6.2: Order of Operations
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Overview
The order of operations (or operator precedence) is a collection of rules that reflect conventions about which procedures to perform first in order to evaluate a given mathematical expression. [1]
Discussion
Single values by themselves are important; however, we need a method of manipulating values (processing data). Scientists wanted an accurate machine for manipulating values. They wanted a machine to process numbers or calculate answers (that is, compute the answer). Prior to 1950, dictionaries listed the definition of computers as ” humans that do computations”. Thus, all of the terminology for describing data manipulation is math oriented. Additionally, the two fundamental data type families (the integer family and floating-point family) consist entirely of number values.
An Expression Example with Evaluation
Let’s look at an example: 2 + 3 * 4 + 5 is our expression but what does it equal?
- the symbols of + meaning addition and * meaning multiplication are our operators
- the values 2, 3, 4 and 5 are our operands
- precedence says that multiplication is higher than addition
- thus, we evaluate the 3 * 4 to get 12
- now we have: 2 + 12 + 5
- the associativity rules say that addition goes left to right, thus we evaluate the 2 +12 to get 14
- now we have: 14 + 5
- finally, we evaluate the 14 + 5 to get 19; which is the value of the expression
Parentheses would change the outcome. (2 + 3) * (4 + 5) evaluates to 45.
Parentheses would change the outcome. (2 + 3) * 4 + 5 evaluates to 25.
Operator Precedence Chart
Each computer language has some rules that define precedence and associativity. They often follow rules we may have already learned. Multiplication and division come before addition and subtraction is a rule we learned in grade school. This rule still works.
Order of Operations [2]
- Parentheses
- Exponents
- Multiplication / Division
- Addition / Subtraction
A common mnemonic to remember this rule is PEMDAS , or Please Excuse My Dear Aunt Sally . Precedence rules may vary from one programming language to another. You should refer to the reference sheet that summarizes the rules for the language that you are using. It is often called an Operator Precedence, Precedence of Operators, or Order of Operations chart. You should review this chart as needed when evaluating expressions.
A valid expression consists of operand(s) and operator(s) that are put together properly. Why the (s)? Some operators are:
- Unary – only have one operand
- Binary – have two operands, one on each side of the operator
- Trinary – have two operator symbols that separate three operands
Most operators are binary, that is they require two operands. Some precedence charts indicate of which operators are unary and trinary and thus all others are binary.
Key Terms
- associativity
- Determines the order in which the operators of the same precedence are allowed to manipulate the operands.
- evaluation
- The process of applying the operators to the operands and resulting in a single value.
- expression
- A valid sequence of operand(s) and operator(s) that reduces (or evaluates) to a single value.
- operand
- A value that receives the operator’s action.
- operator
- A language-specific syntactical token (usually a symbol) that causes an action to be taken on one or more operands.
- parentheses
- Change the order of evaluation in an expression. You do what’s in the parentheses first.
- precedence
- Determines the order in which the operators are allowed to manipulate the operands.


1.3: Institutional Arrangements for Building Infrastructure
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Buildings may be owned by occupants (or residents) or investors of various kinds. As shown in Table 12.2.2, nearly 30% of US housing units are owned by investors and rented to occupants.
Ownership of buildings commonly changes over the building lifetime. Initially, buildings may be constructed with the intent of re-sale upon completion of construction, as with residential developers. Active real estate markets aid in the transfer of ownership during the lifetime of buildings. Opportunities to gain tax advantages through rapid depreciation of buildings can motivate relatively frequent building sales. With or without ownership changes, buildings typically undergo renovations and changes in function during their lifetime.
Lending institutions often make loans using real estate as collateral. In the event of default on the loans, the lending institution can foreclose and gain possession of the property. During construction of buildings, the value of buildings is problematic, so lending institutions typically charge more for construction loans than for mortgage loans secured by a complete building’s collateral.
Building management can be undertaken by a variety of parties, including owners, occupants or contractors. Automated aids for building management are typically less sophisticated than aids for other infrastructure systems, reflecting in part the diverse ownership of the building infrastructure.


4.8: Electrophilic aromatic substitution reactions
- Understand the difference in the electrophilicity of \(\pi\)-bond of benzene and alkenes.
- Draw the electrophilic aromatic substitution mechanism with curly arrows showing the flow of electrons.
- Apply the electrophilic aromatic substitution to some reactions of benzene, including halogenation, nitration, sulfonation, alkylation, and acylation reactions.
Which electrophiles can react with an aromatic substrate?
The \(\pi\)-bonds in a benzene ring of aromatic compounds are weaker nucleophiles than the \(\pi\)-bonds in alkenes. This is because breaking a \(\pi\)-bond of alkene costs about 260 kJ/mole energy, but breaking a \(\pi\)-bond in an aromatic substrate costs an additional 208 kJ/mol because the aromatic stabilization is lost. Unlike alkenes, the aromatic substrates do not react with partial positive (\(\ce{\overset{\delta{+}}{A}{-}\overset{\delta{-}}{B}}\)) electrophiles. The aromatic substrates react with electrophiles in their most reactive cation \(\ce{E^{+}}\) form. The cation electrophiles are generated in situ by acid-base or Lewis acid-Lewis base reactions. For example, halogens (\(\ce{X-X}\) react as Lewis bases with Lewis acids like \(\ce{AlX3}\) or \(\ce{FeX3}\), where \(\ce{X}\) is a halogen atom (\(\ce{Cl}\) or \(\ce{Br}\)). The Lewis acid receives a lone pair from one halogen atom causing a heterolytic breaking of \(\ce{X-X}\). The other halogen leaves as \(\ce{X^{+}}\), as shown below.
Similar reaction happens when an alkylhalide (\(\ce{R-X}\) or an acyl halide ( \(\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-X}\)) reacts with Lewis acid, as shown below.
An \(\ce{-OH}\) bonded with a potential electrophile \(\ce{E^{+}}\) can be converted into a better leaving \(\ce{-\overset{+}{O}H2}\) group by adding a stronger acid to the substrate. The \(\ce{-{\overset{+}{O}}H2}\) leaves as neutral neucleophile \(\ce{H2O}\), leaving behind the \(\ce{E^{+}}\). For example, protonation of nitric acid with sulfuric acid generated nitronium ion (\(\ce{\overset{+}{N}O2}\)), as shown below.
Like the auto-ionization of water, the autoionization of sulfuric acid followed by elimination of \(\ce{H2O}\) generates protonated sulfur trioide (\(\ce{\overset{+}{S}O3H}\)), as shown below.
Electrophilic aromatic substitution mechanism
The nucleophilic \(\pi\)-bond of an aromatic compound attacks the cation electrophile (\(\ce{E^{+}}\)), as shown in step#1 in the mechanism illustrated below. Any base group in the medium removes the acidic proton that re-establishes the \(\pi\)-bond in Step#2.
Removal of the proton by a base is preferred over electrophile attacking the carbonation intermediate in step#2, because aromatic stabilization decreases the energy barrier for the former. It is called electrophilic aromatic substitution reaction because an electrophile \(\ce{E^{+}}\) substitutes another electrophile \(\ce{H^{+}}\) from an aromatic substrate.
Examples of electrophilic aromatic substitution reactions
Some fo the important electrophilic aromatic substitution reactions of benzene are listed below.
- Halogenation of benzene substitutes a \(\ce{-H}\) with a halogen (\(\ce{-Cl}\) or \(\ce{-Br}\)), as shown below.
- Nitration of benzene substitutes a \(\ce{-H}\) with nitro group (\(\ce{-NO2}\)) by the following reaction.
- Sulfonation of benzene substitutes a \(\ce{-H}\) with sulfonic acid group (\(\ce{-SO3H}\)) by the following reaction.
- Alkylation of benzene substitutes a \(\ce{-H}\) with alkyl acid group (\(\ce{-R}\)), e.g.,:
- Acylation of benzene substitutes a \(\ce{-H}\) with acyl group (\(\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-R}\)), e.g.,:


16.6: Creating a Paradigm Shift
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A paradigm is the assumptions, concepts, values, and practices that influence how people view the world. Paradigms have been shaped over a lifetime by many influences that include families, schools, social and professional experiences, and the media. Each of us constructs our paradigms. Paradigms around diversity are often deficit-oriented, focusing on problems and vulnerabilities in the population being considered. These place emphasis on what people and families in that population are lacking.
Too often, children and their families in minority populations are considered “at risk” by policy makers, educators, and the media rather than “placed at risk” by a system that has historically viewed them from a deficit paradigm (Boykin, 2013). Scientific paradigms influence the ways that researchers in child development, often done by those in privilege and power, think and conduct their work. For example, research on minority children often employs a deficit perspective, and as a result, the knowledge base highlights the adversity and maladaptation of the minority children (Cabrera, 2013 a). Therefore, more is known about the problems than the positive qualities and strengths of the minority groups being studied. And this biased information leads many professionals to conclude that children from minority families are “at risk” and that they have many deficits to overcome in their learning and development.
But paradigms are not set in stone and for sure, the deficit-oriented paradigm needs to be turned on its head. A shift in mindset – a paradigm shift – is often experienced as a revolution or a transformation. It does not just happen, but rather it is driven by agents of change who might be families, community advocates, political leaders, researchers, educators, as well as other professionals. Ideas and activities in this resource also might trigger a paradigm shift in how you view those that are different from yourself.
As the research community shifts its deficit paradigm and redefines its research agenda, the knowledge base will expand to highlight successes and assets of children and their families. And this will lead high-quality program planning and implementation of effective strategies to promote all children’s learning and development. [268]


1.2: Slope of a Line
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Learning Objectives
In this section, you will learn to:
- Find the slope of a line.
- Graph the line if a point and the slope are given.
In the last section, we learned to graph a line by choosing two points on the line. A graph of a line can also be determined if one point and the "steepness" of the line is known. The number that refers to the steepness or inclination of a line is called the slope of the line.
From previous math courses, many of you remember slope as the "rise over run," or "the vertical change over the horizontal change" and have often seen it expressed as:
\[\frac{\text {rise}}{\text {run}}, \frac{\text {vertical change}}{\text {horizontal change}}, \frac{\Delta y}{\Delta x} \text { etc. } \nonumber \]
We give a precise definition.
Definition: Slope
If (\(x_1\), \(y_1\)) and (\(x_2\), \(y_2\)) are two different points on a line, the slope of the line is
\[\text{slope}=m=\frac{y_2-y_1}{x_2-x_1} \label{slope} \]
Example \(\PageIndex{1}\)
Find the slope of the line passing through points (−2, 3) and (4, −1), and graph the line.
