13.4: Attitudes and Gender Communication
- Compare and contrast how gender might impact communications styles
Starting in childhood, girls and boys are generally socialized to belong to distinct cultures based on their gender and thus speak in ways particular to their own gender’s rules and norms (Fivush; Hohnson; Tannen). This pattern of gendered socialization continues throughout our lives. As a result, men and women often interpret the same conversation differently. Culturally diverse ways of speaking based on gender can cause miscommunication between members of each culture or speech community. These cultural differences are seen in the simple purpose of communication.
Although gender roles are changing and gender itself is becoming a more fluid concept, traditional roles still influence our communication behaviors. For those socialized to traditional female gender norms, an important use of communication is to create and foster relational connections with other people (Johnson; Stamou). In contrast, the goal of men’s communication is primarily to establish identity. This is accomplished by demonstrating independence and control and entertaining or performing for others.
Deborah Tannen, professor of linguistics and the author of multiple books on gender and language, provides the following examples of differences in men’s and women’s communication:
-
“Men engage in report talk, women in rapport talk.”
- Report talk is used to demonstrate one’s knowledge and expertise.
- Rapport talk is used to share and cultivate relationships.
-
Women request; men direct.
- For example, in communicating a request, a female manager might say: “Could you do this by 5 PM?” A male manager would typically phrase it: “This needs to be done by 5 PM.”
-
Women are information focused; men are image focused.
- For example, women are willing to ask questions to clarify understanding. Men tend to avoid asking clarifying questions in order to preserve their reputation.
-
Empathy is not apology.
- Women often use the phrase “I’m sorry” to express concern or empathy. Men tend to interpret this phrase as an acceptance of responsibility for the situation, which it is not.
- Women are judged by their appearance; men are judged by what they say and do. [1]
As in all things, it’s important to remember that while these differences exist between groups, all individuals will fall somewhere along a spectrum of these tendencies. Additionally, you may run into men who demonstrate more “feminine” tendencies in their speech or vice versa.
Gender in the World
Traditional gender roles also influence how women are heard, as Tannen alluded to above. The Oxford Handbook of Leadership and Organization notes that the historical marginalization of women is still in practice today, with media coverage of women leaders often focusing on fashion sensibility rather than on the strength of their leadership. There is a “Catch-22” for women: “to be ‘too feminine’ is to risk being perceived as weak and emotional or as manipulative and devious when exercising leadership; to be ‘insufficiently feminine’ generally results in being labeled as masculine, abrasive or pushy.” [2]
Thus, gender not only impacts the language we use but the language used to describe us.
Although changing demographics and social trends have begun to erode the base of white male privilege, there are still strong cultural norms that resist this change in the status quo. Additionally, the composition of executive leadership still remains predominantly white male, and organizational culture and communications are largely designed to support that dominance. We see the legacy of that dysfunction in a variety of modes, from pussy hats to the #metoo movement.
We see this struggle playing out at Google, where efforts to include more women in technical roles are meeting with some resistance. The conflict surfaced when James Damore, a white male engineer, posted a ten page critique of Google’s diversity efforts titled “ Google’s Ideological Echo Chamber” on an internal discussion board. One of the most inflammatory points made was that “biological differences between men and women might explain why we don’t see equal representation of women in tech and leadership.” In his memo, Damore states his belief that women are better attuned to aesthetics and people rather than ideas and that this, as well as their “higher agreeableness” (versus aggressiveness) and “neuroticism,” rather than sexism accounts for gender gaps. The “manifesto,” as some call it, resulted in Damore being fired for violating Google’s code of conduct by “advancing harmful gender stereotypes in our workplace.”
Google CEO Sundar Pichai responded to the memo. Read Pichai’s note to employees here. His memo includes this excerpt: “To suggest a group of our colleagues have traits that make them less biologically suited to that work is offensive and not OK. It is contrary to our basic values and our Code of Conduct, which expects ‘each Googler to do their utmost to create a workplace culture that is free of harassment, intimidation, bias and unlawful discrimination.’”
In a development that reflects the nation’s sociopolitical polarization, it appears Damore’s firing, rather than ending the issue, has turned him into what a USA Today writer terms a “hero of a resurgent conservative movement.” Damore has since filed a lawsuit against Google , claiming the search giant discriminates against white, conservative men. In a development worth watching, Damore and David Gudeman, another former Google engineer, are being represented by Harmeet Dhillon, the Republican National Committee’s committeewoman for California. Her law firm is seeking class action status for the plaintiffs.
On the opposite end of the spectrum, Princeton University is an example of an organization that has a clear commitment to inclusivity in both policy and practice. However, it has also met with resistance in moving toward an inclusive campus. In a rather controversial 2015 memo announcing its new communication policy, Princeton drew the distinction between gendered and gender inclusive language, explaining that “gender binary is the traditional view on human gender, which does not take into consideration individuals who identify as otherwise, including and not limited to transgender, genderqueer, gender nonconforming and or intersex.” In contrast, “gender-inclusive language is writing and speaking about people in a manner that does not use gender-based words.” [3]
Some media interpreted the guidelines as an attempt to suppress free speech. Princeton’s clarification: “No words or phrases have been banned at the University, which places a high value (on) free expression.” [4]
Conservative factions also interpreted this statement as an attack. For example, CampusReform.org, a conservative blog, presented the college’s new gender policy as another example of liberal bias and “abuse against conservatives on America’s colleges and universities.” In a post titled “Princeton students can choose any—or every—gender identity” the author, Matthew Penza, closed with a call for donations to support Campus Reform’s “investigative journalism,” stating that “College campuses are no longer bastions of higher learning. Professors indoctrinate students with their agendas. They even silence conservative students with their attempts to suppress free speech.” [5] For perspective, Campus Reform is a project of Leadership Institute , an organization whose mission is to teach conservative Americans how to influence policy through direct participation, activism, and leadership. [6]
- Bucher, Richard D. Diversity Consciousness Opening Our Minds to People, Cultures, and Opportunities . Pearson, 2015, p 130.
- David D. Day, ed., The Oxford Handbook of Leadership and Organization. Oxford: Oxford University Press, 2014. https://books.google.com/books?id=_iqTAwAAQBAJ&pg=PA735&lpg=PA735&dq=gender+bias+language+nancy+pelosi&source=bl&ots=SzNkjbkn7-&sig=kQoD7GfBYJDX_fujtCZXqkV8t2c&hl=en&sa=X&ved=0ahUKEwiGh7qcj8faAhULbK0KHXg7D9kQ6AEIRzAF#v=onepage&q=gender%20bias%20language%20nancy%20pelosi&f=false
- Curran, Philip Sean. "Princeton: University's Gender-Neutral Language Guidelines Come to Light." Centraljersey.com. August 18, 2016. Accessed August 5, 2019. www.centraljersey.com/news/princeton-university-s-gender-neutral-language-guidelines-come-to-light/article_b7567958-657d-11e6-b018-438c69ef92ac.html
- Ibid.
- Penza, Matthew. "Princeton Students Can Choose Any – or Every – Gender Identity." Campus Reform. July 31, 2017. Accessed August 5, 2019. https://www.campusreform.org/?ID=9513
- "About the Leadership Institute." Leadership Institute. Accessed August 5, 2019. https://www.leadershipinstitute.org/aboutus/
Contributors and Attributions
- Attitudes and Gender Communication. Authored by : Nina Burokas. Provided by : Lumen Learning. License : CC BY: Attribution
- M17-115_1cm7933 - . Authored by : Oregon Department of Transportation . Provided by : flickr. License : CC BY-SA: Attribution-ShareAlike
- Survey of Communication Study.. Authored by : Scott T Paynton and Linda K Hahn. Provided by : Humboldt State University. Located at : https://en.wikibooks.org/wiki/Survey_of_Communication_Study/Preface . License : CC BY-SA: Attribution-ShareAlike


22.3: Hindu Art
Architecture of Hindu Temples
Hindu architecture, representing a plurality of beliefs, is commonly divided into the northern Nagara and southern Dravidian styles.
Differentiate between the northen Nagara and southern Dravidian styles of Hindu temple architecture
Key Points
- Hindu art represents a plurality of beliefs and has deeply influenced the painting, sculpture, and architecture of the Indian subcontinent. Architecture and sculpture are inextricably linked in Hindu temples, which are usually devoted to a number of different deities .
- A Hindu temple generally consists of an inner sanctum, in which the idol of the deity is housed; a congregation hall; and sometimes an antechamber and porch.
- Two main styles of temples exist in India: the north Indian Nagara style, characterized by a beehive shaped central tower, and the south Indian Dravidia style, characterized by a graduated tower with multiple layered pavilions.
- The period between the 6th and 12th centuries was marked by the appearance of a large number of Hindu states and was a productive and creative period for Hindu temple architecture.
Key Terms
- shikhara : The tower surmounting a Hindu temple.
- om : A sacred, mystical syllable used to invoke God in Hindu prayer and meditation.
Introduction: Hinduism and Hindu Art
Hinduism is the predominant religion of the Indian subcontinent. Dating back to the Iron Age , it is often called the oldest living religion in the world. Hinduism has no single founder and is a conglomeration of diverse traditions and philosophies rather than a rigid set of beliefs. Most Hindus believe in a single supreme God who appears in many different manifestations as devas (celestial beings or deities), and they may worship specific devas as individual facets of the same God.
Hindu art reflects this plurality of beliefs, and Hindu temples, in which architecture and sculpture are inextricably connected, are usually devoted to different deities. Deities commonly worshiped include Shiva the Destroyer; Vishnu in his incarnations as Rama and Krishna; Ganesha, the elephant god of prosperity; and different forms of the goddess Shakti (literally meaning “power”), the primordial feminine creative principle. These deities are often portrayed with multiple limbs and heads, demonstrating the extent of the god’s power and ability. Hindu art is also characterized by a number of recurring holy symbols, including the om , an invocation of the divine consciousness of God; the swastika, a symbol of auspiciousness; and the lotus flower, a symbol of purity, beauty, fertility, and transcendence.
