solid dot to represent a solution: •
— a hollow dot to represent a non-solution: o
• To name the solution subset, we will use, just as for data sets, two curly brackets, { }, and write the solutions in-between the curly brackets.
1. Usually, a problem has both non-solutions and solutions.
Example 31.
I. In the real world, Jack has the following collection of one-dollar bills
Hi
CHAPTER 7. BASIC PROBLEMS 1 COUNTING NUMERATORS
empty
So the bids that he can at all make in an auction (set of selectable collections) are:
»1 "«;.©. fct*
°. fc.1. «
If the starting bid for a particular object is three dollars (a selectable collection), then the bids that he could make (select subset) would be:
i im iiiQ
II. On paper, we represent this by the following problem:
■ We represent the set of selectable collections by the data set:
{1,2,3,4,5} Dollars
■ We represent the requirement that the bid must be no less than three dollars by the formula
x ^ 3
III. To determine the solution subset we check each and every numerator in the data set. The verb ^ requires that, from the numerator that replaces the unspecified numerator to the gauge numerator, we must count down or must not count.
x = 3|_._i is false because, from 1 to 3, we must count up _ 2 is false because, from 2 to 3, we must count up „ is true because, from 3 to 3, we must not count
a^3| x>3\
x:—5
So:
is true because, from 4 to 3, we must count down is true because, from 5 to 3, we must count down
1 is a non-solution
2 is a non-solution
3 is a solution
4 is a solution
5 is a solution
IV. We represent the solution subset ■ The graph of the solution subset is:
-> Dollars
/ ^ *? r o
■ The name of the solution subset is:
{3,4,5} Dollars 2. Occasionally, it can happen that there is no solution in which case we say that the solution subset is empty. Example 32.
I. In the real world, Jack has the following collection of one-dollar bills
7.4. BASIC PROBLEMS
85
full
So the bids that he can at all make in an auction (set of selectable collections) are:
III HH] lm Wm
If the starting bid for a particular object is seven dollars (a selectable collection), then he would not be able to make any bid (the select subset is empty):
II. On paper, we represent this by the following problem:
■ We represent the set of selectable collections by the data set:
{1,2,3,4,5} Dollars
■ We represent the requirement that the bid must be no less than three dollars by the formula
III. To determine the solution subset we check each and every numerator in the data set. The verb ^ requires that, from the numerator that replaces the unspecified numerator to the gauge numerator, we must count down or must not count.
x ^ 7| . =1 is false because, from 1 to 7, we must count up x = 7| 2 ' s false because, from 2 to 7, we must count up x = 7| x ._ 3 is false because, from 3 to 7, we must count up
x = 71 a is false because, from 4 to 7, we must count up
x ^ 7| a ,._ 5 is false because, from 5 to 7, we must count up
1 is a non-solution
2 is a non-solution
3 is a non-solution
4 is a non-solution
5 is a non-solution
So:
IV. We represent the solution subset ■ The graph of the solution subset is:
-O > Dollars
■T
/ x> J V
■ The name of the solution subset is:
{ } Dollars 3. Occasionally, it can happen that there is no non-solution in which case we say that the solution subset is full. Example 33.
I. In the real world, Jack has the following collection of one-dollar bills
CHAPTER 7. BASIC PROBLEMS 1 COUNTING NUMERATORS
infinite
So the bids that he can at all make in an auction (set of selectable collections) are: -: I 1
If the starting bid for a particular object is one dollars (a selectable collection), then he can any bid any selectable collection (the select subset is full):
II. On paper, we represent this by the following problem:
■ We represent the set of selectable collections by the data set:
{1,2,3,4,5} Dollars
■ We represent the requirement that the bid must be no less than three dollars by the formula
x ^ 1
III. To determine the solution subset we check each and every numerator in the data set. The verb ^ requires that, from the numerator that replaces the unspecified numerator to the gauge numerator, we must count down or must not count.
x ^ 1
x ^ 1
x ^ 1
x^ 1
X > 1
j is true because, from 1 to 1, we must not count 2 is true because, from 2 to 1, we must count down x -=3 ' s TR UE because, from 3 to 1, we must count down „._ A is true because, from 4 to 1, we must count down
x:= 5
So:
is true because, from 5 to 1, we must count down
1 is a solution
2 is a solution
3 is a solution
4 is a solution
5 is a solution
IV. We represent the solution subset ■ The graph of the solution subset is:
• • • • • > Dollars
/ •? v? ? J-
■ The name of the solution subset is:
{1,2,3,4,5} Dollars 4. When the data set is infinite, we cannot check every numerator in the data set and we must make the case that beyond a certain numerator, the numerators are all solutions or all non-solutions.