Solution
Let (\(x_1\), \(y_1\)) = (−2, 3) and (\(x_2\), \(y_2\)) = (4, −1), then the slope (via Equation \ref{slope}) is
| \(m = \frac{-1-3}{4-(-2)} = \frac{-4}{6} = -\frac{2}{3}\) |
To give the reader a better understanding, both the vertical change, -4, and the horizontal change, 6, are shown in the above figure.
When two points are given, it does not matter which point is denoted as (\(x_1\), \(y_1\)) and which (\(x_2\), \(y_2\)). The value for the slope will be the same.
In Example \(\PageIndex{1}\), if we instead choose (\(x_1\), \(y_1\)) = (4, −1) and (\(x_2\), \(y_2\)) = (−2, 3), then we will get the same value for the slope as we obtained earlier.
The steps involved are as follows.
\[m = \frac{3-(-1)}{-2-4} = \frac{4}{-6} = -\frac{2}{3} \nonumber \]
The student should further observe that
- if a line rises when going from left to right, then it has a positive slope. In this situation, as the value of \(x\) increases, the value of \(y\) also increases
- if a line falls going from left to right, it has a negative slope; as the value of \(x\) increases, the value of \(y\) decreases.
Example \(\PageIndex{2}\)
Find the slope of the line that passes through the points (2, 3) and (2, −1), and graph.
Solution
Let (\(x_1\), \(y_1\)) = (2, 3) and (\(x_2\), \(y_2\)) = (2, −1) then the slope is
\[m = \frac{-1-3}{2-2}=\frac{4}{0}=\text{undefined.} \nonumber \]
Note: The slope of a vertical line is undefined.
Example \(\PageIndex{3}\)
Find the slope of the line that passes through the points (−1, −4) and (3, −4)
Solution
Let (\(x_1\), \(y_1\)) = (−1, −4) and (\(x_2\), \(y_2\)) = (3, −4), then the slope is
\[ m = \frac{-4-(-4)}{3-(-1)} = \frac{0}{4} = 0 \nonumber \]
Note: The slope of a horizontal line is 0
Example \(\PageIndex{4}\)
Graph the line that passes through the point (1, 2) and has slope \(-\frac{3}{4}\).
Solution
Slope equals \(\frac{\text{rise}}{\text{run}}\). The fact that the slope is \(\frac{-3}{4}\), means that for every rise of -3 units (fall of 3 units) there is a run of 4. So if from the given point (1, 2) we go down 3 units and go right 4 units, we reach the point (5, -1). The graph is obtained by connecting these two points.
Alternatively, since \(\frac{3}{-4}\) represents the same number, the line can be drawn by starting at the point (1,2) and choosing a rise of 3 units followed by a run of -4 units. So from the point (1, 2), we go up 3 units, and to the left 4, thus reaching the point (-3, 5) which is also on the same line. See figure below.
Example \(\PageIndex{5}\)
Find the slope of the line \(2x + 3y = 6\).
Solution
In order to find the slope of this line, we will choose any two points on this line.
Again, the selection of \(x\) and \(y\) intercepts seems to be a good choice. The \(x\)-intercept is (3, 0), and the \(y\)-intercept is (0, 2). Therefore, the slope is
\[m = \frac{2-0}{0-3} =-\frac{2}{3}. \nonumber \]
The graph below shows the line and the \(x\)-intercepts and \(y\)-intercepts:
Example \(\PageIndex{6}\)
Find the slope of the line \(y = 3x + 2\).
Solution
We again find two points on the line, e.g., (0, 2) and (1, 5). Therefore, the slope is
\[m =\frac{5-2}{1-0} = \frac{3}{1} = 3. \nonumber \]
Look at the slopes and the \(y\)-intercepts of the following lines.
| The line | slope | \(y\)-intercept |
|---|---|---|
| \(y=3x+2\) | 3 | 2 |
| \(y=-2x+5\) | -2 | 5 |
| \(y=\frac{3}{2}x-4\) | \(\frac{3}{2}\) | -4 |
It is no coincidence that when an equation of the line is solved for \(y\), the coefficient of the \(x\) term represents the slope, and the constant term represents the \(y\)-intercept.
In other words, for the line \(y = mx + b\), \(m\) is the slope, and \(b\) is the \(y\)-intercept.
Example \(\PageIndex{7}\)
Determine the slope and \(y\)-intercept of the line \(2x + 3y = 6\).
Solution
We solve for \(y\):
\begin{align*}
&2 x+3 y=6 \nonumber \\
&3 y=-2 x+6 \nonumber \\
&y=(-2 / 3) x+2 \nonumber
\end{align*}
The slope = the coefficient of the \(x\) term = − 2/3.
The \(y\)-intercept = the constant term = 2.


1.3: Colons
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Many writers think the colon is such a confusing piece of punctuation that they simply avoid it altogether, but it can be fun to use a colon every now and then. Plus, the colon can add some important variety and excitement to your writing. So why not review this list of uses for the colon and give it a try?
In most cases, essentially, a colon signals “anticipation”—the reader knows that what follows the colon will define, illustrate, or explain what preceded it. This is certainly the case in the colon’s first three uses, as described below:
example
Road construction in New York City might pose a problem if there is a zombie attack: It is best to know which streets are closed, as you do not want to end up lost during a dire situation.
2. Use a colon to separate an independent clause from a list that follows the independent clause.
example
I have collected a wide variety of important items in case there is a zombie attack: canned food, bottled water, and wood for boarding my windows.
Tip!
You shouldn’t use a colon when the introductory portion of the sentence is a dependent clause (incomplete thought). The first part of the sentence must be an independent clause or a complete sentence. So the following would NOT work:
I have collected: canned food, bottled water, and wood for boarding my windows.
example
I have the perfect solution to your problems with bullies at work: Chuck Norris.
4. You should also use a colon at the end of a formal, business letter greeting.
example
To Whom It May Concern:
5. And, of course, you should use a colon to separate the hour from the minutes when writing numerical time.
example
3:00 a.m.


7.3: A Systematic Approach to Interpreting an ECG
A systematic approach to interpreting an ECG improves the speed and reliability of the assessment, especially if a dysrhythmia is present. This section outlines a systematic approach to interpreting an ECG, as well as common findings in ECG waveforms that occur during dysrhythmias. Review basic information about ECG waveforms and intervals in Table 7.2a in the “Basic Concepts” section of this chapter. Supplementary videos demonstrating how to interpret an ECG are provided at the end of this section, and an online cardiac rhythm game in the “Learning Activities” section provides practice in interpreting ECG strips.
Calculate the Rate
Calculate the ventricular and atrial rates.
There are several methods to calculate the rate. The simplest method to obtain a ventricular rate is to count the number of R waves in a 6-second strip (i.e., over 30 large boxes) and multiply this number by 10. See Figure 7.10 [1] for an image of counting the R waves on a 6-second strip for a ventricular heart rate of 80.
In a similar manner, the atrial rate can be calculated (if P waves are present) by counting the number of P waves in a 6-second strip and multiplying this number by 10. In order to ensure accuracy of this method, please make sure that the rhythm strip that is used is a 6-second strip.
Determine the Regularity of the Rhythm
Determine if the rhythm is regular or irregular.
View leads I, II, aVF, and V1 for the most accurate interpretation of the rhythm. [2] Assess the distances between the R waves to determine the regularity of the rhythm using a caliper or ruler. It is important to assess the distances across the whole strip to account for any potential abnormality. See Figure 7.11 [3] for an image depicting the R to R distance on an ECG strip. Cardiac rhythms are categorized into four types of rhythms:
- Regular Rhythm: The R to R distances are always equal distance apart. For example, normal sinus rhythm is a regular rhythm.
- Irregularly Irregular Rhythm: The R to R distances are never equal distances apart. For example, atrial fibrillation is an irregularly irregular rhythm.
- Regularly Irregular: The R to R distances are unequal, but there is a repetitive pattern to the unequal distances. For example, second-degree heart block type II may have a regularly irregular rhythm.
- Occasionally Irregular: The R to R distances are equal except for an occasional out of place R. For example, sporadic premature ventricular contractions (PVCs) are an occasionally irregular rhythm.
Assess the P Waves
The P waves are critical to determining the origin of a heartbeat. Assess the following characteristics related to the P waves:
- Are P waves present? The absence of P waves indicates the heartbeat did not originate in the SA node.
- Do all of the P waves look the same? When a heartbeat originates somewhere other than the SA node, the contour of the P wave changes.
- Do the P waves occur at a regular rate? Using a caliper or ruler, assess the distance from “P to P” waves. If this distance varies, there is an issue in conduction from the SA node to the AV node. See Figure 7.12 [4] for an image of assessing the distance between P waves.
- Is there one P wave with every QRS complex? This helps to further assess conduction. Is the electrical communication between the atria and the ventricles working correctly?
An example of a dysrhythmia that lacks P waves is atrial fibrillation. The presence of “irregularly irregular” narrow QRS complexes with no distinct P waves or a wavy baseline is the hallmark feature in the identification of atrial fibrillation. [5]
Determine the PR Interval
Measure the distance from the beginning of the P wave to the beginning of the QRS complex.
Recall that the normal PR interval is 0.12 to 0.2 seconds (i.e., 3-5 small boxes). When the PR interval is too long, there is a problem with conduction from the SA node to the AV node. A delay in SA to AV node conduction is representative of a heart block. See Figure 7.13 [6] for an image of measuring the PR interval.
The PR interval represents the time from the beginning of atrial depolarization to the start of ventricular depolarization and includes the delay at the AV node. Variations in the PR interval can be seen in various disorders. For example, a long PR interval may indicate first-degree AV block. Short PR intervals are present in conditions with accelerated AV conduction, such as Wolf-Parkinson-White syndrome, a disorder that is present at birth. [7]
Determine the QRS Duration
Measure the distance from the beginning to the end of the QRS complex.
Recall that normal QRS duration is 0.04 to 0.12 seconds (i.e., 1-3 small boxes). When the QRS complex is too wide, it indicates the heartbeat did not originate in the atria. See Figure 7.14 [8] for an image of measuring the QRS duration.
The QRS complex represents ventricular depolarization as the electrical signal passes down from the AV node. A prolonged QRS may indicate hyperkalemia or bundle branch block. A premature ventricular contraction, intraventricular conduction delay (bundle branch block), or other ventricular dysrhythmias are associated with a wide QRS. [9]
Evaluate the T Waves
Evaluate the T waves. Do they follow the QRS complex? Are they upright and rounded?
Recall the T wave represents ventricular repolarization. Its morphology is highly susceptible to cardiac and noncardiac influences such as hormonal and neurological factors. The size of a typical T wave is between one eighth and two thirds of the size of the R wave and a height of less than 10 mm.