Hindu Temples
A Hindu temple generally consists of a garba griha (“womb chamber”), the inner sanctum in which the murti , or idol of the deity, is housed; a congregation hall; and sometimes an antechamber and porch. The garba griha is surmounted by a shikhara , or tower. Two main styles of temples exist in India: the northern Nagara style and the southern Dravida style.
The Nagara Style
In the northern Nagara style, the shikhara takes the shape of a curvilinear beehive. The temple is a square with a number of graduated projections in the middle of each side, giving a cruciform shape with a number of re-entrant angles on each side. The projections in the plan are also carried upwards to the top of the shikhara, giving a strong emphasis on vertical lines in elevation .
The Dravidian Style
In the southern Dravidian style, the tower or gopuram consists of progressive smaller stories of pavilions. These temples were square in plan and pyramidal in shape; included porches ( mandapams) and pillard halls ( chaultris or chawadis); and contained tanks or wells for water to be used for sacred purposes or the convenience of the priests.
History and Well-Known Temples
The earliest Hindu temples found in India date back to the Gupta period (ca. 320–550 CE); one of these is the Dashavatara Vishnu Temple in Deogarh in central India, built ca. 500 CE. The period between the 6th and 12th centuries was marked by the appearance of a large number of states, most of which were ruled by Hindu dynasties . This was a deeply productive and creative period for Hindu temple architecture, and many beautiful examples survive to the present day. Some surviving works include the monumental , rock-cut Kailashnath Temple (754–774 CE) dedicated to Shiva at Ellora in the western state of Maharashtra; the 11th century Brihadeeswarar Temple in Tanjore in the southern state of Tamil Nadu, which is India’s largest temple; and the Sun Temple (1238–1250 CE) at Konarak in Orissa. Although many Hindu temples were destroyed during the period of Muslim rule in India (12th to 18th centuries), Hindu influence on Indian art and architecture has withstood the test of time and continues to shape works of art.
Hindu Sculpture
Hindu sculpture represents the themes of its religion through its depiction of deities and recurring symbols, such as the lotus flower.
Evaluate the religious themes that contribute to an understanding of Hindu sculpture
Key Points
- Most Hindus believe in a single supreme God who appears in many different manifestations as devas, or celestial beings or deities ; Hindu sculpture reflects this plurality of beliefs.
- Because religion and culture are inseparable with Hinduism , recurring symbols such as the gods and their reincarnations, the lotus flower, and extra limbs make their appearances in many sculptures of Hindu origin.
- Deities are often portrayed with multiple limbs and heads, demonstrating the extent of the God’s power and ability.
- Hindu sculpture is characterized by recurring holy symbols such as the om , an invocation of the divine consciousness of God; the swastika, a symbol of auspiciousness; and the lotus flower, a symbol of purity, beauty, fertility, and transcendence.
- Sculpture is inextricably linked with architecture in Hindu temples, which are usually devoted to a number of different deities.
Key Terms
- Hinduism : A religion or a way of life found most notably in India and Nepal; with over one billion followers, it is the world’s third largest religion by population.
- dharma : A key concept with multiple meanings in the Indian religions Hinduism, Buddhism, Sikhism, and Jainism; in Hinduism, it signifies behaviors that are considered to be in accord with the order that makes life and universe possible, including duties, rights, laws, conduct, virtues, and “right way of living.”
Themes of Hindu Art and Sculpture
Hinduism is a conglomeration of diverse traditions and philosophies rather than a rigid set of beliefs. Most Hindus believe in a single supreme God who appears in many different manifestations as devas (celestial beings or deities), and they may worship specific devas as individual facets of the same God. Hindu sculpture, as seen in other forms of Hindu art, reflects this plurality of beliefs. Because religion and culture are inseparable with Hinduism, recurring symbols such as the gods and their reincarnations, the lotus flower, extra limbs, and even the traditional arts make their appearances in many sculptures of Hindu origin.
Depictions of Deities
Deities commonly worshiped and portrayed through sculpture include Shiva the Destroyer; Vishnu in his incarnations as Rama and Krishna; Ganesha, the elephant god of prosperity; and different forms of the goddess Shakti (literally meaning “power”), the primordial feminine creative principle. These deities are often portrayed with multiple limbs and heads, demonstrating the extent of the god’s power and ability. For example, the goddess Sarasvati is always depicted with a minimum of four arms: two of the arms will be playing a vina, representing the tuning of her knowledge; her other two hands often hold prayer beads and a scripture, both of which represent her devotion to her spirituality. As the goddess of learning and art, she is depicted in this way as very capable and powerful in her area of expertise.
Symbols in Hindu Sculpture
Hindu sculpture is also characterized by a number of recurring holy symbols, including the om , an invocation of the divine consciousness of God; the swastika, a symbol of auspiciousness; and the lotus flower, a symbol of purity, beauty, prosperity, fertility, and transcendence. The lotus flower is associated with these attributes due to its own process of blossoming: the flower grows out of mud and rests atop the water in which it grows, illustrating the hardship it must endure in order to achieve maximum beauty. Many deities have their name based on the Sanskrit word for lotus, such as Lakshimi. In addition to these symbols, flowers, birds, animals, instruments, symmetric mandala drawings, objects, and idols are all part of symbolic iconography in Hinduism.
Connection to Architecture
Sculpture is inextricably linked with architecture in Hindu temples, which are usually devoted to a number of different deities. The Hindu temple style reflects a synthesis of arts, the ideals of dharma , beliefs, values , and the way of life cherished under Hinduism.
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15.7D: The Posterior Pituitary
The posterior pituitary secretes two important endocrine hormones—oxytocin and antidiuretic hormone.
- Identify the location of the posterior pituitary and the hormones associated with it
Key Points
- The posterior pituitary (or neurohypophysis) comprises the posterior lobe of the pituitary gland and is part of the endocrine system.
- Hormones known as posterior pituitary hormones are synthesized by the hypothalamus, and include oxytocin and antidiuretic hormone.
- The hormones are then stored in neurosecretory vesicles (Herring bodies) before being secreted by the posterior pituitary into the bloodstream.
Key Terms
- oxytocin : A hormone that stimulates contractions during labor.
- posterior pituitary : The posterior pituitary (or neurohypophysis) comprises the posterior lobe of the pituitary gland and is part of the endocrine system. Despite its name, the posterior pituitary gland is not a true gland; rather, it is largely a collection of axonal projections from the hypothalamus that terminate behind the anterior pituitary gland.
- Antidiuretic hormone : A hormone that stimulates water re-absorption in the kidneys.
Posterior Pituitary Gland
Pituitary : Pituitary gland representation.
The posterior pituitary (or neurohypophysis) comprises the posterior lobe of the pituitary gland and is part of the endocrine system. Despite its name, the posterior pituitary gland is not a gland; rather, it is largely a collection of axonal projections from the hypothalamus that terminate behind the anterior pituitary gland.
The posterior pituitary consists mainly of neuronal projections ( axons ) extending from the supraoptic and paraventricular nuclei of the hypothalamus. These axons release peptide hormones into the capillaries of the hypophyseal circulation. These are then stored in neurosecretory vesicles (Herring bodies) before being secreted by the posterior pituitary into the systemic bloodstream.
Anatomy of the Posterior Pituitary Gland
The posterior pituitary is derived from the hypothalamus and is distinct from the more fleshy, vascularized anterior lobe. The posterior pituitary is composed of two parts:
- The pars nervosa, also called the neural lobe or posterior lobe, constitutes the majority of the posterior pituitary and is the storage site of oxytocin and vasopressin.
- The infundibular stalk, also known as the infundibulum or pituitary stalk, bridges the hypothalamic and hypophyseal systems.
Major Hormones Secreted by the Posterior Pituitary Gland
The posterior pituitary stores two hormones secreted by the hypothalamus for later release:
- Oxytocin, most of which is released from the paraventricular nucleus in the hypothalamus. Oxytocin is one of the few hormones that create a positive feedback loop.
- Antidiuretic hormone (ADH, also known as vasopressin), the majority of which is released from the supraoptic nucleus in the hypothalamus. ADH acts on the collecting ducts of the kidney to facilitate the reabsorption of water into the blood.
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2.3: Spherical Mirrors
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Learning Objectives
By the end of this section, you will be able to:
- Describe image formation by spherical mirrors.
- Use ray diagrams and the mirror equation to calculate the properties of an image in a spherical mirror.
The image in a plane mirror has the same size as the object, is upright, and is the same distance behind the mirror as the object is in front of the mirror. A curved mirror, on the other hand, can form images that may be larger or smaller than the object and may form either in front of the mirror or behind it. In general, any curved surface will form an image, although some images make be so distorted as to be unrecognizable (think of fun house mirrors). Because curved mirrors can create such a rich variety of images, they are used in many optical devices that find many uses. We will concentrate on spherical mirrors for the most part, because they are easier to manufacture than mirrors such as parabolic mirrors and so are more common.
Curved Mirrors
We can define two general types of spherical mirrors. If the reflecting surface is the outer side of the sphere, the mirror is called a convex mirror . If the inside surface is the reflecting surface, it is called a concave mirror .
Symmetry is one of the major hallmarks of many optical devices, including mirrors and lenses. The symmetry axis of such optical elements is often called the principal axis or optical axis. For a spherical mirror, the optical axis passes through the mirror’s center of curvature and the mirror’s vertex, as shown in Figure \(\PageIndex{1}\).
Consider rays that are parallel to the optical axis of a parabolic mirror, as shown in Figure \(\PageIndex{2a}\). Following the law of reflection, these rays are reflected so that they converge at a point, called the focal point . Figure \(\PageIndex{2b}\) shows a spherical mirror that is large compared with its radius of curvature. For this mirror, the reflected rays do not cross at the same point, so the mirror does not have a well-defined focal point. This is called spherical aberration and results in a blurred image of an extended object. Figure \(\PageIndex{2c}\) shows a spherical mirror that is small compared to its radius of curvature. This mirror is a good approximation of a parabolic mirror, so rays that arrive parallel to the optical axis are reflected to a well-defined focal point. The distance along the optical axis from the mirror to the focal point is called the focal length of the mirror.