Example 34.
I. On paper, we represent such a situation by the following problem:
■ We represent the set of selectable collections by the data set:
{1,2,3,4,5, ...} Dollars where ... is read "and so on".
■ We represent the requirement that the bid must be no less than three dollars by the formula
x ^ 3
II. To determine the solution subset we are supposed to check each and every numerator in the data set. The verb ^ requires that, from the numerator that replaces the unspecified numerator to the gauge numerator, we must count down or must not count.
i. We start by checking each and every numerator in the data set until we pass the gauge numerator 3:
x = 3L. =1 is false because, from 1 to 3, we must count up
x = 3L._2 is false because, from 2 to 3, we must count up
x ^ 3| ._ 3 is true because, from 3 to 3, we must not count
x ^ 3| 4 is true because, from 4 to 3, we must count down
So:
1 is a non-solution
2 is a non-solution
3 is a solution
4 is a solution
ii. We now make the case that any numerator beyond 4, that is 5,6,7, ..., is a solution:
■ Since, from any numerator beyond 4, that is 5,6,7, ..., to 4, we must count down,
■ And since, from 4 to the gauge 3, we must count down,
■ It follows that from any numerator beyond 4, that is 5, 6, 7, ..., to the gauge 3, we must count down.
So, any numerator beyond 4, that is 5,6,7, ... is also going to be a solution.
III. We represent the solution subset
■ The graph of the solution subset is:
• • • • • > Dollars
/ sP J •? -J" and so on
where we actually write "and so on" because . . .would run the risk of not being seen.
■ The name of the solution subset is:
{1,2,3,4,5, ...} Dollars where we use .. . to mean "and so on".
5. When the data set involves signed numerators, we proceed essentially in the same manner as with plain numerators.
Example 35.
I. On paper, we represent such a situation by the following problem:
■ We represent the set of selectable collections by the data set:
{-5, -4, -3, -2, -1,0, +1, +2, +3, +4, +5,... } Dollars where ... is read "and so on".
■ We represent the requirement that the balance must be more than a three dollar debt by the formula
x > -3 II. i. We start by checking each and every numerator in the data set until we pass the gauge numerator 3:
x ^ — 3| x .__ 5 is false because, from —5 to —3, we must count up
x ^ — 3| x .__ 4 is false because, from —4 to —3, we must count up
x ^ — 3| x .__ 3 is true because, from —3 to —3, we must not count
= ~~ ^| -__ 2 ' s TRUE because, from —2 to —3, we must count down
So:
— 5 is a non-solution —4 is a non-solution —3 is a solution —2 is a solution
ii. We now make the case that any numerator beyond —2, that is —1,0,-1-1, +2,... is a solution:
■ Since, from any numerator beyond —2, that is —1, 0,-1-1, +2,..., to —2, we must count down,
■ And since, from —2 to the gauge —3, we must count down,
■ It follows that from any numerator beyond —2, that is — 1, 0, +1,-1-2,... to the gauge —3, we must count down.
So, any numerator beyond —2, that is —1,0, +1, +2,..., is also going to be a solution. III. We represent the solution subset
■ The graph of the solution subset is:
-> Dollars
\? V V V ° V **? x s ™d so on where we actually write "and so on" because ...would run the risk of not being seen. The name of the solution subset is:
{-3, -2,-1,0, +1, +2, +3,... } Dollars where we use ... to mean "and so on".
Chapter 8
Basic Problems 2:
Decimal Numerators
We continue our investigation of BASIC PROBLEMS in the case when the numerators are decimal numerators rather than counting numerators as was the case in the previous chapter.