Abnormalities in the T wave morphology include inverted, flat, biphasic, or tall tented or peaked T waves. T wave changes can occur with a variety of conditions. Tall T waves in an anterior chest lead III, aVR, and V1 with a negative QRS complex may suggest acute myocardial ischemia. Other causes of T wave abnormalities are caused by temporary physiological factors (for example, after eating a high carbohydrate meal), endocrine or electrolyte imbalances, myocarditis, pericarditis, cardiomyopathy, post-cardiac surgery state, pulmonary embolism, fever, infection, anemia, acid-base disorders, drugs, endogenous catecholamines, metabolic changes, acute abdominal processes, and intracranial pathology. [10]
Evaluate the ST Segment
What is the appearance of the S-T segment? Is it isoelectric (i.e., is it on the same level of the PR segment)?
Recall the ST segment depicts the interval between the end of ventricular depolarization and the beginning of ventricular repolarization. A normal ST segment falls along the isoelectric line and lies at the same level as the PR interval. See Figure 7.15 [11] to review an illustration of ST segments, waves, and other intervals discussed in this section.
Elevation or depression of the ST segment by 1 mm or more, measured at the J point, is abnormal. The J point is a region between the QRS complex and the ST segment. ST elevation is highly specific if present in two or more contiguous leads in the setting of an acute myocardial infarction and usually indicates a complete blockage of the coronary artery. If ST elevation on the ECG tracing above the baseline after the J point is at least 1 mm in a limb lead or 2 mm in a precordial lead, it is clinically significant for diagnosing acute myocardial infarction. ST depression greater than 1 mm below the baseline is often a sign of myocardial ischemia or angina. It can appear as a downsloping, upsloping, or a horizontal segment on the ECG. ST depressions are also associated with nonischemic causes, including digoxin toxicity, hypokalemia, hypothermia, and tachycardia. [12]
Determine the QT Interval Duration
Is the QT (or QTc) interval duration within normal limits?
Recall the QT interval, measured from the Q wave to the end of the T wave, represents the start of depolarization to the end of the repolarization of the ventricles. Generally, the normal QT interval is less than 0.4 to 0.44 seconds. Women usually have a slightly longer QT interval than men, and heart rate also impacts the QT interval, so typically the corrected QT interval (QTc) is monitored.
A short QTc (less than 0.36 seconds) may be associated with hypercalcemia, acidosis, hyperkalemia, hyperthermia, or short QT syndrome. [13] A prolonged QTc presents an imminent risk for serious ventricular arrhythmias, including torsades de pointes, ventricular tachycardia, and ventricular fibrillation. Common causes of QTc prolongation include side effects of medications, electrolyte abnormalities such as hypocalcemia and hypomagnesemia, and congenital long QT syndrome. [14]
Evaluate Other ECG Components
Are there extra beats on the waveform? Is anything else abnormal?
The U wave is a small wave that can follow the T wave, though it is not always visible. It represents the delayed repolarization of the papillary muscles or Purkinje fibers and is commonly associated with hypokalemia. [15]
The J wave is an abnormal finding typically found during hypothermia. It appears as an extra deflection on an ECG at the QRS complex and ST segment junction. [16]
Clinically Observe the Client
When analyzing an ECG strip, it is vital to clinically observe the client for signs and symptoms that determine the significance of the dysrhythmia. For example, the more abnormal a rhythm becomes, the less blood the heart pumps, causing symptoms due to low blood pressure such as dizziness, confusion, and loss of consciousness. [17]
View a supplementary YouTube video [18] on ECG interpretation basics:
View a supplementary YouTube video [19] on PQRST EKG rhythm:
- “Counting the R Waves” by Chippewa Valley Technical College is licensed by CC BY 4.0 ↵
- This work is a derivative of StatPearls by Sattar and Chhabra and is licensed under CC BY 4.0 ↵
- “Assessing Distance Between R Waves” by Chippewa Valley Technical College is licensed by CC BY 4.0 ↵
- “Assessing Distance Between P Waves” by Chippewa Valley Technical College is licensed by CC BY 4.0 ↵
- This work is a derivative of StatPearls by Sattar and Chhabra and is licensed under CC BY 4.0 ↵
- “Measuring the PR Interval” by Chippewa Valley Technical College is licensed by CC BY 4.0 ↵
- This work is a derivative of StatPearls by Sattar and Chhabra and is licensed under CC BY 4.0 ↵
- “Measuring the QRS Duration” by Chippewa Valley Technical College is licensed by CC BY 4.0 ↵
- This work is a derivative of StatPearls by Sattar and Chhabra and is licensed under CC BY 4.0 ↵
- This work is a derivative of StatPearls by Sattar and Chhabra and is licensed under CC BY 4.0 ↵
- “ECG-waves-segments-and-intervals-LITFL-ECG-library-3.jpg” by unknown author at LITFL is licensed under CC BY-NC-SA 4.0 . Access the original image at https://litfl.com/st-segment-ecg-library/ ↵
- This work is a derivative of StatPearls by Sattar and Chhabra and is licensed under CC BY 4.0 ↵
- This work is a derivative of StatPearls by Sattar and Chhabra and is licensed under CC BY 4.0 ↵
- This work is a derivative of StatPearls by Sattar and Chhabra and is licensed under CC BY 4.0 ↵
- This work is a derivative of StatPearls by Sattar and Chhabra and is licensed under CC BY 4.0 ↵
- This work is a derivative of StatPearls by Sattar and Chhabra and is licensed under CC BY 4.0 ↵
- Physical Therapy Reviewer. (n.d.). How to read an ECG . https://ptreviewer.com/electrocardiogram-ecg-2/reading-an-ecg/ ↵
- RegisteredNurseRN. (2022, March 7). EKG/ECG interpretation basics nursing NCLEX | QRS complex, P wave, T wave, PR interval [Video]. YouTube. All rights reserved. https://youtu.be/bUF12VIgzP0 ↵
- RegisteredNurseRN. (2015, May 21). How to memorize the PQRST EKG rhythm strip wave for anatomy & pathophysiology [Video]. YouTube. All rights reserved. https://youtu.be/QAQiK-zRtl0 ↵


9.3: Simplification of Denominate Numbers
- be able to convert an unsimplified unit of measure to a simplified unit of measure
- be able to add and subtract denominate numbers
- be able to multiply and divide a denominate number by a whole number
Converting to Multiple Units
Numbers that have units of measure associated with them are called denominate numbers . It is often convenient, or even necessary, to simplify a denominate number.
A denominate number is simplified when the number of standard units of measure associated With it does not exceed the next higher type of unit.
The denominate number 55 min is simplified since it is smaller than the next higher type of unit, 1 hr. The denominate number 65 min is not simplified since it is not smaller than the next higher type of unit, 1 hr. The denominate number 65 min can be simplified to 1 hr 5 min. The denominate number 1 hr 5 min is simplified since the next higher type of unit is day, and 1 hr does not exceed 1 day.
Simplify 19 in.
Solution
Since \(\text{12 in. = 1 ft.}\), and \(19 = 12 + 7\).
\(\begin{array} {rcl} {\text{19 in.}} & = & {\text{12 in. + 7 in.}} \\ {} & = & {\text{1 ft + 7 in.}} \\ {} & = & {\text{1 ft 7 in.}} \end{array}\)
Simplify 4 gal 5 qt.
Solution
Since \(\text{4 qt = 1 gal}\), and \(5 = 4 + 1\).
\(\begin{array} {rcl} {\text{4 gal 5 qt}} & = & {\text{4 gal + 4 qt + 1 qt}} \\ {} & = & {\text{4 gal + 1 gal + 1 qt}} \\ {} & = & {\text{5 gal + 1 qt}} \\ {} & = & {\text{5 gal 1 qt}}\end{array}\)
Simplify 2 hr 75 min.
Solution
Since \(\text{60 min = 1 hr}\), and \(75 = 60 + 15\).
\(\begin{array} {rcl} {\text{2 hr 75 min}} & = & {\text{2 hr + 60 min + 15 min}} \\ {} & = & {\text{2 hr + 1 hr + 15 min}} \\ {} & = & {\text{3 hr + 15 min}} \\ {} & = & {\text{3 hr 15 min}}\end{array}\)
Simplify 43 fl oz.
Solution
Since \(\text{8 fl oz = 1 c}\) (1 cup), and \(43 \div 8 = \text{5R3}\).
\(\begin{array} {rcl} {\text{43 fl oz}} & = & {\text{40 fl oz + 3 fl oz}} \\ {} & = & {5 \cdot 8 \text{ fl oz + 3 fl oz}} \\ {} & = & {5 \cdot 1 \text{ c + 3 fl oz}} \\ {} & = & {\text{5 c + 3 fl oz}}\end{array}\)
But, \(\text{2c = 1 pt}\) and \(5 \div 2 = \text{2R1}\). So,
\(\begin{array} {rcl} {\text{5 c + 3 fl oz}} & = & {2 \cdot 2 \text{ c + 1 c + 3 fl oz}} \\ {} & = & {2 \cdot 1 \text{ pt + 1 c + 3 fl oz}} \\ {} & = & {\text{2 pt + 1 c + 3 fl oz}} \end{array}\)
But, \(\text{2 pt = 1 qt}\), so
\(\text{2 pt + 1 c + 3 fl oz = 1 qt 1 c 3 fl oz}\)
Practice Set A
Simplify each denominate number. Refer to the conversion tables given in [link] , if necessary.
18 in.
- Answer
-
1 ft 6 in.
Practice Set A
8 gal 9 qt
- Answer
-
10 gal 1 qt
Practice Set A
5 hr 80 min
- Answer
-
6 hr 20 min
Practice Set A
8 wk 11 da
- Answer
-
9 wk 4 da
Practice Set A
86 da
- Answer
-
12 wk 2 da
Adding and Subtracting Denominate Numbers
Adding and Subtracting Denominate Numbers
Denominate numbers can be added or subtracted by:
- writing the numbers vertically so that the like units appear in the same column.
- adding or subtracting the number parts, carrying along the unit.
- simplifying the sum or difference.
Add 6 ft 8 in. to 2 ft 9 in.
Solution
\(\begin{array} {r} {\text{6 ft 8 in.}} \\ {\underline{\text{+ 2 ft 9 in.}}} \\ {\text{8 ft 17 in.}} \end{array}\) Simplify this denominate number.