A convex spherical mirror also has a focal point, as shown in Figure \(\PageIndex{3}\). Incident rays parallel to the optical axis are reflected from the mirror and seem to originate from point \(F\) at focal length \(f\) behind the mirror. Thus, the focal point is virtual because no real rays actually pass through it; they only appear to originate from it.
How does the focal length of a mirror relate to the mirror’s radius of curvature? Figure \(\PageIndex{4}\) shows a single ray that is reflected by a spherical concave mirror. The incident ray is parallel to the optical axis. The point at which the reflected ray crosses the optical axis is the focal point. Note that all incident rays that are parallel to the optical axis are reflected through the focal point—we only show one ray for simplicity. We want to find how the focal length \(FP\) (denoted by \(f\)) relates to the radius of curvature of the mirror, \(R\), whose length is
\[R=CF+FP. \label{eq31} \]
The law of reflection tells us that angles \(\angle OXC\) and \(\angle CXF\) are the same, and because the incident ray is parallel to the optical axis, angles \(\angle OXC\) and \(\angle XCP\) are also the same. Thus, triangle \(CXF\) is an isosceles triangle with \(CF=FX\). If the angle \(θ\) is small then
\[\sin θ≈ θ \label{sma} \]
which is called the “ small-angle approximation ”), then \(FX≈FP\) or \(CF≈FP\). Inserting this into Equation \ref{eq31} for the radius \(R\), we get
\[\begin{align} R &=CF+FP \nonumber \\[4pt] &=FP+FP \nonumber \\[4pt] &=2FP\nonumber \\[4pt] &=2f \end{align} \nonumber \]
In other words, in the small-angle approximation, the focal length \(f\) of a concave spherical mirror is half of its radius of curvature, \(R\):
\[f=\dfrac{R}{2}. \nonumber \]
In this chapter, we assume that the small-angle approximation (also called the paraxial approximation ) is always valid. In this approximation, all rays are paraxial rays, which means that they make a small angle with the optical axis and are at a distance much less than the radius of curvature from the optical axis. In this case, their angles \(θ\) of reflection are small angles, so
\[\sin θ≈ \tan θ≈ θ. \label{smallangle} \]
Using Ray Tracing to Locate Images
To find the location of an image formed by a spherical mirror, we first use ray tracing, which is the technique of drawing rays and using the law of reflection to determine the reflected rays (later, for lenses, we use the law of refraction to determine refracted rays). Combined with some basic geometry, we can use ray tracing to find the focal point, the image location, and other information about how a mirror manipulates light. In fact, we already used ray tracing above to locate the focal point of spherical mirrors, or the image distance of flat mirrors. To locate the image of an object, you must locate at least two points of the image. Locating each point requires drawing at least two rays from a point on the object and constructing their reflected rays. The point at which the reflected rays intersect, either in real space or in virtual space, is where the corresponding point of the image is located. To make ray tracing easier, we concentrate on four “principal” rays whose reflections are easy to construct.
Figure \(\PageIndex{5}\) shows a concave mirror and a convex mirror, each with an arrow-shaped object in front of it. These are the objects whose images we want to locate by ray tracing. To do so, we draw rays from point \(Q\) that is on the object but not on the optical axis. We choose to draw our ray from the tip of the object. Principal ray 1 goes from point \(Q\) and travels parallel to the optical axis. The reflection of this ray must pass through the focal point, as discussed above. Thus, for the concave mirror, the reflection of principal ray 1 goes through focal point \(F\), as shown in Figure \(\PageIndex{5b}\). For the convex mirror, the backward extension of the reflection of principal ray 1 goes through the focal point (i.e., a virtual focus). Principal ray 2 travels first on the line going through the focal point and then is reflected back along a line parallel to the optical axis. Principal ray 3 travels toward the center of curvature of the mirror, so it strikes the mirror at normal incidence and is reflected back along the line from which it came. Finally, principal ray 4 strikes the vertex of the mirror and is reflected symmetrically about the optical axis.
The four principal rays intersect at point \(Q′\), which is where the image of point \(Q\) is located. To locate point \(Q′\), drawing any two of these principle rays would suffice. We are thus free to choose whichever of the principal rays we desire to locate the image. Drawing more than two principal rays is sometimes useful to verify that the ray tracing is correct.
To completely locate the extended image, we need to locate a second point in the image, so that we know how the image is oriented. To do this, we trace the principal rays from the base of the object. In this case, all four principal rays run along the optical axis, reflect from the mirror, and then run back along the optical axis. The difficulty is that, because these rays are collinear, we cannot determine a unique point where they intersect. All we know is that the base of the image is on the optical axis. However, because the mirror is symmetrical from top to bottom, it does not change the vertical orientation of the object. Thus, because the object is vertical, the image must be vertical. Therefore, the image of the base of the object is on the optical axis directly above the image of the tip, as drawn in the figure.
For the concave mirror, the extended image in this case forms between the focal point and the center of curvature of the mirror. It is inverted with respect to the object, is a real image, and is smaller than the object. Were we to move the object closer to or farther from the mirror, the characteristics of the image would change. For example, we show, as a later exercise, that an object placed between a concave mirror and its focal point leads to a virtual image that is upright and larger than the object. For the convex mirror, the extended image forms between the focal point and the mirror. It is upright with respect to the object, is a virtual image, and is smaller than the object.
Ray-Tracing Rules
Ray tracing is very useful for mirrors. The rules for ray tracing are summarized here for reference:
- A ray traveling parallel to the optical axis of a spherical mirror is reflected along a line that goes through the focal point of the mirror (ray 1 in Figure \(\PageIndex{5}\)).
- A ray traveling along a line that goes through the focal point of a spherical mirror is reflected along a line parallel to the optical axis of the mirror (ray 2 in Figure \(\PageIndex{5}\)).
- A ray traveling along a line that goes through the center of curvature of a spherical mirror is reflected back along the same line (ray 3 in Figure \(\PageIndex{5}\)).
- A ray that strikes the vertex of a spherical mirror is reflected symmetrically about the optical axis of the mirror (ray 4 in Figure \(\PageIndex{5}\)).
We use ray tracing to illustrate how images are formed by mirrors and to obtain numerical information about optical properties of the mirror. If we assume that a mirror is small compared with its radius of curvature, we can also use algebra and geometry to derive a mirror equation, which we do in the next section. Combining ray tracing with the mirror equation is a good way to analyze mirror systems.
Image Formation by Reflection—The Mirror Equation
For a plane mirror, we showed that the image formed has the same height and orientation as the object, and it is located at the same distance behind the mirror as the object is in front of the mirror. Although the situation is a bit more complicated for curved mirrors, using geometry leads to simple formulas relating the object and image distances to the focal lengths of concave and convex mirrors.
Consider the object \(OP\) shown in Figure \(\PageIndex{6}\). The center of curvature of the mirror is labeled \(C\) and is a distance \(R\) from the vertex of the mirror, as marked in the figure. The object and image distances are labeled \(d_o\) and \(d_i\), and the object and image heights are labeled \(h_o\) and \(h_i\), respectively. Because the angles \(ϕ\) and \(ϕ′\) are alternate interior angles, we know that they have the same magnitude. However, they must differ in sign if we measure angles from the optical axis, so \(ϕ=−ϕ′\). An analogous scenario holds for the angles \(θ\) and \(θ′\). The law of reflection tells us that they have the same magnitude, but their signs must differ if we measure angles from the optical axis. Thus, \(θ=−θ′\). Taking the tangent of the angles \(θ\) and \(θ′\), and using the property that \(\tan(−θ)=−\tan θ\), gives us
\[\left. \begin{array}{rcl} \tanθ=\dfrac{h_o}{d_o} \\ \tanθ′=−\tanθ=\dfrac{h_i}{d_i} \end{array}\right\} =\dfrac{h_o}{d_o}=−\dfrac{h_i}{d_i} \label{eq51} \]
or
\[- \dfrac{h_o}{h_i}=\dfrac{d_o}{d_i}.\label{eq52} \]
Similarly, taking the tangent of \(ϕ\) and \(ϕ′\) gives
\[\left. \begin{array}{rcl} \tanϕ=\dfrac{h_o}{d_o-R} \\ \tanϕ′=−\tanϕ=\dfrac{h_i}{R-d_i} \end{array}\right\} =\dfrac{h_o}{d_o-R}=−\dfrac{h_i}{R-d_i} \nonumber \]
or
\[−\dfrac{h_o}{h_i}=\dfrac{d_o-R}{R-d_i}. \label{eq55} \]
Combining Equation \ref{eq51} and \ref{eq55} gives
\[\dfrac{d_o}{d_i}=\dfrac{d_o−R}{R−d_i}. \nonumber \]
After a little algebra, this becomes
\[\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{2}{R}. \label{eq57} \]
No approximation is required for this result, so it is exact. However, as discussed above, in the small-angle approximation, the focal length of a spherical mirror is one-half the radius of curvature of the mirror, or \(f=R/2\). Inserting this into Equation \ref{eq57} gives the mirror equation :
\[\underbrace{ \dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f}}_{\text{mirror equation}}. \label{mirror equation} \]
The mirror equation relates the image and object distances to the focal distance and is valid only in the small-angle approximation (Equation \ref{sma}). Although it was derived for a concave mirror, it also holds for convex mirrors (proving this is left as an exercise). We can extend the mirror equation to the case of a plane mirror by noting that a plane mirror has an infinite radius of curvature. This means the focal point is at infinity, so the mirror equation simplifies to
\[d_o=−d_i \nonumber \]
which is the same equation obtained earlier .
Notice that we have been very careful with the signs in deriving the mirror equation. For a plane mirror, the image distance has the opposite sign of the object distance. Also, the real image formed by the concave mirror in Figure \(\PageIndex{6}\) is on the opposite side of the optical axis with respect to the object. In this case, the image height should have the opposite sign of the object height. To keep track of the signs of the various quantities in the mirror equation, we now introduce a sign convention.