The reason we are investigating the case of decimal numerators separately is that we cannot compare decimal numerators just by counting up or counting down as we did in the previous chapter where the numerators were counting numerators. While there is of course a procedure for comparing decimal numerators, we will not use it here for two reasons:
• We have not discussed in this book the comparison procedures for decimal numerators since, for reasons of space and time, we have had to take decimal numerators for granted,
• As it happens, we will not need to use any comparison procedure because we will introduce a general procedure that is extremely powerful in that it will allow us to investigate not only BASIC PROBLEMS in the case when the numerators are decimal but also many other types of problems.
So, this chapter is turned towards the chapters to follow for which it is in fact a direct preparation as well as a foundation. Finally, we shall use
DEFAULT RULE # 3. When no data set is declared, it will go without saying that the data set consists of all signed decimal numerators.
But, of course, in order to make sense in terms of the real world, we will still have to declare the denominator.
Also, to graph solution subsets, we will use rulers that have no tick-mark other than the ones directly relevant to the problem at hand but will have the symbol for minus infinity, -co, and the symbol for plus infinity, +oo, at the corresponding ends of the ruler:
-^ Dollars
+oo
8.1 Basic Equation Problems
When a problem involves an equation with decimal number-phrases, things remain pretty much the same as with counting number-phrases because equations usually do not have many solutions.
In the present case of a basic equation,
i. We determine the solution subset from the fact that the one and only one solution is the gauge numerator.
ii. We represent the solution subset just as in the case of counting numerators, namely:
• To graph the solution subset, we will use:
— a solid dot to represent a solution: •
— Since, here, there is no reason to consider any numerator aside from the gauge, there is no non-solution and so no need for hollow dots.
• To name the solution subset, just as for data sets, we will use two curly brackets,{ }, and write the solution in-between the curly brackets.
EXAMPLE 1. Given the problem in Dollars in which
■ the data set consists of all signed decimal numerators
■ the formula is the basic equation
a; = -13.72 we proceed as follows:
■ The only solution is —13.72
■ The graph of the solution subset is
• > Dollars
■ The name of the solution subset is
{-13.72} Dollars
8.2 Basic Inequation Problems
In the case of an inequation, though, things are very different with decimal numerators from what they were with counting numerators because
8.2. BASIC INEQUATION PROBLEMS
91
inequations can have too many solutions for us to handle them individually boundary
and we will develop and use a general procedure which we will call Pasch Procedure.
1. Roughly, to determine the solution subset of a given inequation problem with decimal numerators, we will proceed in two stages:
I. We will locate the boundary of its solution subset, that is the solution subset of the associated equation problem.
II. We will locate the interior of its solution subset, that is the solution subset of the associated strict inequation problem.
EXAMPLE 2. Given the problem in Dollars in which
■ the data set consists of all signed decimal numerators
■ the formula is the lenient inequation
x^ -13.72 we will locate separately:
i. the boundary of the solution subset, that is the solution subset of the associated equation
x = -13.72 ii. the interior of the solution subset, that is the solution subset of the associated strict inequation
x > -13.72 As already noted in the previous chapter, when the problem involves a strict inequation in the first place, this would appear rather senseless but, in fact, it is precisely by distinguishing the boundary from the interior that we will be able us to develop a general procedure.
EXAMPLE 3. Given the problem in Dollars in which
■ the data set consists of all signed decimal numerators
■ the formula is the basic inequation
x < -55.06 we will locate separately:
i. the boundary of the solution subset, that is the solution subset of the associated equation
x= -55.06 ii. the interior of the solution subset, that is the solution subset of the associated strict inequation
x < -55.06
2. More precisely, in the case of a given basic inequation problem, I. We locate the boundary as follows:
i. There is only one boundary point namely the gauge.
ii. The boundary point, though, may be a solution or a non-solution of the given inequation problem and we must check which it is: • If the basic inequation is strict, then the boundary point is a non-solution
and is therefore non-included in the solution set.
interior boundary point
CHAPTER 8. BASIC PROBLEMS 2 (DECIMAL NUMERATORS)
section half-line ray, solid ray, hollow
• If the basic inequation is lenient, then the boundary point is a solution and is therefore included in the solution set.
II. We locate the interior as follows:
i. The boundary point separates the data set in two sections, Section A and Section B.
ii. We pick a test numerator in