Since \(\text{12 in. = 1 ft.}\)
\(\begin{array} {rcl} {\text{8 ft + 12 in. + 5 in.}} & = & {\text{8 ft + 1 ft + 5 in.}} \\ {} & = & {\text{9 ft + 5 in.}} \\ {} & = & {\text{9 ft 5 in.}} \end{array}\)
Subtract 5 da 3 hr from 8 da 11 hr.
Solution
\(\begin{array} {r} {\text{8 da 11 hr}} \\ {\underline{\text{- 5 da 3 hr}}} \\ {\text{3 da 8 hr}} \end{array}\)
Subtract 3 lb 14 oz from 5 lb 3 oz.
Solution
\(\begin{array} {r} {\text{5 lb 3 oz}} \\ {\underline{\text{- 3 lb 14 oz}}} \end{array}\)
We cannot directly subtract 14 oz from 3 oz, so we must borrow 16 oz from the pounds.
\(\begin{array} {rcl} {\text{5 lb 3 oz}} & = & {\text{5 lb + 3 oz}} \\ {} & = & {\text{4 lb + 1 lb + 3 oz}} \\ {} & = & {\text{4 lb + 16 oz + 3 oz (Since 1 lb = 16 oz.)}} \\ {} & = & {\text{4 lb + 19 oz}} \\ {} & = & {\text{4 lb 19 oz}} \end{array}\)
\(\begin{array} {r} {\text{4 lb 19 oz}} \\ {\underline{\text{- 3 lb 14 oz}}} \\ {\text{1 lb 5 oz}} \end{array}\)
Subtract 4 da 9 hr 21 min from 7 da 10 min.
Solution
\(\begin{array} {r} {\text{7 da 0 hr 10 min}} \\ {\underline{\text{- 4 da 9 hr 21 min}}} \end{array}\) Borrow 1 da from the 7 da.
\(\begin{array} {r} {\text{6 da 24 hr 10 min}} \\ {\underline{\text{- 4 da 9 hr 21 min}}} \end{array}\) Borrow 1 hr from the 24 hr.
\(\begin{array} {r} {\text{6 da 23 hr 70 min}} \\ {\underline{\text{- 4 da 9 hr 21 min}}} \\ {\text{2 da 14 hr 49 min}} \end{array}\)
Practice Set B
Perform each operation. Simplify when possible.
Add 4 gal 3 qt to 1 gal 2 qt.
- Answer
-
6 gal 1 qt
Practice Set B
Add 9 hr 48 min to 4 hr 26 min.
- Answer
-
14 hr 14 min
Practice Set B
Subtract 2 ft 5 in. from 8 ft 7 in.
- Answer
-
6 ft 2in.
Practice Set B
Subtract 15 km 460 m from 27 km 800 m.
- Answer
-
12 km 340 m
Practice Set B
Subtract 8 min 35 sec from 12 min 10 sec.
- Answer
-
3 min 35 sec
Practice Set B
Add 4 yd 2 ft 7 in. to 9 yd 2 ft 8 in.
- Answer
-
14 yd 2 ft 3 in
Practice Set B
Subtract 11 min 55 sec from 25 min 8 sec.
- Answer
-
13 min 13 sec
Multiplying a Denominate Number by a Whole Number
Let's examine the repeated sum
\(\underbrace{\text{4 ft 9 in. + 4 ft 9 in. + 4 ft 9 in.}}_{\text{3 times}} = \text{12 ft 27 in.}\)
Recalling that multiplication is a description of repeated addition, by the distributive property we have
\(\begin{array} {rcl} {\text{3(4 ft 9 in.)}} & = & {\text{3 (4ft + 9 in.)}} \\ {} & = & {3 \cdot 4 \text{ ft } + 3 \cdot 9 \text{ in.}} \\ {} & = & {\text{12 ft + 27 in. Now, 27 in. = 2 ft 3 in.}} \\ {} & = & {\text{12 ft + 2 ft + 3 in.}} \\ {} & = & {\text{14 ft + 3 in.}} \\ {} & = & {\text{14 ft 3 in.}} \end{array}\)
From these observations, we can suggest the following rule.
Multiplying a Denominate Number by a Whole Number
To multiply a denominate number by a whole number, multiply the number part of each unit by the whole number and affix the unit to this product.
Perform the following multiplications. Simplify if necessary.
\(\begin{array} {rcl} {6 \cdot \text{(2 ft 4 in.)}} & = & {6 \cdot 2 \text{ ft + 6} \cdot 4 \text{in.}} \\ {} & = & {\text{12 ft + 24 in.}} \end{array}\)
Since \(\text{3 ft = 1 yd}\) and \(\text{12 in. = 1 ft.}\)
\(\begin{array} {rcl} {\text{12 ft + 24 in.}} & = & {\text{4 yd + 2 ft}} \\ {} & = & {\text{4 yd 2 ft}} \end{array}\)
\(\begin{array} {rcl} {8 \cdot \text{(5 hr 21 min 55 sec)}} & = & {8 \cdot 5 \text{ hr} + 8 \cdot 21 \text{ min} + 8 \cdot 55 \text{ sec}} \\ {} & = & {\text{40 hr + 168 min + 440 sec}} \\ {} & = & {\text{40 hr + 168 min + 7 min + 20 sec}} \\ {} & = & {\text{40 hr + 175 min + 20 sec}} \\ {} & = & {\text{40 hr + 2 hr + 55 min + 20 sec}} \\ {} & = & {\text{42 hr + 55 min + 20 sec}} \\ {} & = & {\text{24 hr + 18 hr + 55 min + 20 sec}} \\ {} & = & {\text{1 da + 18 hr + 55 min + 20 sec}} \\ {} & = & {\text{1 da 18 hr 55 min 20 sec}} \end{array}\)
Practice Set C
Perform the following multiplications. Simplify.
\(2 \cdot \text{(10 min)}\)
- Answer
-
20 min
Practice Set C
\(5 \cdot \text{(3 qt)}\)
- Answer
-
\(\text{15 qt = 3 gal 3 qt}\)
Practice Set C
\(4 \cdot \text{(5 ft 8 in.)}\)
- Answer
-
\(\text{20 ft 32 in. = 7 yd 1 ft 8 in.}\)
Practice Set C
\(10 \cdot \text{(2 hr 15 min 40 sec)}\)
- Answer
-
\(\text{20 hr 150 min 400 sec = 22 hr 36 min 40 sec}\)
Dividing a Denominate Number by a Whole Number
Dividing a Denominate Number by a Whole Number
To divide a denominate number by a whole number, divide the number part of each unit by the whole number beginning with the largest unit. Affix the unit to this quotient. Carry any remainder to the next unit.
Perform the following divisions. Simplify if necessary.
\(\text{(12 min 40 sec)} \div 4\)
Solution
Thus \(\text{(12 min 40 sec)} \div 4 = \text{3 min 10 sec}\)
\(\text{(5 yd 2 ft 9 in.)} \div 3\)
Solution
\(\begin{array} {c} {\text{Convert to feet: 2 yd 2 ft = 8 ft}} \\ {\text{Convert to inches: 2 ft 9 in. = 33 in.}} \end{array}\)
Thus \(\text{(5 yd 2 ft 9 in.)} \div 3 = \text{1 yd 2 ft 11 in.}\)
Practice Set D
Perform the following divisions. Simplify if necessary.
\(\text{(18 hr 36 min)} \div 9\)
- Answer
-
2 hr 4 min
Practice Set D
\(\text{(36 hr 8 min)} \div 8\)
- Answer
-
4 hr 18 min
Practice Set D
\(\text{(13 yd 7 in.)} \div 5\)
- Answer
-
2 yd 1 ft 11 in
Practice Set D
\(\text{(47 gal 2 qt 1 pt)} \div 3\)
- Answer
-
15 gal 3 qt 1 pt
Exercises
For the following 15 problems, simplify the denominate numbers.
Exercise \(\PageIndex{1}\)
16 in.
- Answer
-
1 foot 4 inches
Exercise \(\PageIndex{2}\)
19 ft
Exercise \(\PageIndex{3}\)
85 min
- Answer
-
1 hour 25 minutes
Exercise \(\PageIndex{4}\)
90 min
Exercise \(\PageIndex{5}\)
17 da
- Answer
-
2 weeks 3 days
Exercise \(\PageIndex{6}\)
25 oz
Exercise \(\PageIndex{7}\)
240 oz
- Answer
-
15 pounds
Exercise \(\PageIndex{8}\)
3,500 lb
Exercise \(\PageIndex{9}\)
26 qt
- Answer
-
6 gallons 2 quarts
Exercise \(\PageIndex{10}\)
300 sec
Exercise \(\PageIndex{11}\)
135 oz
- Answer
-
8 pounds 7 ounces
Exercise \(\PageIndex{12}\)
14 tsp
Exercise \(\PageIndex{13}\)
18 pt
- Answer
-
2 gallons 1 quart
Exercise \(\PageIndex{14}\)
3,500 m
Exercise \(\PageIndex{15}\)
16,300 mL
- Answer
-
16 liters 300 milliliters (or 1daL 6 L 3dL)
For the following 15 problems, perform the indicated operations and simplify the answers if possible.
Exercise \(\PageIndex{16}\)
Add 6 min 12 sec to 5 min 15 sec.
Exercise \(\PageIndex{17}\)
Add 14 da 6 hr to 1 da 5 hr.
- Answer
-
15 days 11 hours
Exercise \(\PageIndex{18}\)
Add 9 gal 3 qt to 2 gal 3 qt.
Exercise \(\PageIndex{19}\)
Add 16 lb 10 oz to 42 lb 15 oz.
- Answer
-
59 pounds 9 ounces
Exercise \(\PageIndex{20}\)
Subtract 3 gal 1 qt from 8 gal 3 qt.
Exercise \(\PageIndex{21}\)
Subtract 3 ft 10 in. from 5 ft 8 in.
- Answer
-
1 foot 10 inches
Exercise \(\PageIndex{22}\)
Subtract 5 lb 9 oz from 12 lb 5 oz.
Exercise \(\PageIndex{23}\)
Subtract 10 hr 10 min from 11 hr 28 min.
- Answer
-
1 hour 18 minutes
Exercise \(\PageIndex{24}\)
Add 3 fl oz 1 tbsp 2 tsp to 5 fl oz 1 tbsp 2 tsp.
Exercise \(\PageIndex{25}\)
Add 4 da 7 hr 12 min to 1 da 8 hr 53 min.
- Answer
-
5 days 16 hours 5 minutes
Exercise \(\PageIndex{26}\)
Subtract 5 hr 21 sec from 11 hr 2 min 14 sec.
Exercise \(\PageIndex{27}\)
Subtract 6 T 1,300 lb 10 oz from 8 T 400 lb 10 oz.