Sign convention for spherical mirrors
Using a consistent sign convention is very important in geometric optics. It assigns positive or negative values for the quantities that characterize an optical system. Understanding the sign convention allows you to describe an image without constructing a ray diagram. This text uses the following sign convention:
- The focal length \(f\) is positive for concave mirrors and negative for convex mirrors.
- The image distance \(d_i\) is positive for real images and negative for virtual images.
Notice that rule 1 means that the radius of curvature of a spherical mirror can be positive or negative. What does it mean to have a negative radius of curvature? This means simply that the radius of curvature for a convex mirror is defined to be negative.
Image Magnification
Let’s use the sign convention to further interpret the derivation of the mirror equation. In deriving this equation, we found that the object and image heights are related by
\[−\dfrac{h_o}{h_i}=\dfrac{d_o}{d_i}. \label{eq61} \]
See Equation \ref{eq52}. Both the object and the image formed by the mirror in Figure \(\PageIndex{6}\) are real, so the object and image distances are both positive. The highest point of the object is above the optical axis, so the object height is positive. The image, however, is below the optical axis, so the image height is negative. Thus, this sign convention is consistent with our derivation of the mirror equation.
Equation \ref{eq61} in fact describes the linear magnification (often simply called “ magnification ”) of the image in terms of the object and image distances. We thus define the dimensionless magnification \(m\) as follows:
\[\underbrace{m=\dfrac{h_i}{h_o}}_{\text{linear magnification}}. \label{mag} \]
If \(m\) is positive, the image is upright, and if \(m\) is negative, the image is inverted. If \(|m|>1\), the image is larger than the object, and if \(|m|<1\), the image is smaller than the object. With this definition of magnification, we get the following relation between the vertical and horizontal object and image distances:
\[m=\dfrac{h_i}{h_o}=−\dfrac{d_i}{d_o}. \nonumber \]
This is a very useful relation because it lets you obtain the magnification of the image from the object and image distances, which you can obtain from the mirror equation.
Example \(\PageIndex{1}\): Solar Electric Generating System
One of the solar technologies used today for generating electricity involves a device (called a parabolic trough or concentrating collector) that concentrates sunlight onto a blackened pipe that contains a fluid. This heated fluid is pumped to a heat exchanger, where the thermal energy is transferred to another system that is used to generate steam and eventually generates electricity through a conventional steam cycle. Figure \(\PageIndex{7}\) shows such a working system in southern California. The real mirror is a parabolic cylinder with its focus located at the pipe; however, we can approximate the mirror as exactly one-quarter of a circular cylinder.
- If we want the rays from the sun to focus at 40.0 cm from the mirror, what is the radius of the mirror?
- What is the amount of sunlight concentrated onto the pipe, per meter of pipe length, assuming the insolation (incident solar radiation) is 900 W/m 2 ?
- If the fluid-carrying pipe has a 2.00-cm diameter, what is the temperature increase of the fluid per meter of pipe over a period of 1 minute? Assume that all solar radiation incident on the reflector is absorbed by the pipe, and that the fluid is mineral oil.
Strategy
First identify the physical principles involved. Part (a) is related to the optics of spherical mirrors. Part (b) involves a little math, primarily geometry. Part (c) requires an understanding of heat and density.
Solution
a. The sun is the object, so the object distance is essentially infinity: \(d_o=\infty\). The desired image distance is \(d_i=40.0\,cm\). We use the mirror equation (Equation \ref{mirror equation}) to find the focal length of the mirror:
\[\begin{align*} \dfrac{1}{d_o}+\dfrac{1}{d_i} &=\dfrac{1}{f} \nonumber \\[4pt] f &= \left(\dfrac{1}{d_o}+\dfrac{1}{d_i}\right)^{−1} \\[4pt] &= \left(\dfrac{1}{\infty}+\dfrac{1}{40.0\,cm}\right)^{−1} \\[4pt] &= 40.0 \,cm \end{align*} \nonumber \]
Thus, the radius of the mirror is
\[R=2f=80.0\,cm. \nonumber \]
b. The insolation is 900 W/m 2 . You must find the cross-sectional area \(A\) of the concave mirror, since the power delivered is \(900\, W/m^2×A\). The mirror in this case is a quarter-section of a cylinder, so the area for a length \(L\) of the mirror is \(A=\frac{1}{4}(2πR)L\). The area for a length of 1.00 m is then
\[\begin{align*} A&=\dfrac{\pi}{2}R(1.00m) \\[4pt] &=\dfrac{(3.14)}{2}(0.800\,m)(1.00\,m) \\[4pt] &=1.26\,m^2. \end{align*} \nonumber \]
The insolation on the 1.00-m length of pipe is then
\[(9.00×10^2\dfrac{W}{m^2})(1.26\,m^2)=1130\,W. \nonumber \]
c. The increase in temperature is given by \(Q=mcΔT\). The mass \(m\) of the mineral oil in the one-meter section of pipe is
\[ \begin{align*} m &= ρV = ρπ\left(\dfrac{d}{2}\right)^2(1.00\,m) \nonumber \\[4pt] &=(8.00×10^2kg/m^3)(3.14)(0.0100\,m)^2(1.00\,m) \nonumber \\[4pt] &=0.251\,kg \end{align*} \nonumber \]
Therefore, the increase in temperature in one minute is
\[ \begin{align*} \Delta T&= \dfrac{Q}{mc} \nonumber \\[4pt] &=\dfrac{(1130\,W)(60.0\,s)}{(0.251\,kg)(1670\,J⋅kg/°C)} \nonumber \\[4pt] &=162°\end{align*} \nonumber \]
Significance
An array of such pipes in the California desert can provide a thermal output of 250 MW on a sunny day, with fluids reaching temperatures as high as 400°C. We are considering only one meter of pipe here and ignoring heat losses along the pipe.
Example \(\PageIndex{2}\): Image in a Convex Mirror
A keratometer is a device used to measure the curvature of the cornea of the eye, particularly for fitting contact lenses. Light is reflected from the cornea, which acts like a convex mirror, and the keratometer measures the magnification of the image. The smaller the magnification, the smaller the radius of curvature of the cornea. If the light source is 12 cm from the cornea and the image magnification is 0.032, what is the radius of curvature of the cornea?
Strategy
If you find the focal length of the convex mirror formed by the cornea, then you know its radius of curvature (it’s twice the focal length). The object distance is d o =12cm and the magnification is m=0.032. First find the image distance \(d_i\) and then solve for the focal length \(f\).
Solution
Start with the equation for magnification (Equation \ref{mag}) and solving for \(d_i\) and inserting the given values yields
\[d_i=−m d_o=−(0.032)(12\,cm)=−0.384\,cm \nonumber \]
where we retained an extra significant figure because this is an intermediate step in the calculation. Solve the mirror equation for the focal length \(f\) and insert the known values for the object and image distances. The result is
\[ \begin{align*} \dfrac{1}{d_o}+\dfrac{1}{d_i} &= \dfrac{1}{f} \\[4pt] f &= \left(\dfrac{1}{d_o}+\dfrac{1}{d_i}\right)^{−1} \\[4pt] &= \left(\dfrac{1}{12cm}+\dfrac{1}{-0.384cm}\right)^{−1} \\[4pt] &=-40.0 \,cm \end{align*} \nonumber \]
The radius of curvature is twice the focal length, so
\[R=2f=−0.80\,cm \nonumber \]
Significance
The focal length is negative, so the focus is virtual, as expected for a concave mirror and a real object. The radius of curvature found here is reasonable for a cornea. The distance from cornea to retina in an adult eye is about 2.0 cm. In practice, corneas may not be spherical, which complicates the job of fitting contact lenses. Note that the image distance here is negative, consistent with the fact that the image is behind the mirror. Thus, the image is virtual because no rays actually pass through it. In the problems and exercises, you will show that, for a fixed object distance, a smaller radius of curvature corresponds to a smaller the magnification.
PROBLEM-SOLVING STRATEGY: SPHERICAL MIRRORS
- Step 1. First make sure that image formation by a spherical mirror is involved.
- Step 2. Determine whether ray tracing, the mirror equation, or both are required. A sketch is very useful even if ray tracing is not specifically required by the problem. Write symbols and known values on the sketch.
- Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns).
- Step 4. Make a list of what is given or can be inferred from the problem as stated (identify the knowns).
- Step 5. If ray tracing is required, use the ray-tracing rules listed near the beginning of this section.
- Step 6. Most quantitative problems require using the mirror equation. Use the examples as guides for using the mirror equation.
- Step 7. Check to see whether the answer makes sense. Do the signs of object distance, image distance, and focal length correspond with what is expected from ray tracing? Is the sign of the magnification correct? Are the object and image distances reasonable?
Departure from the Small-Angle Approximation
The small-angle approximation (Equation \ref{smallangle}) is a cornerstone of the above discussion of image formation by a spherical mirror. When this approximation is violated, then the image created by a spherical mirror becomes distorted. Such distortion is called aberration. Here we briefly discuss two specific types of aberrations: spherical aberration and coma.
Spherical aberration
Consider a broad beam of parallel rays impinging on a spherical mirror, as shown in Figure \(\PageIndex{8}\). The farther from the optical axis the rays strike, the worse the spherical mirror approximates a parabolic mirror. Thus, these rays are not focused at the same point as rays that are near the optical axis, as shown in the figure. Because of spherical aberration, the image of an extended object in a spherical mirror will be blurred. Spherical aberrations are characteristic of the mirrors and lenses that we consider in the following section of this chapter (more sophisticated mirrors and lenses are needed to eliminate spherical aberrations).