- Answer
-
1 ton 1,100 pounds (or 1T 1,100 lb)
Exercise \(\PageIndex{28}\)
Subtract 15 mi 10 in. from 27 mi 800 ft 7 in.
Exercise \(\PageIndex{29}\)
Subtract 3 wk 5 da 50 min 12 sec from 5 wk 6 da 20 min 5 sec.
- Answer
-
2 weeks 23 hours 29 minutes 53 seconds
Exercise \(\PageIndex{30}\)
Subtract 3 gal 3 qt 1 pt 1 oz from 10 gal 2 qt 2 oz.
Exercises for Review
Exercise \(\PageIndex{31}\)
Find the value: \((\dfrac{5}{8})^2 + \dfrac{39}{64}\).
- Answer
-
1
Exercise \(\PageIndex{32}\)
Find the sum: \(8 + 6 \dfrac{3}{5}\).
Exercise \(\PageIndex{33}\)
Convert \(2.05 \dfrac{1}{11}\) to a fraction.
- Answer
-
\(2 \dfrac{14}{275}\)
Exercise \(\PageIndex{34}\)
An acid solution is composed of 3 parts acid to 7 parts water. How many parts of acid are there in a solution that contains 126 parts water?
Exercise \(\PageIndex{35}\)
Convert 126 kg to grams.
- Answer
-
126,000 g


3.14: Friedel's Law
Friedel's law states that the intensities of the \(h\), \(k\), \(l\) and \(\overline{h}\), \(\overline{k}\), \(\overline{l}\) reflections are equal. This is true either if the crystal is centrosymmetric or if no resonant scattering is present. It is in that case not possible to tell by diffraction whether an inversion center is present or not. The apparent symmetry of the crystal is then one of the eleven Laue classes .
The reason for Friedel's rule is that the diffracted intensity is proportional to the square of the modulus of the structure factor, | F h | 2 , according to the geometrical, or kinematical theory of diffraction. It depends similarly on the modulus of the structure factor according to the dynamical theory of diffraction. The structure factor is given by:
\[F_h = \sum_j f_j (exp - 2 \pi i) \textbf {h} \cdot \textbf{r_j} \nonumber \]
where \(f_j\) is the atomic scattering factor of atom j , \(\textbf{h}\) the reflection vector and \(\textbf{r_j}\) the position vector of atom \(j\). There comes:
\[|F_h|^2 = F_h F_h^* = F_h F_{\bar h} = |F_{\bar h}|^2 \nonumber \]
if the atomic scattering factor, f j , is real. The intensities of the \(h\), \(k\), \(l\) and \(\overline{h}\), \(\overline{k}\), \(\overline{l}\) reflections are then equal. If the crystal is absorbing, however, due to resonant scattering, the atomic scattering factor is complex and
\[F_{\bar h} \ne F_h^* \nonumber \]
The reflections h , k , l and \({\bar h}\), \({\bar k}\), \({\bar l}\) are called a Friedel pair . They are used in the resolution of the phase problem for the solution of crystal structures and in the determination of absolute structure.
History
Friedel's law was stated by G. Friedel (1865-1933) in 1913 (Friedel G., 1913, Sur les symétries cristallines que peut révéler la diffraction des rayons X. , C.R. Acad. Sci. Paris , 157 , 1533-1536.
See also
Section 3.1 of International Tables of crystallography, Volume A


14.1: Introduction
-
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- John Banks & Stuart Cunningham
- University of California Press
In this chapter, we pay full attention to the structural conditions and human cost of precarious labor in a particular local instance of the games industry. But at the same time, we attempt to shift the debate on precarity from the existential (the creative individual attracted to industries promising autonomy and meaningful work and finding only casualization, no work/life balance, and poor management) and the totalizing (all work under regimes of neoliberal hypercapitalism is increasingly characterized by precarity; indeed a whole new class—the precariat 1 —is posited as emerging) to a focus on analysis for actionable reform.
Significant “creative destruction” 2 through the global financial crisis (GFC) led to games industry restructuring and consolidation, including withdrawal of major publisher investment in many dispersed regional hubs of games production. More fundamentally, major platform shifts and new business models started before the global downturn and continue through this contemporary period of slowdown in the world economy. There has been major consolidation at the console production end of the games industry, with more expensive blockbuster or AAA titles, a hollowing out of the midrange games market, and rapid growth and proliferation of casual gaming and mobile applications with unprecedentedly lower production costs and barriers to entry.
What has happened to one such regional hub, the Australian games industry, spatially remote from the centers of publisher power and hubs of creative ferment?
A recent “perfect storm” of factors has arisen to change the face of the Australian games industry. The industry had grown on the model of work for hire producing “catalog fillers” for the major publishers; very little original IP was produced. And while very few AAA titles were made in Australia, games companies had a reputation for quality. However, the business proposition was buttressed by more than a decade of favorable exchange rates, which (literally) underwrote international investment. The industry by 2007 was structured around approximately forty-five midsize small businesses. 3 Notable companies included Krome, Pandemic, THQ StudioOz, Creative Assembly, Torus, and 2K.
The global financial crisis saw higher-end production scaled back, a withdrawal by the major publishers from spatially distended supply chains, and a new preference for formally affiliated production companies. At the beginning of 2007, the Australian dollar was 75 cents on the U.S. dollar. During the GFC, the Australian dollar became a “currency haven,” such that by the start of 2012 it was worth US$1.02, gutting the industry of its price advantage. Of even greater structural consequence for the industry was the simultaneous explosion of apps-based mobile casual games play based on the smartphone platform and later the tablet.
Official statistics tell a stark story of destruction of value. Of the 1,431 reported employees in 2007, only 581 remained by mid-2012, and reported game development income had dropped from A$116.9 million to just A$44.4 million. 4 The industry’s spatial pattern in 2007 evidenced a significant presence in Queensland and Victoria, with additional studios in New South Wales, the Australian Capital Territory, and South Australia. By 2012, the majority of the bigger studios had closed, and the industry had retreated to be concentrated in Victoria. Those whose doors had closed or who had radically downsized included Krome, Pandemic, THQ StudioOz, BlueTongue, Team Bondi, SEGA Creative Assembly, and Tantalus Media Brisbane. The major studios remaining included Halfbrick (Brisbane), 2K Australia (Canberra), and in Melbourne, Big Ant, Torus Games, Tantalus, and Wicked Witch. According to the Games Developers Association of Australia (GDAA), the main advocacy and professional association for the industry, somewhere between 60 and 70 percent of industry workers had either moved to another industry (many skills, preeminently programming skills, are very transferable) or had left Australia for more resilient industry locations or those better supported by government policy and programs. 5
In 2014, the GDAA characterized the industry as composed of two hundred formally registered businesses, of which 92 percent are considered to be independents. 6 It defines independent as a typically small-scale enterprise that concentrates exclusively on original IP and self-publishes on the new digital platforms (Apple App Store, Android, Steam). It estimates about eight hundred workers now in the industry. This is a recent history of an industry much reduced in turnover and traditional employment, but which has transformed its revenue base from 80 percent work for hire to 75 percent original IP—an almost complete reversal in the balance between business models. 7
But, invoking Joseph Schumpeter, how “creative” has this destruction been? A rigorous critical organizational-studies analysis of the Australian industry advances the argument that severe power differentials between publisher and producer/developer have persisted across this momentous industry restructure and continue to compromise local agency in global supply chains. 8 An equally rigorous media-studies argument anatomizing poor labor conditions in the industry globally is nevertheless clear that “the most plentiful and well-paying jobs in the video game industry continue to be those provided by major video game publishers either directly or indirectly.” 9 Neither view offers much comfort for the idea that this destruction could be in any way “creative.”
These perspectives, however, contrast with the self-understanding of many of those games workers (whom we have interviewed for the research that supports this chapter) who survived the shakeout or are sufficiently new to the industry to know no other conditions. Culturally and industrially, original IP—and the conditions under which it can be prioritized—tends to be championed by these developers against fee-for-service and as a normative aspiration. Industrially, a dominant narrative in the industry has been the desire to move from fee-for-service (where the company is a price taker and doesn’t control its own destiny) to original IP. Culturally, this aspiration also speaks to many developers’ creative impulse and is actually enshrined in the advocacy and the definition of indie established by the representative body, the GDAA. It is reinforced by normative criteria built into state policy and program funding support.
Given the degree to which higher-end fee-for-service business has dried up, while essentially self-publication on major digital distribution platforms (Apple’s App Store, the Google Play Store, Steam, and so on) has grown exponentially, necessity has become a virtue. Conditions have crafted an industry that is much reduced in terms of turnover and traditional employment but now operates within a disintermediated value chain that radically forces the pace of innovation. Despite much commentary that treats Apple, for example, as basically yet another global corporation “taking their (un)fair share of financial profits,” 10 near-global dissemination via the digital platforms on a 30/70 split of income derived represents an ostensibly better deal than the power asymmetries enshrined in dealing with the major publishers. 11
Australian companies, in particular Halfbrick after its huge success with Fruit Ninja (2010), made hay while the sun shone in the early days of apps-driven games and became a sort of template for national ruminations on how to succeed in the new environment. 12 It is distinctly harder now to capture attention: massively lower barriers to entry create conditions in which it is estimated that more than 1.3 million apps are now available on the App Store with duplication across the platforms, of which around 20 percent are games. 13 Mobile games production is markedly less driven by the crunch associated with games development under the dominant business model of fee-for-service work, in which development schedules were driven by milestones at the behest of large international publishers. This has led, Antony Reed suggests, to a situation where the industry has seen much less attrition in last few years. Furthermore, there is arguably a great deal more innovation activity in original IP. Indeed, there is runaway innovation, 14 with the rapid shift from games as a product to games as a service driving the mobile apps purchase price points to zero, accompanied by the proliferation of in-app purchasing. And these rapid shifts have in turn been challenged by a return by some to premium mobile app pricing as well as premium pricing for games released through Steam.


35.4: Urinary Stimulants
By the end of this section, you should be able to:
- 35.4.1 Identify the characteristics of urinary stimulant drugs used for urinary and bladder disorders.
- 35.4.2 Explain the indications, actions, adverse reactions, contraindications, and interactions of urinary stimulant drugs used for urinary and bladder disorders.
- 35.4.3 Describe nursing implications of urinary stimulant drugs used for urinary and bladder disorders.
- 35.4.4 Explain the client education related to urinary stimulant drugs used for urinary and bladder disorders.
Urinary stimulants stimulate the smooth muscles of the urinary bladder. These drugs work by activating the muscarinic receptors of the bladder, leading to increased bladder contractions and improved bladder emptying. They are primarily prescribed for urinary retention or to improve bladder function in individuals with certain neurologic conditions, such as a spinal cord injury.
Bethanechol Chloride
Bethanechol chloride ( Duvoid ) is indicated for acute postoperative and postpartum nonobstructive (functional) urinary retention and for neurogenic atony of the urinary bladder with retention (Padda & Derain, 2022).
Determine the minimum effective dose by giving 5–10 mg initially and repeating the same amount at hourly intervals until a satisfactory response occurs or until a maximum of 50 mg has been given (DailyMed, Bethanechol chloride , 2022).
The drug’s effects sometimes appear within 30 minutes, usually peak between 60 and 90 minutes, and persist for approximately 1 hour (DailyMed, Bethanechol chloride , 2022).
Adverse Effects and Contraindications
Common adverse effects include malaise, urinary urgency, headache, vasomotor response , sweating, and miosis. Contraindications include hyperthyroidism, peptic ulcer disease, asthma, pronounced bradycardia, hypotension, coronary artery disease, seizure, Parkinson’s disease, and hypersensitivity to the drug or any of its components.
Table 35.6 is a drug prototype table for urinary stimulants featuring bethanechol chloride. It lists drug class, mechanism of action, adult dosage, indications, therapeutic effects, drug and food interactions, adverse effects, and contraindications.
|
Drug Class
Urinary stimulant, cholinergic Mechanism of Action Stimulates activation of the parasympathetic nervous system, causing increased tone of the detrusor muscle of the bladder, producing contraction and emptying of the bladder |
Drug Dosage
10–50 mg orally 3–4 times a day. |
|
Indications
Treatment of acute postoperative and postpartum nonobstructive (functional) urinary retention Neurogenic atony of the urinary bladder with retention Therapeutic Effects Minimizes urinary retention |
Drug Interactions
Atropine Diphenhydramine Donepezil Glutamine Hyoscyamine Neostigmine Procainamide Propantheline Scopolamine Food Interactions No significant interactions |
|
Adverse Effects
Malaise Headache Urinary urgency Sweating Vasomotor response Miosis Monitor closely for bacteremia due to reflex infection from incomplete bladder emptying |
Contraindications
Hypersensitivity Hyperthyroidism Peptic ulcer disease Asthma Bradycardia Hypotension Coronary artery disease Seizures Parkinson’s disease |
Nursing Implications
The nurse should do the following for clients who are taking urinary stimulants:
- Before administering the drug, check the client’s medical history, current drug list (including over-the-counter medications), and allergies.
- Monitor the client closely for vasomotor responses, such as hypotension, bradycardia, and orthostatic hypotension.
- For a client taking bethanechol chloride, monitor closely for bronchospasm, wheezing, and tachycardia because these can be serious adverse effects of the drug.
- Monitor the client’s urine output closely.
- Provide client teaching regarding the drug and when to call the health care provider. See below for additional client teaching guidelines.
Client Teaching Guidelines
The client taking a urinary stimulant should:
- Take the medication as prescribed by their health care provider.
- Take bethanechol chloride on an empty stomach, at least 1 hour before or 2 hours after a meal, for better absorption.
- Notify their health care provider if symptoms do not improve within 90 minutes after taking bethanechol chloride because the treatment may need to be modified.
- Store bethanechol chloride at room temperature and away from moisture and heat.
- Avoid getting up too fast from a sitting or lying position because this drug may cause a vasomotor response that includes dizziness.
- Take two or more doses at one time because this may cause a significant vasomotor response such as severe hypotension and bradycardia.


1.3: The Cell
Life takes many forms, from giant redwood trees towering hundreds of feet in the air to the tiniest known microbes, which measure only a few billionths of a meter. Humans have long pondered life’s origins and debated the defining characteristics of life, but our understanding of these concepts has changed radically since the invention of the microscope. In the 17th century, observations of microscopic life led to the development of the cell theory: the idea that the fundamental unit of life is the cell, that all organisms contain at least one cell, and that cells only come from other cells.
Despite sharing certain characteristics, cells may vary significantly. The two main types of cells are prokaryotic cells (lacking a nucleus) and eukaryotic cells (containing a well-organized, membrane-bound nucleus). Each type of cell exhibits remarkable variety in structure, function, and metabolic activity. This chapter will focus on the historical discoveries that have shaped our current understanding of microbes, including their origins and their role in human disease. We will then explore the distinguishing structures found in prokaryotic and eukaryotic cells.
-
- 1.3.1: Spontaneous Generation
- The theory of spontaneous generation states that life arose from nonliving matter. It was a long-held belief dating back to Aristotle and the ancient Greeks. Experimentation by Francesco Redi in the 17th century presented the first significant evidence refuting spontaneous generation by showing that flies must have access to meat for maggots to develop on the meat. Louis Pasteur is credited with conclusively disproving the theory and proposed that “life only comes from life.”
-
- 1.3.2: Foundations of Modern Cell Theory
- Although cells were first observed in the 1660s by Robert Hooke, cell theory was not well accepted for another 200 years. The work of scientists such as Schleiden, Schwann, Remak, and Virchow contributed to its acceptance. Endosymbiotic theory states that mitochondria and chloroplasts, organelles found in many types of organisms, have their origins in bacteria. Significant structural and genetic information support this theory. The miasma theory was widely accepted until the 19th century.
Thumbnail: A 3D rendering of an animal cell cut in half. (CC -BY-SA 4.0; Zaldua I., Equisoain J.J., Zabalza A., Gonzalez E.M., Marzo A., Public University of Navarre).


7.23C: Biochemical Products of Recombinant DNA Technology
- Describe the advances made possible by recombinant DNA technology
Recombinant DNA technology is the latest biochemical analysis that came about to satisfy the need for specific DNA segments. In this process, surrounding DNA from an existing cell is clipped in the desired amount of segments so that it can be copied millions of times.
Recombinant DNA technology engineers microbial cells for producing foreign proteins, and its success solely depends on the precise reading of equivalent genes made with the help of bacterial cell machinery. This process has been responsible for fueling many advances related to modern molecular biology. The last two decades of cloned-DNA sequence studies have revealed detailed knowledge about gene structure as well as its organization. It has provided hints to regulatory pathways with the aid of which gene expression in myriad cell types is controlled by the cells, especially in those organisms having body plan with basic vertebrae structure.
Recombinant DNA technology, apart from being an important tool of scientific research, has also played a vital role in the diagnosis and treatment of various diseases, especially those belonging to genetic disorders.
Some of the recent advances made possible by recombinant DNA technology are:
1. Isolating proteins in large quantities: many recombinant products are now available, including follicle stimulating hormone (FSH), Follistim AQ vial, growth hormone, insulin and some other proteins.
2. Making possible mutation identification: due to this technology, people can be easily tested for mutated protein presence that can lead to breast cancer, neurofibromatosis, and retinoblastoma.
3. Hereditary diseases carrier diagnosis: tests now available to determine if a person is carrying the gene for cystic fibrosis, the Tay-Sachs diseases, Huntington’s disease or Duchenne muscular dystrophy.
4. Gene transfer from one organism to other: the advanced gene therapy can benefit people with cystic fibrosis, vascular disease, rheumatoid arthritis and specific types of cancers.
Key Points
- Recombinant DNA (rDNA) is widely used in biotechnology, medicine and research. Proteins and other products that result from the use of rDNA technology are found in essentially every western pharmacy, doctor’s or veterinarian’s office, medical testing laboratory, and biological research laboratory.
- Organisms that have been manipulated using recombinant DNA technology, and products derived from those organisms have found their way into many farms, supermarkets, home medicine cabinets, and even pet shops.
- Biochemical products of recombinant DNA technology in medicine and research include: human recombinant insulin, growth hormone, blood clotting factors, hepatitis B vaccine, and diagnosis of HIV infection.
- Biochemical products of recombinant DNA technology in agriculture include: golden rice, herbicide-resistant crops, and insect-resistant crops.
Key Terms
- retinoblastoma : A malignant tumour of the retina; a hereditary condition found mostly in children.
- neurofibromatosis : A genetic disorder characterized by the presence of multiple neurofibromas under the skin
- cystic fibrosis : An inherited condition in which the exocrine glands produce abnormally viscous mucus, causing chronic respiratory and digestive problems.
- recombinant DNA technology : the process of taking a gene from one organism and inserting it into the DNA of another


5.1: Gaussian Elimination Revisited
\(\newcommand{\twovec}[2]{\begin{pmatrix} #1 \\ #2 \end{pmatrix} } \)
\(\newcommand{\threevec}[3]{\begin{pmatrix} #1 \\ #2 \\ #3 \end{pmatrix} } \)
\(\newcommand{\fourvec}[4]{\begin{pmatrix} #1 \\ #2 \\ #3 \\ #4 \end{pmatrix} } \)
\(\newcommand{\fivevec}[5]{\begin{pmatrix} #1 \\ #2 \\ #3 \\ #4 \\ #5 \end{pmatrix} } \)
In this section, we revisit Gaussian elimination and explore some problems with implementing it in the straightforward way that we described back in Section 1.2 . In particular, we will see how the fact that computers only approximate arithmetic operations can lead us to find solutions that are far from the actual solutions. Second, we will explore how much work is required to implement Gaussian elimination and devise a more efficient means of implementing it when we want to solve equations \(A\mathbf x = \mathbf b\) for several different vectors \(\mathbf b\text{.}\)
Preview Activity 5.1.1.
To begin, let's recall how we implemented Gaussian elimination by considering the matrix
- What is the first row operation we perform? If the resulting matrix is \(A_1\text{,}\) find a matrix \(E_1\) such that \(E_1A = A_1\text{.}\)
- What is the matrix inverse \(E_1^{-1}\text{?}\) You can find this using your favorite technique for finding a matrix inverse. However, it may be easier to think about the effect that the row operation has and how it can be undone.
-
Perform the next two steps in the Gaussian elimination algorithm to obtain \(A_3\text{.}\) Represent these steps using multiplication by matrices \(E_2\) and \(E_3\) so that
\begin{equation*} E_3E_2E_1A = A_3\text{.} \end{equation*}
- Suppose we need to scale the second row by \(-2\text{.}\) What is the \(3\times3\) matrix that perfoms this row operation by left multiplication?
- Suppose that we need to interchange the first and second rows. What is the \(3\times3\) matrix that performs this row operation by left multiplication?
Partial pivoting
The first issue that we will address is the fact that computers do not perform arithemtic operations exactly. For instance, if we ask Python to evaluate
0.1 + 0.2
, it reports
0.30000000000000004
though we know that the true value is 0.3. There are a couple of reasons for this.