Coma or Comatic Aberration
Coma is similar to spherical aberration, but arises when the incoming rays are not parallel to the optical axis, as shown in Figure \(\PageIndex{8b}\). Recall that the small-angle approximation holds for spherical mirrors that are small compared to their radius. In this case, spherical mirrors are good approximations of parabolic mirrors. Parabolic mirrors focus all rays that are parallel to the optical axis at the focal point. However, parallel rays that are not parallel to the optical axis are focused at different heights and at different focal lengths, as show in Figure \(\PageIndex{8b}\). Because a spherical mirror is symmetric about the optical axis, the various colored rays in this figure create circles of the corresponding color on the focal plane.
Although a spherical mirror is shown in Figure \(\PageIndex{8b}\), comatic aberration occurs also for parabolic mirrors—it does not result from a breakdown in the small-angle approximation (Equation \ref{smallangle}). Spherical aberration, however, occurs only for spherical mirrors and is a result of a breakdown in the small-angle approximation. We will discuss both coma and spherical aberration later in this chapter, in connection with telescopes.


17.7: Sage
Sage is particularly adept at building, analyzing and manipulating polynomial rings. We have seen some of this in the previous chapter. Let's begin by creating three polynomial rings and checking some of their basic properties. There are several ways to construct polynomial rings, but the syntax used here is the most straightforward.
Polynomial Rings and their Elements
Basic properties of rings are availble for these examples.
With the construction syntax used above, the variables can be used to create elements of the polynomial ring without explicit coercion (though we need to be careful about constant polynomials).
Polynomials can be evaluated like they are functions, so we can mimic the evaluation homomorphism.
Notice that
p
is a degree two polynomial, yet through a brute-force examination we see that the polynomial only has one root, contrary to our usual expectations. It can be even more unusual.
Sage can create and manipulate rings of polynomials in more than one variable, though we will not have much occasion to use this functionality in this course.
Irreducible Polynomials
Whether or not a polynomial factors, taking into consideration the ring used for its coefficients, is an important topic in this chapter and many of the following chapters. Sage can factor, and determine irreducibility, over the integers, the rationals, and finite fields.
First, over the rationals.
Factoring over the integers is really no different than factoring over the rationals. This is the content of Theorem \(17.14\) — finding a factorization over the integers can be converted to finding a factorization over the rationals. So it is with Sage, there is little difference between working over the rationals and the integers. It is a little different working over a finite field. Commentary follows.
To check these factorizations, we need to compute in the finite field,
F
, and so we need to know how the symbol
a
behaves. This symbol is considered as a root of a degree two polynomial over the integers mod 5, which we can get with the
.modulus()
method.
So \(a^2+4a+2=0\text{,}\) or \(a^2=-4a-3=a+2\text{.}\) So when checking the factorizations, anytime you see an \(a^2\) you can replace it by \(a+2\text{.}\) Notice that by Corollary \(17.8\) we could find the one linear factor of
r
, and the four linear factors of
s
, through a brute-force search for roots. This is feasible because the field is finite.
However,
q
factors into a pair of degree 2 polynomials, so no amount of testing for roots will discover a factor.
With Eisenstein's Criterion, we can create irreducible polynomials, such as in Example \(17.18\).
Over the field \({\mathbb Z}_p\text{,}\) the field of integers mod a prime \(p\text{,}\) Conway polynomials are canonical choices of a polynomial of degree \(n\) that is irreducible over \({\mathbb Z}_p\text{.}\) See the exercises for more about these polynomials.
Polynomials over Fields
If \(F\) is a field, then every ideal of \(F[x]\) is principal (Theorem \(17.20\)). Nothing stops you from giving Sage two (or more) generators to construct an ideal, but Sage will determine the element to use in a description of the ideal as a principal ideal.
Theorem \(17.22\) is the key fact that allows us to easily construct finite fields. Here is a construction of a finite field of order \(7^5=16\,807\text{.}\) All we need is a polynomial of degree \(5\) that is irreducible over \({\mathbb Z}_7\text{.}\)
The symbol
xbar
is a generator of the field, but right now it is not accessible.
xbar
is the coset \(x + \langle x^5+ x + 4\rangle\text{.}\) A better construction would include specifying this generator.


10.3: Dichotomous Keys
Diagnostics is a practical science which helps to determine living organisms. One of the best way of determining was invented in the end of 18 century by famous French naturalist, Jean-Baptiste Lamarck. He created the dichotomous key (sometimes called descriptive key, or descriptive table). The legend says that when Lamarck demonstrated this key for the first time, he gave it to the random stranger (who had no idea about plants and their names), and plant were determined without problems! How to make such a key? The example is below:
1.
We need to start with “players”. In this example, it will be same three plant families:
- Alphaceae
- Betaceae
- Gammaceae
2.
Assess descriptions of these three groups (we copy this from the above):
- Alphaceae : Flowers red, petioles short, leaves whole, spines absent
- Betaceae : Flowers red, petioles long, leaves whole, spines absent
- Gammaceae : Flowers green, petioles short, leaves dissected, spines present
3
Start with a character which let to split the list into two nearly equal groups. Then add other character(s). It is always good to use more characters!
1. Petioles long......................................................................................... Betaceae .
-Petioles short ........................................................................................................2.
2. Flowers red, leaves whole, spines absent.......................................... Alphaceae .
Flowers green, leaves dissected, spines present ................................ Gammaceae .
As you see here, key consists of steps. Every step has a number and typically two choices. Number is attached to the first choice whereas the second choice is marked with minus “–”. The choice will lead either to the name, or to another step. The choice sentence might contain several phrases, the first is the most important and the last is the least important.


3.3: Transition Elements and Ionic Compounds
- Name and write formula of the ions of the transition elements
- Name and write the formulas of simple ionic compounds from monoatomic ions
After learning a few more details about the names of individual ions, you will be a step away from knowing how to write the formula of ionic compounds and name them. This section begins the formal study of writing formula and nomenclature (the systematic naming of ionic compounds).
Naming Ions
The names of many monatomic cation is simply the name of the element followed by the word ion . Thus, Na + is the sodium ion, Al 3 + is the aluminum ion, Ca 2 + is the calcium ion, and so forth.
However, most transitional elements, tin, and lead lose different numbers of electrons, producing ions of different charges. Iron, for example, can form two cations, each of which, when combined with the same anion, makes a different compound with unique physical and chemical properties. Thus, we need a different name for each iron ion to distinguish Fe 2 + from Fe 3 + .
There are two ways to make this distinction. In the simpler, more modern approach, called the Stock system , an ion’s positive charge is indicated by a roman numeral in parentheses after the element name, followed by the word ion. Thus, Fe 2 + is called the iron(II) ion, while Fe 3 + is called the iron(III) ion. This system is used only for elements that form more than one common positive ion. We do not call the Na + ion the sodium(I) ion because (I) is unnecessary. Sodium forms only a 1+ ion, so there is no ambiguity about the name sodium ion.
The second system, called the common system , is not conventional but is still prevalent and used in the health sciences. This system recognizes that many metals have two common cations. The common system uses two suffixes (- ic and - ous ) that are appended to the stem of the element name. The -ic suffix represents the greater of the two cation charges, and the -ous suffix represents the lower one. In many cases, the stem of the element name comes from the Latin name of the element. This system does not work when there are more than two cations for a metal. We simply run out of names. The element manganese is an example. See table \(\PageIndex{1}\) for the list.
The element zinc in group 2B is an exception to the transitional elements. The metal loses two valence electrons to form the zinc ion (Zn 2 + ). The most common ion for the element silver in group 1B is the silver ion (Ag + ).
| Element | Stem/Root | Charge | Name | ||||||
|---|---|---|---|---|---|---|---|---|---|
| iron | fe | rr-2+ |
ferrous ion or iron(II) ion |
||||||
| 3+ | ferric ion or iron(III) ion | ||||||||
| copper | cupr- | 1+ | cuprous ion or copper(I) ion | ||||||
| 2+ | cupric ion or copper(II) ion | ||||||||
| tin | stann- | 2+ | stannous ion or tin(II) ion | ||||||
| 4+ | stannic ion or tin(IV) ion | ||||||||
| lead | plumb- | 2+ | plumbous ion or lead(II) ion | ||||||
| 4+ | plumbic ion or lead(IV) ion | ||||||||
| chromium | chrom- | 2+ | chromous ion or chromium(II) ion | ||||||
| 3+ | chromic ion or chromium(III) ion | ||||||||
| cobalt | colbalt- | 2+ | colbaltous ion or colbalt(II) ion | ||||||
| 3+ | Cobaltic ion or cobalt(III) ion | ||||||||
| manganese | - |
|
|
||||||
| nickel | - |
|
|
The name of a monatomic anion consists of the stem of the element name, the suffix - ide , and then the word ion . Thus, as we have already seen, Cl − is “chlor-” + “-ide ion,” or the chloride ion. Similarly, O 2− is the oxide ion, Se 2 − is the selenide ion, and so forth. Table \(\PageIndex{2}\) lists the names of some common monatomic ions.
| Ion | Name |
|---|---|
| F − | fluoride ion |
| Cl − | chloride ion |
| Br − | bromide ion |
| I − | iodide ion |
| O 2− | oxide ion |
| S 2− | sulfide ion |
| P 3− | phosphide ion |
| N 3− | nitride ion |
The polyatomic ions have their own characteristic names. They will be discussed in the next section.
Name each ion.
- Ca 2 +
- S 2−
- Br -
- K +
- Cu +
- Answer a
-
the calcium ion
- Answer b
-
the sulfide ion (from Table \(\PageIndex{2}\) )
- Answer c
-
the bromide ion
- Answer d
-
the potassium ion
- Answer e
-
the copper(I) ion or the cuprous ion (copper can form cations with either a 1+ or 2+ charge, so we have to specify which charge this ion has
Name each ion.
- Fe 2 +
- Fe 3 +
- N 3−
- Ba 2 +
- F -
- Answer a
-
the iron (II) or ferrous ion
- Answer b
-
the iron (III) or ferric ion
- Answer c
-
the nitride ion
- Answer d
-
the barium ion
- Answer e
-
the fluoride ion
Write the formula for each ion.
- the bromide ion
- the phosphide ion
- the cupric ion
- the magnesium ion
- Answer a
-
Br −
- Answer b
-
P 3 −
- Answer c
-
Cu 2 +
- Answer d
-
Mg 2 +
Write the formula for each ion.