First, computers perform arithmetic using base 2 numbers, which means that numbers we enter in decimal form, such as \(0.1\text{,}\) must be converted to base 2. Even though 0.1 has a simple decimal form, its representation in base 2 is the repeating decimal
To accurately represent this number inside a computer would require an infinite number of digits. Since a computer can only hold a finite number of digits, we are necessarily using an approximation when simply representing this number in a computer.
In addition, arithmetic operations, such as addition, are prone to error. To keep things simple, suppose we have a computer that represents numbers using only three decimal digits. For instance, the number 1.023 would be represented as
1.02
while 0.023421 would be
0.0234
. If we add these numbers, we have 1.023 + 0.023421 = 1.046421; the computer reports this sum as
1.02 + 0.0234 = 1.04
, whose last digit is not correctly rounded. Generally speaking, we will see this problem, which is called
round off error
, whenever we add numbers of signficantly different magnitudes.
Remember that Gaussian elimination, when applied to an \(n\times n\) matrix, requires approximately \(\frac 23 n^3\) operations. If we have a \(1000\times1000\) matrix, performing Gaussian elimination requires roughly a billion operations, and the errors introduced in each operation could accumulate. How can we have confidence in the final result? We can never completely avoid these errors, but we can take steps to mitigate them. The next activity will introduce one such technique.
Activity 5.1.2.
Suppose we have a hypothetical computer that represents numbers using only three decimal digits, as we have been discussing. We will consider the linear system
-
Show that this system has the unique solution
\begin{equation*} \begin{aligned} x & {}={} \frac{10000}{9999} = 1.00010001\ldots, \\ y & {}={} \frac{9998}{9999} = 0.99989998\ldots\text{.} \end{aligned} \end{equation*}
- If we represent this solution inside our computer that only holds 3 decimal digits, what do we find for the solution? This is the best that we can hope to find using our computer.
- Let's imagine that we use our computer to find the solution using Gaussian elimination; that is, after every arithmetic operation, we keep only three decimal digits. Our first step is to multiply the first equation by 10000 and subtract it from the second equation. If we represent numbers using only three decimal digits, what does this give for the value of \(y\text{?}\)
- By substituting our value for \(y\) into the first equation, what do we find for \(x\text{?}\)
- Compare the solution we find on our computer with the actual solution and assess the quality of the approximation.
-
Let's now modify the linear system by simplying interchanging the equations:
\begin{equation*} \begin{alignedat}{3} x & {}+{} & y & {}={} & 2 \\ 0.0001x & {}+{} & y & {}={} & 1\text{.} \\ \end{alignedat} \end{equation*}
Of course, this doesn't change the actual solution. Let's imagine we use our computer to find the solution using Gaussian elimination. Perform the first step where we multiply the first equation by 0.0001 and subtract from the second equation. What does this give for \(y\) if we represent numbers using only three decimal digits?
- Substitute the value you found for \(y\) into the first equation and solve for \(x\text{.}\) Then compare the approximate solution found with our hypothetical computer to the exact solution.
- Which approach produces the most accurate approximation?
This activity demonstrates how the practical aspects of computing differ from the theoretical. We know that the order in which we write the equations has no effect on the solution space; row interchange is one of our three allowed row operations in the Gaussian elimination algorithm. However, when we are only able to perform arithmetic operations approximately, applying row interchanges can dramatically improve the accuracy of our approximations.
If we could compute the solution exactly, we find
Since our hypothetical computer represents numbers using only three decimal digits, our computer finds
This is the best we can hope to do with our computer since it is impossible to represent the solution exactly.
When the equations are written in their original order and we multiply the first equation by 10000 and subtract from the second, we find
In fact, we find the same value for \(y\) when we interchange the equations. Here we multiply the first equation by 0.0001 and subtract from the second equation. We then find
The difference occurs when we substitute \(y\approx 1\) into the first equation. When the equations are written in their original order, we have
When the equations are written in their original order, we find the solution \(x\approx 0.00, y \approx 1.00\text{.}\)
When we write the equation in the opposite order, however, substituting \(y\approx 1\) into the first equation gives
In this case, we find the approximate solution \(x\approx 1.00, y\approx1.00\text{,}\) which is the most accurate solution that our hypothetical computer can find. Simply interchanging the order of the equation produces a much more accurate solution.
We can understand why this works graphically. Each equation represents a line in the plane, and the solution is the intersection point. Notice that the slopes of these lines differ considerably.
When the equations are written in their original order, we substitute \(y\approx1\) into the equation \(0.00001x + y = 1\text{,}\) which is a nearly horizontal line. Along this line, a small change in \(y\) leads to a large change in \(x\text{.}\) The slight difference in our approximation \(y\approx 1\) from the exact value \(y=0.9998999\ldots\) leads to a large difference in the approximation \(x\approx0\) from the exact value \(x=1.00010001\ldots\text{.}\)
If we exchange the order in which the equations are written, we substitute our approximation \(y\approx 1\) into the equation \(x+y=2\text{.}\) Notice that the slope of the associated line is \(-1\text{.}\) On this line, a small change in \(y\) leads to a relatively small change in \(x\) as well. Therefore, the difference in our approximation \(y\approx1\) from the exact value leads to only a small difference in the approximation \(x\approx1\) from the exact value.
This example motivates the technique that computers usually use to perform Gaussian elimation. In practice, we only need to perform a row interchange when a zero occurs in a pivot position, such as
Instead, we will perform a row interchange to put the entry having the largest possible absolute value entry into the pivot position. For instance, when performing Gaussian elimination on the following matrix, we begin by interchanging the first and third rows so that the upper left entry has the largest possible absolute value.
This technique is called partial pivoting , and it means that, in practice, we will perform many more row interchange operations than we typically do when computing exactly by hand.
\(LU\) factorizations
In Subsection 1.3 , we saw that the number of arithmetic operations needed to perform Gaussian elimination on an \(n\times n\) matrix is about \(\frac23 n^3\text{.}\) This means that a \(1000\times1000\) matrix, requires about two thirds of a billion operations.
Suppose that we have two equations, \(A\mathbf x = \mathbf b_1\) and \(A\mathbf x = \mathbf b_2\text{,}\) that we would like to solve. Usually, we would form augmented matrices \(\left[\begin{array}{c|c} A & \mathbf b_1 \\ \end{array}\right]\) and \(\left[\begin{array}{c|c} A & \mathbf b_2 \\ \end{array}\right]\) and apply Gaussian elimination. Of course, the steps we perform in these two computations are nearly identical. Is there a way to store some of the computation we perform in reducing \(\left[\begin{array}{c|c} A & \mathbf b_1 \\ \end{array}\right]\) and reuse it in solving subsequent equations? The next activity will point us in the right direction.
Activity 5.1.3.
We will consider the matrix
and begin performing Gaussian elimination.
-
Perform two row replacement operations to find the row equivalent matrix
\begin{equation*} A' = \left[\begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \\ \end{array}\right]\text{.} \end{equation*}
Find elementary matrices \(E_1\) and \(E_2\) that perform these two operations so that \(E_2E_1 A = A'\text{.}\)
-
Perform a third row replacement to find the upper triangular matrix
\begin{equation*} U = \left[\begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]\text{.} \end{equation*}
Find the elementary matrix \(E_3\) such that \(E_3E_2E_1A = U\text{.}\)
-
We can write \(A=E_1^{-1}E_2^{-1}E_3^{-1} U\text{.}\) Find the inverse matrices \(E_1^{-1}\text{,}\) \(E_2^{-1}\text{,}\) and \(E_3^{-1}\) and the product \(L = E_1^{-1}E_2^{-1}E_3^{-1}\text{.}\) Then verify that \(A=LU\text{.}\)
-
Suppose that we want to solve the equation \(A\mathbf x = \mathbf b = \threevec4{-7}{12}\text{.}\) We will write
\begin{equation*} A\mathbf x = LU\mathbf x = L(U\mathbf x) = \mathbf b \end{equation*}
and introduce an unknown vector \(\mathbf c\) such that \(U\mathbf x = \mathbf c\text{.}\) Find \(\mathbf c\) by noting that \(L\mathbf c = \mathbf b\) and solving this equation.
- Now that we have found \(\mathbf c\text{,}\) find \(\mathbf x\) by solving \(U\mathbf x = \mathbf c\text{.}\)
- Use the factorization \(A=LU\) and this two-step process, solve the equation \(A\mathbf x = \threevec{2}{-2}{7}\text{.}\)
This activity introduces a method for factoring a matrix \(A\) as a product of two triangular matrices, \(A=LU\text{,}\) where \(L\) is lower triangular and \(U\) is upper triangular. The key to finding this factorization is to represent the row operations that we apply in the Gaussian elimination algorithm through multiplication by elementary matrices.
Beginning with the matrix \(A = \left[\begin{array}{rrr} 1 & 2 & 1 \\ -2 & -3 & -2 \\ 3 & 7 & 4 \\ \end{array}\right] \text{,}\) we multiply the first row by \(2\) and add the result to the second row. The result is obtained by multiplying \(A\) by the matrix \(E_1 = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right] \text{.}\) Notice that the inverse of \(E_1\) is \(E_1^{-1} = \left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 0 \\ \end{array}\right] \text{.}\) There is no computation required to find \(E_1^{-1}\text{;}\) to undo the effect of \(E_1\text{,}\) we simply multiply the first row by \(-2\text{,}\) rather than \(2\text{,}\) and add it to the second row. In practice, to find the inverse of an elementary matrix corresponding to a row replacement operation, we simply change the sign of the number below the diagonal. That's easy!
Likewise, the next two row replacements are performed by
From here, we find \(A = E_1^{-1}E_2^{-1}E_3^{-1} U\) so that
If we want to solve the equation \(A\mathbf x = LU\mathbf x = \mathbf b\text{,}\) we first solve for \(U\mathbf x = \mathbf c\) by solving \(L\mathbf c = \mathbf b\text{.}\) Once we have \(\mathbf c\text{,}\) we solve \(U\mathbf x = \mathbf c\) for the vector \(\mathbf x\text{,}\) which is the solution to the original equation \(A\mathbf x = \mathbf b\text{.}\) In our example,
If we want to solve \(A\mathbf x = \mathbf b\) for a different right-hand side \(\mathbf b\text{,}\) we can simply repeat this two-step process.