- the fluoride ion
- the oxide ion
- the ferrous ion
- the potassium ion
- Answer a
-
F −
- Answer b
-
O 2 -
- Answer c
-
Fe 2 +
- Answer d
-
K +
Formulae and Names for Ionic Compounds from Monoatomic Ions
The formula for an ionic compound follows several conventions. First, the cation is written before the anion . Because most metals form cations and most nonmetals form anions, formulas typically list the metal first and then the nonmetal. Second, charges are not written in a formula. Remember that in an ionic compound, the components are ions, not neutral atoms, even though the formula does not contain charges. Finally, the proper formula for an ionic compound always has a net zero charge, meaning the total positive charge must equal the total negative charge. To determine the proper formula of any combination of ions, determine how many of each ion is needed to balance the total positive and negative charges in the compound.
This rule is ultimately based on the fact that matter is, overall, electrically neutral.
If we look at the ionic compound consisting of lithium ions and bromide ions, we see that the lithium ion has a 1+ charge and the bromide ion has a 1− charge. Only one ion of each is needed to balance these charges. The formula for lithium bromide is \(\ce{LiBr}\). By convention, assume that there is only one atom if a subscript is not present. We do not use 1 as a subscript.
When an ionic compound is formed from magnesium ion and oxide ion, the magnesium ion has a 2+ charge, and the oxide has a 2− charge. Although both of these ions have higher charges than the ions in lithium bromide, they still balance each other in a one-to-one ratio. Therefore, the proper formula for this magnesium oxide is \(\ce{MgO}\).
Now consider the ionic compound formed by magnesium ion and chloride ion. A magnesium ion has a 2+ charge, while a chloride ion has a 1− charge:
\[\ce{Mg^{2+}Cl^{−}} \nonumber \]
Combining one ion of each does not completely balance the positive and negative charges. The easiest way to balance these charges is to assume the presence of two chloride ions for each magnesium ion:
\[\ce{Mg^{2+} Cl^{−} Cl^{−}} \nonumber \]
Now the positive and negative charges are balanced, the convention is to use a numerical subscript when there is more than one ion of a given type: so \(\ce{MgCl2}\). This chemical formula says that there are one magnesium ion and two chloride ions in the formula for magnesium chloride . (Do not read the “Cl 2 ” part of the formula as a molecule of the diatomic elemental chlorine. (Chlorine does not exist as a diatomic element in this compound. Rather, it exists as two individual chloride ions.) By convention, the lowest whole number ratio is used in the formulas of ionic compounds. The formula \(\ce{Mg2Cl4}\) has balanced charges with the ions in a 1:2 ratio, but it is not the lowest whole number ratio.
By convention, the lowest whole-number ratio of the ions is used in ionic formulas. There are exceptions for certain ions, such as \(\ce{Hg2^{2+}}\) known as the mercurous ion.
For compounds in which the ratio of ions is not as obvious, the subscripts in the formula can be obtained by crossing charges : use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically in Figure \(\PageIndex{1}\).
When crossing charges, it is sometimes necessary to reduce the subscripts to their simplest ratio to write the formula. Consider, for example, the compound formed by Pb 4 + and O 2− . Using the absolute values of the charges on the ions as subscripts gives the formula Pb 2 O 4 . This simplifies to its correct formula PbO 2 for lead(IV) oxide. The formula has one Pb 4 + ion and two O 2− ions.
Write the chemical formula and name for an ionic compound composed of each pair of ions.
- the sodium ion and the sulfide ion
- the aluminum ion and the fluoride ion
- the iron(III) ion and the oxide ion
Solution
- To obtain an octet, sodium forms an ion with a 1+ charge, while the sulfide ion has a 2− charge. Two sodium 1+ ions are needed to balance the 2− charge on the sulfide ion. Rather than writing the formula as \(\ce{NaNaS}\), we shorten it by convention to \(\ce{Na2S}\). The name of the compound is sodium sulfide.
- The aluminum ion has a 3+ charge, while the fluoride ion formed by fluorine has a 1− charge. Three fluoride ions are needed to balance the 3+ charge on the aluminum ion. This combination is written as \(\ce{AlF3}\). The name of the compound is aluminum fluoride.
- Iron can form two possible ions, but the ion with a 3+ charge is specified here. The oxide has a 2− charge as an ion. To balance the positive and negative charges, we look to the least common multiple (6): two iron 3+ ions will give 6+, while three oxide ions will give 6−, thereby balancing the overall positive and negative charges. Thus, the formula for this ionic compound is \(\ce{Fe2O3}\). Alternatively, use the crossing charges method shown in Figure 3.3.1.
Write the chemical formula for an ionic compound composed of each pair of ions.
- the calcium ion and the oxide ion
- the copper(II) ion and the sulfide ion
- the copper(I) ion and the sulfide ion
- Answer a:
-
CaO
- Answer b:
-
CuS
- Answer c:
-
Cu 2 S
Name each ionic compound, using both Stock and common systems if necessary.
- Ca 3 P 2
- Fe 3 N 2
- KCl
- CuCl
- SnF 2
- Answer a
-
calcium phosphide
- Answer b
-
iron(II) nitride or ferrous nitride
- Answer c
-
potassium chloride
- Answer d
-
copper(I) chloride or cuprous chloride
- Answer e
-
tin(II) fluoride or stannous fluoride
Name each ionic compound, using both Stock and common systems if necessary.
- ZnBr 2
- FeN
- Al 2 O 3
- CuF 2
- AgF
- Answer a
-
zinc bromide
- Answer b
-
iron (III) nitride or ferric nitride
- Answer c
-
aluminum oxide
- Answer d
-
copper (II) fluoride or cupric fluoride
- Answer e
-
silver fluoride
KEY TAKEAWAY
- Each ionic compound has its own unique name and formula that comes from the names and formulae of the ions.
EXERCISES
- Briefly describe the process for naming an ionic compound.
2. In what order do the names of ions appear in the names of ionic compounds?
3. Which ionic compounds can be named using two different systems? Give an example.
4. Name each ion.
- Ba 2 +
- P 3−
- S 2-
- Sn 4 +
5. Name each ion.
- Cs +
- Al 3 +
- I -
- Sn 2 +
6. Name the ionic compound formed by each pair of ions.
- Na + and Br −
- Mg 2 + and Br −
- Mg 2 + and S 2−
7. Name the ionic compound formed by each pair of ions.
- K + and Cl −
- Mg 2 + and Cl −
- Mg 2 + and S 2 −
8. Name the ionic compound formed by each pair of ions.
- Na + and N 3−
- Mg 2 + and N 3−
- Al 3 + and S 2−
9. Name the ionic compound formed by each pair of ions.
- Li + and N 3−
- Mg 2 + and P 3−
- Li + and P 3−
10. Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate.
- Fe 3 + and Br −
- Fe 2 + and Br −
- Au 3 + and S 2−
- Au + and S 2−
11. Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate.
- Cr 3 + and O 2−
- Cr 2 + and O 2−
- Pb 2 + and Cl −
- Pb 4 + and Cl −
Answers
- Name the cation and then the anion but don’t use numerical prefixes.
- the cation name followed by the anion name
- Ionic compounds in which the cation can have more than one possible charge have two naming systems. FeCl 3 is either iron(III) chloride or ferric chloride (answers will vary).
4.
- the barium ion
- the phosphide ion
- the sulfide ion
- the tin(IV) ion or the stannic ion
5.
- the cesium ion
- the aluminum ion
- the iodide ion
- the tin(II) ion or the stannous ion
6.
- sodium bromide
- magnesium bromide
- magnesium sulfide
7.
- potassium chloride
- magnesium chloride
- magnesium sulfide
8.
- sodium nitride
- magnesium nitride
- aluminum sulfide
9.
- lithium nitride
- magnesium phosphide
- lithium phosphide
10.
- iron(III) bromide or ferric bromide
- iron(II) bromide or ferrous bromide
- gold(III) sulfide or auric sulfide
- gold(I) sulfide or aurous sulfide


1.9: Learning Activities
Learning Activities
(Answers to “Learning Activities” can be found in the “Answer Key” at the end of the book. Answers to interactive activity elements will be provided within the element as immediate feedback.)
Apply what you have learned from this chapter by completing the following learning activities:
1.  You have been assisting a critical care nurse with the care of a patient who has been experiencing significantly low blood pressures throughout the day. The nurse has to step away from the bedside to take a phone call and instructs you to increase the intravenous (IV) medication if the patient’s systolic blood pressure drops below 90 mmHg. What is the appropriate response to this instruction?
2. You are completing a clinical rotation on a medical surgical unit and are invited to join a few staff nurses in the breakroom for a lunch break. While you are in the breakroom, you notice one of the staff nurses complaining loudly about a patient and discussing sensitive patient care information. What is an appropriate response to this situation?
Query \(\PageIndex{1}\)
Query \(\PageIndex{2}\)
Query \(\PageIndex{3}\)
Query \(\PageIndex{4}\)


Explain the
relationship between education and work in early adulthood
Education in Early Adulthood
According to the U.S. Census Bureau (2017), 90 percent of the American population 25 and older have completed high school or higher level of education—compare this to just 24 percent in 1940!
[1]
Each generation tends to earn (and perhaps need) increased levels of formal education. As we can see in the graph, approximately one-third of the American adult population has a bachelor’s degree or higher, as compared with less than 5 percent in 1940. Educational attainment rates vary by gender and race. All races combined, women are slightly more likely to have graduated from college than men; that gap widens with graduate and professional degrees. However, wide racial disparities still exist. For example, 23 percent of African-Americans have a college degree and only 16.4 percent of Hispanic Americans have a college degree, compared to 37 percent of non-Hispanic white Americans. The college graduation rates of African-Americans and Hispanic Americans have been growing in recent years, however (the rate has doubled since 1991 for African-Americans and it has increased 60 percent in the last two decades for Hispanic-Americans).