We have now introduced \(LU\) factorizations as a technique for solving equations \(A\mathbf x = \mathbf b\text{,}\) but we have yet to explain why they are useful. It becomes apparent when we consider the type of equations that we are solving; for instance, \(L\mathbf c = \mathbf b\) in our example has the form
Because \(L\) is a lower-triangular matrix, it contains lots of zeros. In fact, we are able to read off the first component of \(\mathbf c\) directly from the equations: \(c_1 = 4\text{.}\) We then have \(-2c_1+c_2 = -7\text{,}\) which gives \(c_2 = 1\text{,}\) and \(3c_1 + c_2 + c_3 = 12\text{,}\) which gives \(c_3=-1\text{.}\) As this shows, solving equations involving triangular matrices is quite easy because we essentially only need to perform a sequence of substitutions.
In fact, solving an equation with an \(n\times n\) triangular matrix requires approximately \(\frac 12 n^2\) operations. Once we have the factorization \(A=LU\text{,}\) we solve the equation \(A\mathbf x=\mathbf b\) by solving two equations involving triangular matrices, which requires about \(n^2\) operations. For example, if \(A\) is a \(1000\times1000\) matrix, we solve the equation \(A\mathbf x = \mathbf b\) using about one million steps. The compares with the third of a billion operations we need to perform Gaussian elimination. That represents a significant savings. Of course, we have to first find the \(LU\) factorization of \(A\) and this requires roughly the same amount of work as performing Gaussian elimination. However, once we have the \(LU\) factorization, we can use it to solve \(A\mathbf x=\mathbf b\) for different right hand sides \(\mathbf b\text{.}\)
Our discussion so far has ignored one issue, however. Remember that we sometimes have to perform row interchange operations in addition to row replacement. A typical row interchange is represented by multiplication by a matrix such as
which has the effect of interchanging the first and third rows. Notice that this matrix is not triangular so performing a row interchange will ruin the \(LU\) factorization we seek. We will not give the details here, but linear algebra software packages provide a matrix \(P\) that describes how the rows are permuted in the Gaussian elimination process. In particular, we will write \(PA = LU\text{,}\) where \(P\) is a permutation matrix, \(L\) is lower triangular, and \(U\) is upper triangular.
Therefore, to solve the equation \(A\mathbf x = \mathbf b\text{,}\) we first multiply both sides by \(P\) to obtain
That is, we multiply \(\mathbf b\) by \(P\) and then find \(\mathbf x\) using the factorization: \(L\mathbf c = P\mathbf b\) and \(U\mathbf x = \mathbf c\text{.}\)
Activity 5.1.4.
Sage will create \(LU\) factorizations; once we have a matrix
A
, we write
P, L, U = A.LU()
to obtain the matrices \(P\text{,}\) \(L\text{,}\) and \(U\) such that \(PA = LU\text{.}\)
-
In the previous activity, we found the \(LU\) factorization
\begin{equation*} A = \left[\begin{array}{rrr} 1 & 2 & 1 \\ -2 & -3 & -2 \\ 3 & 7 & 4 \\ \end{array}\right] = LU = \left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 3 & 1 & 1 \end{array}\right] \left[\begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]\text{.} \end{equation*}
Using Sage, define the matrix \(A\) and then ask Sage for the \(LU\) factorization. What are the matrices \(P\text{,}\) \(L\text{,}\) and \(U\text{?}\)
Notice that Sage finds a different \(LU\) factorization than we found in the previous activity. This is because Sage uses partial pivoting, as described in the previous section, when it performs Gaussian elimination. This is reflected by the fact that the permutation \(P\) is not the identity.
- Define the vector \(\mathbf b = \threevec4{-7}{12}\) in Sage and compute \(P\mathbf b\text{.}\)
-
Use the matrices
L
andU
to solve \(L\mathbf c = P\mathbf b\) and \(U\mathbf x = \mathbf c\text{.}\) You should find the same solution \(\mathbf x\) that you found in the previous activity. - Use the factorization to solve the equation \(A\mathbf x = \threevec{11}{-17}{35}\text{.}\)
- How does the factorization show us that \(A\) is invertible and that, therefore, every equation \(A\mathbf x=\mathbf b\) has a unique solution?
-
Suppose that we have the matrix
\begin{equation*} A = \left[\begin{array}{rrr} 3 & -1 & 2 \\ 2 & -1 & 1 \\ 2 & 1 & 3 \\ \end{array}\right]\text{.} \end{equation*}
Use Sage to find the \(LU\) factorization. Explain how the factorization shows that \(A\) is not invertible.
-
Consider the matrix
\begin{equation*} A = \left[\begin{array}{rrrr} -2 & 1 & 2 & -1 \\ 1 & -1 & 0 & 2 \\ 3 & 2 & -1 & 0 \\ \end{array}\right] \end{equation*}
and find its \(LU\) factorization. Explain why \(A\) and \(U\) have the same null space and use this observation to find a basis for \(Nul(A)\text{.}\)
Summary
We returned to Gaussian elimination, which we have used as a primary tool for finding solutions to linear systems, and explored its practicality as a computation tool, both in terms of numerical accuracy and computational effort.
- We saw that the accuracy of computations implemented on a computer could be improved using partial pivoting , a technique that performs row interchanges so that the entry in a pivot position has the largest possible magnitude.
- Beginning with a matrix \(A\text{,}\) we used the Gaussian elimination algorithm to write \(PA = LU\text{,}\) where \(P\) is a permutation matrix, \(L\) is lower triangular, and \(U\) is upper triangular.
- Finding this factorization involves roughly as much work as performing Gaussian elimination. However, once we have the factorization, we are able to quickly solve equations of the form \(A\mathbf x = \mathbf b\) by first solving \(L\mathbf c = P\mathbf b\) and then \(U\mathbf x = \mathbf c\text{.}\)
Exercises 5.1.4Exercises
In this section, we saw that errors made in computer arithmetic can produce approximate solutions that are far from the exact solutions. Here is another example in which this can happen. Consider the matrix
- Find the exact solution to the equation \(A\mathbf x = \twovec{2}{2}\text{.}\)
- Suppose that this linear system arises in the midst of a larger computation except that, due to some error in the computation of the right hand side of the equation, our computer thinks we want to solve \(A \mathbf{x}=\left[\begin{array}{c}2 \\ 2.0001\end{array}\right]\). Find the solution to this equation and compare it to the solution of the equation in the previous part of this exericse.
Notice how a small change in the right hand side of the equation leads to a large change in the solution. In this case, we say that the matrix \(A\) is ill-conditioned because the solutions are extremely sensitive to small changes in the right hand side of the equation. Though we will not do so here, it is possible to create a measure of the matrix that tells us when a matrix is ill-conditioned. Regrettably, there is not much we can do to remedy this problem.
In this section, we found the \(LU\) factorization of the matrix
in one of the activities, without using partial pivoting. Apply a sequence of row operations, now using partial pivoting, to find an upper triangular matrix \(U\) that is row equivalent to \(A\text{.}\)
In the following exercises, use the given \(LU\) factorizations to solve the equations \(A\mathbf x = \mathbf b\text{.}\)
-
Solve the equation
\begin{equation*} A\mathbf x = \left[\begin{array}{rr} 1 & 0 \\ -2 & 1 \\ \end{array}\right] \left[\begin{array}{rr} 3 & 1 \\ 0 & -2 \\ \end{array}\right]\mathbf x = \twovec{-3}{0}\text{.} \end{equation*}
-
Solve the equation
\begin{equation*} A\mathbf x = \left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 2 & 1 \\ \end{array}\right] \left[\begin{array}{rrr} 2 & 1 & 0 \\ 0 & -1 & 3 \\ 0 & 0 & 1 \\ \end{array}\right]\mathbf x = \threevec{5}{-5}{7}\text{.} \end{equation*}
Use Sage to solve the following equation by finding an \(LU\) factorization:
Here is another problem with approximate computer arithmetic that we will encounter in the next section. Consider the matrix
- Notice that this is a positive stochastic matrix. What do we know about the eigenvalues of this matrix?
-
Use Sage to define the matrix \(A\) using decimals such as
0.2
and the \(3\times3\) identity matrix \(I\text{.}\) Ask Sage to compute \(B = A-I\) and find the reduced row echelon form of \(B\text{.}\) - Why is the computation that Sage performed incorrect?
- Explain why using a computer to find the eigenvectors of a matrix \(A\) by finding a basis for \(Nul(A-\lambda I)\) is problematic.
In practice, one rarely finds the inverse of a matrix \(A\text{.}\) It requires considerable effort to compute, and we can solve any equation of the form \(A\mathbf x = \mathbf b\) using an \(LU\) factorization, which means that the inverse isn't necessary. In any case, the best way to compute an inverse is using an \(LU\) factorization, as this exericse demonstrates.
-
Suppose that \(PA = LU\text{.}\) Explain why \(A^{-1} = U^{-1}L^{-1}P\text{.}\)
Since \(L\) and \(U\) are triangular, finding their inverses is relatively efficient. That makes this an effective means of finding \(A^{-1}\text{.}\)
-
Consider the matrix
\begin{equation*} A = \left[\begin{array}{rrr} 3 & 4 & -1 \\ 2 & 4 & 1 \\ -3 & 1 & 4 \\ \end{array}\right]\text{.} \end{equation*}
Find the \(LU\) factorization of \(A\) and use it to find \(A^{-1}\text{.}\)
Consider the matrix
- Find the \(LU\) factorization of \(A\text{.}\)
- What conditions on \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) guarantee that \(A\) is invertible?
In the \(LU\) factorization of a matrix, the diagonal entries of \(L\) are all \(1\) while the diagonal entries of \(U\) are not necessarily \(1\text{.}\) This exercise will explore that observation by considering the matrix
- Perform Gaussian elimination without partial pivoting to find \(U\text{,}\) an upper triangular matrix that is row equivalent to \(A\text{.}\)
- The diagonal entries of \(U\) are called pivots . Explain why \(\det A\) equals the product of the pivots.
- What is \(\det A\) for our matrix \(A\text{?}\)
- More generally, if we have \(PA=LU\text{,}\) explain why \(\det A\) equals plus or minus the product of the pivots.
Please provide a justification to your responses to these questions.
- In this section, our hypothetical computer could only store numbers using 3 decimal places. Most computers can store numbers using 15 or more decimal places. Why do we still need to be concerned about the accuracy of our computations when solving systems of linear equations?
- Finding the \(LU\) factorization of a matrix \(A\) is roughly the same amount of work as finding its reduced row echelon form. Why is the \(LU\) factorization useful then?
- How can we detect whether a matrix is invertible from its \(LU\) factorization?
Consider the matrix
- Find the \(LU\) factorization of \(A\text{.}\)
- Use the factorization to find a basis for \(Nul(A)\text{.}\)
- We have seen that \(Nul(A) = Nul(U)\text{.}\) Is it true that \(Col(A) = Col(L)\text{?}\)