What about those young or emerging adults graduating high school today—is the majority of that group going to college? According to the U.S. Bureau of Labor Statistics (2017), 66.7 percent of youth ages 16-24 who graduated high school between January and October 2017 were enrolled in colleges or universities in October 2017. There were gender differences (71.7 percent of females vs. 61.1 percent of males) and racial differences (83 percent of Asians, 67.1 percent of non-Hispanic whites, 61 percent Hispanics, and 59.4 percent Blacks). Not all of these students will persist and earn college degrees, however.
[2]
With the rising costs of higher education, various news headlines have asked if a college education is worth the cost. One way to address this question is in terms of the earning potential associated with various levels of educational achievement. In 2016, the average earnings for Americans 25 and older with only a high school education was $35,615, compared with $65,482 for those with a bachelor’s degree, compared with $92,525 for those with more advanced degrees. Average earnings vary by gender, race, and geographical location in the United States.
[3]
Of concern in recent years is the relationship between higher education and the workplace. In 2005, American educator and then Harvard University President, Derek Bok, called for a closer alignment between the goals of educators and the demands of the economy. Companies outsource much of their work, not only to save costs but to find workers with the skills they need. What is required to do well in today’s economy? Colleges and universities, he argued, need to promote global awareness, critical thinking skills, the ability to communicate, moral reasoning, and responsibility in their students. Regional accrediting agencies and state organizations provide similar guidelines for educators. Workers need skills in listening, reading, writing, speaking, global awareness, critical thinking, civility, and computer literacy—all skills that enhance success in the workplace.
More than a decade later, the question remains: does formal education prepare young adults for the workplace? It depends on whom you ask. In an article referring to information from the National Association of Colleges and Employers’ 2018 Job Outlook Survey, Bauer-Wolf (2018) explains that employers perceive gaps in students’ competencies but many graduating college seniors are overly confident. The biggest difference was in perceived professionalism and work ethic (only 43 percent of employers thought that students are competent in this area compared to 90 percent of the students).
[4]
Similar differences were also found in terms of oral communication, written communication, and critical thinking skills. Only in terms of digital technology skills were more employers confident about students’ competencies than were the students (66 percent compared to 60 percent).
It appears that students need to learn what some call “soft skills,” as well as the particular knowledge and skills within their college major. As education researcher Loni Bordoloi Pazich (2018) noted, most American college students today are enrolling in business or other pre-professional programs and to be effective and successful workers and leaders, they would benefit from the communication, teamwork, and critical thinking skills, as well as the content knowledge, gained from liberal arts education.
[5]
In fact, two-thirds of children starting primary school now will be employed in jobs in the future that currently do not exist. Therefore, students cannot learn every single skill or fact that they may need to know, but they can learn how to learn, think, research, and communicate well so that they are prepared to continually learn new things and adapt effectively in their careers and lives since the economy, technology, and global markets will continue to evolve.
[6]
Career Choices in Early Adulthood
Hopefully, we are each becoming lifelong learners, particularly since we are living longer and will most likely change jobs multiple times during our lives. However, for many, our job changes will be within the same general occupational field, so our initial career choice is still significant. We’ve seen with Erikson that identity largely involves occupation and, as we will learn in the next section, Levinson found that young adults typically form a dream about work (though females may have to choose to focus relatively more on work or family initially with “split” dreams). The American School Counselor Association recommends that school counselors aid students in their career development beginning as early as kindergarten and continue this development throughout their education.
[7]
One of the most well-known theories about career choice is from John Holland (1985), who proposed that there are six personality types (realistic, investigative, artistic, social, enterprising, and conventional), as well as varying types of work environments.
[8]
The better matched one’s personality is to the workplace characteristics, the more satisfied and successful one is predicted to be with that career or vocational choice. Research support has been mixed and we should note that there is more to satisfaction and success in a career than one’s personality traits or likes and dislikes. For instance, education, training, and abilities need to match the expectations and demands of the job, plus the state of the economy, availability of positions, and salary rates may play practical roles in choices about work.
Link to Learning: What’s Your Right Career?
To complete a free online career questionnaire and identify potential careers based on your preferences, go to:
Did you find out anything intere
sting? Think of this activity as a starting point to your career exploration. Other great ways for young adults to research careers include informational interviewing, job shadowing, volunteering, practicums, and internships. Once you have a few careers in mind that you want to find out more about, go to the
Occupational Outlook Handbook
from the U.S. Bureau of Labor Statistics to learn about job tasks, required education, average pay, and projected outlook for the future.
Henseler, C. (2017, September 6). Liberal arts is the foundation for professional success in the 21st century. Huffington Post.
<EMAIL_ADDRESS>Holland, J.L. (1985). Making vocational choices: A theory of vocational personalities and work environments. Englewood Cliffs, NJ: Prentice-Hall.
<EMAIL_ADDRESS>Modification, adaptation, and original content.
Authored by
: Margaret Clark-Plaskie for Lumen Learning.
Provided by
: Lumen Learning.
License
:
CC BY: Attribution


3.6: Venn Diagrams
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A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space \(S\) together with circles or ovals. The circles or ovals represent events. Venn diagrams also help us to convert common English words into mathematical terms that help add precision.
Venn diagrams are named for their inventor, John Venn, a mathematics professor at Cambridge and an Anglican minister. His main work was conducted during the late 1870's and gave rise to a whole branch of mathematics and a new way to approach issues of logic. We will develop the probability rules just covered using this powerful way to demonstrate the probability postulates including the Addition Rule, Multiplication Rule, Complement Rule, Independence, and Conditional Probability.
Example \(\PageIndex{1}\)
Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an equal chance of occurring. Let event \(A = \{1, 2, 3, 4, 5, 6\}\) and event \(B = \{6, 7, 8, 9\}\). Then \(A\) intersect \(B = A \cap B=\{6\}\) and \(A\) union \(B = A\cup B=\{1, 2, 3, 4, 5, 6, 7, 8, 9\}\). The Venn diagram is as follows:
Figure \(\PageIndex{1}\) shows the most basic relationship among these numbers. First, the numbers are in groups called sets; set A and set B. Some number are in both sets; we say in set A \(\cap \) in set B. The English word "and" means inclusive, meaning having the characteristics of both A and B, or in this case, being a part of both A and B. This condition is called the INTERSECTION of the two sets. All members that are part of both sets constitute the intersection of the two sets. The intersection is written as \(A\cap B\) where \(\cap\) is the mathematical symbol for intersection. The statement \(A\cap B\) is read as "A intersect B." You can remember this by thinking of the intersection of two streets.
There are also those numbers that form a group that, for membership, the number must be in either one or the other group. The number does not have to be in BOTH groups, but instead only in either one of the two. These numbers are called the UNION of the two sets and in this case they are the numbers 1-5 (from A exclusively), 7-9 (from set B exclusively) and also 6, which is in both sets A and B. The symbol for the UNION is \(\cup \), thus \(A\cup B=\) numbers 1-9, but excludes number 10, 11, and 12. The values 10, 11, and 12 are part of the universe, but are not in either of the two sets.
Translating the English word "AND" into the mathematical logic symbol \(\cap\), intersection, and the word "OR" into the mathematical symbol \(\cup\), union, provides a very precise way to discuss the issues of probability and logic. The general terminology for the three areas of the Venn diagram in Figure \(\PageIndex{1}\) is shown in Figure \(\PageIndex{2}\), below.
Exercise \(\PageIndex{1}\)
Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event C = {green, blue, purple} and event P = {red, yellow, blue}. Then \(C\cap P=\{blue\}\) and \(C \cup P=\{\text { green, blue, purple, red, yellow }\}\). Draw a Venn diagram representing this situation.
Example \(\PageIndex{2}\)
Flip two fair coins. Let \(A\) = tails on the first coin. Let \(B\) = tails on the second coin. Then \(A = {TT, TH}\) and \(B = {TT, HT}\). Therefore, \(A\cap B=\{TT\}\). \(A\cup B=\{TH, TT, HT\}\).
The sample space when you flip two fair coins is \(X = {HH, HT, TH, TT}\). The outcome \(HH\) is in NEITHER \(A\) NOR \(B\). The Venn diagram is as follows:
Exercise \(\PageIndex{2}\)
Roll a fair, six-sided die. Let \(A\) = a prime number of dots is rolled. Let \(B\) = an odd number of dots is rolled. Then \(A= {2, 3, 5}\) and \(B = {1, 3, 5}\). Therefore, \(A\cap B=\{3, 5\}\). \(A\cup B=\{1, 2, 3, 5\}\). The sample space for rolling a fair die is \(S = {1, 2, 3, 4, 5, 6}\). Draw a Venn diagram representing this situation.
 Four percent of African Americans have type O blood and a negative RH factor, 5−10% of African Americans have the Rh- factor, and 51% have type O blood.
The “O” circle represents the African Americans with type O blood. The “Rh-“ oval represents the African Americans with the Rh- factor.
We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rh- factor. Let O = African American with Type O blood and R = African American with Rh- factor.
- \(P(O)\) = ___________
- \(P(R)\) = ___________
- \(P(O\cap R)=\) ___________
- \(P(O\cup R)=\) ____________
- In the Venn Diagram, describe the overlapping area using a complete sentence.
- In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence.
- Answer
-
1. 0.51; 2. 0.075; 3. 0.04; 4. 0.545; 5. The area represents the African Americans that have type O blood and the Rh- factor. 6. The area represents the African Americans that have neither type O blood nor the Rh- factor.
Example \(\PageIndex{4}\)
Fifty percent of the workers at a factory work a second job, 25% have a spouse who also works, 5% work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let W = works a second job and S = spouse also works.
- Answer
-
Forty percent of the students at a local college belong to a club and 50% work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let C = student belongs to a club and PT = student works part time.
If a student is selected at random, find
- the probability that the student belongs to a club. \(P(C) = 0.40\)
- the probability that the student works part time. \(P(PT) = 0.50\)
- the probability that the student belongs to a club AND works part time. \(P(C\cap PT)=0.05\)
- the probability that the student belongs to a club given that the student works part time. \(P(C | P T)=\frac{P(C \cap P T)}{P(P T)}=\frac{0.05}{0.50}=0.1\)
- the probability that the student belongs to a club OR works part time. \(P(C \cup P T)=P(C)+P(P T)-P(C \cap P T)=0.40+0.50-0.05=0.85\)
In order to solve Example \(\PageIndex{4}\) we had to draw upon the concept of conditional probability from the previous section. There we used tree diagrams to track the changes in the probabilities, because the sample space changed as we drew without replacement. In short, conditional probability is the chance that something will happen given that some other event has already happened. Put another way, the probability that something will happen conditioned upon the situation that something else is also true. In Example \(\PageIndex{4}\) the probability \(P(C|PT)\) is the conditional probability that the randomly drawn student is a member of the club, conditioned upon the fact that the student also is working part time. This allows us to see the relationship between Venn diagrams and the probability postulates.
Exercise \(\PageIndex{3}\)
In a bookstore, the probability that the customer buys a novel is 0.6, and the probability that the customer buys a non-fiction book is 0.4. Suppose that the probability that the customer buys both is 0.2.
- Draw a Venn diagram representing the situation.
- Find the probability that the customer buys either a novel or a non-fiction book.
- In the Venn diagram, describe the overlapping area using a complete sentence.
- Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event.
Example \(\PageIndex{5}\)
A set of 20 German Shepherd dogs is observed. 12 are male, 8 are female, 10 have some brown coloring, and 5 have some white sections of fur. Answer the following using Venn Diagrams.
Draw a Venn diagram simply showing the sets of male and female dogs.
- Answer
-
The Venn diagram below demonstrates the situation of mutually exclusive events where the outcomes are independent events. If a dog cannot be both male and female, then there is no intersection. Being male precludes being female and being female precludes being male: in this case, the characteristic gender is therefore mutually exclusive. A Venn diagram shows this as two sets with no intersection. The intersection is said to be the null set using the mathematical symbol ∅.
Draw a second Venn diagram illustrating that 10 of the male dogs have brown coloring.
- Answer
-
The Venn diagram below shows the overlap between male and brown where the number 10 is placed in it. This represents \(\text{ Male}\cap \text{Brown }\): both male and brown. This is the intersection of these two characteristics. To get the union of Male and Brown, then it is simply the two circled areas minus the overlap. In proper terms, \( \text{ Male}\cup \text{ Brown }=\text { Male }+\text { Brown }-\text { Male } \cap \text { Brown}\) will give us the number of dogs in the union of these two sets. If we did not subtract the intersection, we would have double counted some of the dogs.
Now draw a situation depicting a scenario in which the non-shaded region represents "No white fur and female," or \(\text{White Fur}^C \cap \text{Female}\) . The \({}^C\) above "fur" indicates "not white fur." The \({}^C\) above a set means not in that set.
- Answer
The Addition Rule of Probability
We met the addition rule earlier but without the help of Venn diagrams. Venn diagrams help visualize the counting process that is inherent in the calculation of probability. To restate the Addition Rule of Probability:
\[P(A \cup B)=P(\mathrm{A})+P(B)-P(A \cap B)\nonumber\]
Remember that probability is simply the proportion of the objects we are interested in relative to the total number of objects. This is why we can see the usefulness of the Venn diagrams. Example \(\PageIndex{5}\) shows how we can use Venn diagrams to count the number of dogs in the union of brown and male by reminding us to subtract the intersection of brown and male. We can see the effect of this directly on probabilities in the addition rule.
Example \(\PageIndex{6}\)
Let's sample 50 students who are in a statistics class. 20 are freshmen and 30 are sophomores. 15 students get a "B" in the course, and 5 students both get a "B" and are freshmen.
Find the probability of selecting a student who either earns a "B" OR is a freshmen. We are translating the word OR to the mathematical symbol for the addition rule, which is the union of the two sets.
- Answer
-
We know that there are 50 students in our sample, so we know the denominator of our fraction to give us probability. We need only to find the number of students that meet the characteristics we are interested in, i.e. any freshman and any student who earned a grade of "B." With the Addition Rule of probability, we can skip directly to probabilities.
Let "A" = the number of freshmen, and let "B" = the grade of "B." Below we can see the process for using Venn diagrams to solve this.
The \(P(A)=\frac{20}{50}=0.40, P(B)=\frac{15}{50}=0.30, \text { and } P(A \cap B)=\frac{5}{50}=0.10\)
Therefore, \(P(A \cap B)=0.40+0.30-0.10=0.60\)
If two events are mutually exclusive, then, like the example where we diagram the male and female dogs, the addition rule is simplified to just \(P(A\cup B)=P(A)+P(B)−0\). This is true because, as we saw earlier, the union of mutually exclusive events is the null set, ∅. The diagrams below demonstrate this.
The Multiplication Rule of Probability
Restating the Multiplication Rule of Probability using the notation of Venn diagrams, we have:
\[P(A\cap B)=P(A|B)⋅P(B)\nonumber\]
The multiplication rule can be modified with a bit of algebra into the following conditional rule. Then Venn diagrams can then be used to demonstrate the process.
The conditional rule: \(P(A | B)=\frac{P(A \cap B)}{P(B)}\)
Using the same facts from Example \(\PageIndex{6}\) above, find the probability that someone will earn a "B" if they are a "freshman."
\[P(A | B)=\frac{0.10}{0.30}=\frac{1}{3}\nonumber\]
The multiplication rule must also be altered if the two events are independent. Independent events are defined as a situation where the conditional probability is simply the probability of the event of interest. Formally, independence of events is defined as \(P(A|B)=P(A)\) or \(P(B|A)=P(B)\). When flipping coins, the outcome of the second flip is independent of the outcome of the first flip; coins do not have memory. The Multiplication Rule of Probability for independent events thus becomes:
\[P(A\cap B)=P(A)⋅P(B)\nonumber\]
One easy way to remember this is to consider what we mean by the word "and." We see that the Multiplication Rule has translated the word "and" to the Venn notation for intersection. Therefore, the outcome must meet the two conditions of freshmen and grade of "B" in the above example. It is harder, less probable, to meet two conditions than just one or some other one. We can attempt to see the logic of the Multiplication Rule of probability due to the fact that fractions multiplied times each other become smaller.
The development of the Rules of Probability with the use of Venn diagrams can be shown to help as we wish to calculate probabilities from data arranged in a contingency table.
Example \(\PageIndex{7}\)
Table \(\PageIndex{1}\) is from a sample of 200 people who were asked how much education they completed. The columns represent the highest education they completed, and the rows separate the individuals by male and female.
| Less than high school grad | High school grad | Some college | College grad | Total | |
|---|---|---|---|---|---|
| Male | 5 | 15 | 40 | 60 | 120 |
| Female | 8 | 12 | 30 | 30 | 80 |
| Total | 13 | 27 | 70 | 90 | 200 |
Now, we can use this table to answer probability questions. The following examples are designed to help understand the format above while connecting the knowledge to both Venn diagrams and the probability rules.
What is the probability that a selected person both finished college and is female?
- Answer
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This is a simple task of finding the value where the two characteristics intersect on the table, and then applying the postulate of probability, which states that the probability of an event is the proportion of outcomes that match the event in which we are interested as a proportion of all total possible outcomes.
\(P(\text {College Grad } \cap \text { Female })=\frac{30}{200}=0.15\)
What is the probability of selecting either a female or someone who finished college?
- Answer
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This task involves the use of the addition rule to solve for this probability.
\(P(\text { College Grad } \cup \text{ Female })=P(F)+P(C G)-P(F \cap C G)\)
\(P(\text { College Grad } \cup \text{ Female }) =\frac{80}{200}+\frac{90}{200}-\frac{30}{200}=\frac{140}{200}=0.70\)
What is the probability of selecting a high school graduate if we only select from the group of males?
- Answer
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Here we must use the conditional probability rule (the modified multiplication rule) to solve for this probability.
\(P (\text{HS Grad } | \text { Male }）=\frac{P(\mathrm{HS} \text { Grad } \cap \mathrm{Male})}{\mathrm{P}(\mathrm{Male})}=\frac{\left(\frac{15}{200}\right)}{\left(\frac{120}{200}\right)}=\frac{15}{120}=0.125\)
Can we conclude that the level of education attained by these 200 people is independent of the gender of the person?
- Answer
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There are two ways to approach this test. The first method seeks to test if the intersection of two events equals the product of the events separately remembering that if two events are independent than \(P(A)^{*} P(B)=P(A \cap B)\). For simplicity's sake, we can use calculated values from above.
Does \(P(\text { College Grad } \cap \text { Female })=P(C G) \cdot P(F)\)?
\(\frac{30}{200} \neq \frac{90}{200} \cdot \frac{80}{200}\) because 0.15 ≠ 0.18.
Therefore, gender and education here are not independent.
The second method is to test if the conditional probability of A given B is equal to the probability of A. Again for simplicity, we can use an already calculated value from above.
Does \(P(H S \text { Grad } | \text { Male })=P(H S \text { Grad) }\)?
\(\frac{15}{120} \neq \frac{27}{200}\) because 0.125 ≠ 0.135.
Therefore, again gender and education here are not independent.


1.0: What Does Psychology Say?
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You’ve been looking forward to your trip to New York City. Upon landing at JFK airport, you take the subway to reach Midtown Manhattan. The train accelerates and you watch the platform disappear as you find a seat and take your first look around the car. There is life here, rocketing through the underground tunnels, and a broad cross-section of human experience. You steal glimpses of your fellow riders’ physical appearance, belongings, interactions, and facial expressions. You wonder where each of them is going and what they hope to accomplish today. At each station, the train stops, and some people exit and even more board, the crowd growing as it nears Manhattan.
Like many students studying lifespan development, you likely share a curiosity about the people around you. You may wonder:
In this course, you can explore questions like these, contemplate what answers can be found through psychological research, and learn more about your own life journey.