2.1.7: Modern Synthesis
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The Modern Synthesis incorporates data from multiple scientific disciplines: biology, the natural sciences, genetics, paleontology, and paleoanthropology. While the basis for evolutionary theory was established by the beginning of the twentieth-century, debate continued as to whether natural selection or mutation was more important in evolution. In the 1930s, these two points of views came together to form the basis of the modern synthesis. Fisher, Haldane, and Wright developed mathematical models that became the foundation of population genetics, which studies the frequency of alleles, genotypes, and phenotypes in populations (Miller 2011: 321). However, it was Theodosius Dobzhansky that married theoretical modeling and empirical research in his seminal work, Genetics and the Origin of Species (1937) and signals the start of the Modern Synthesis (Artmann 2006). Biologist Ernst Mayr was another major contributor to the development of the modern synthesis. His work on the isolating mechanisms of speciation led to new insight into the evolution of new species. Mayr’s Systematics and the Origin of Species (1942) is one of the key works of the Modern Synthesis.
Figure \(\PageIndex{1}\) - Note by Francis Crick.
A New Definition
Emerging from the work of scientists of the Modern Synthesis, we know that evolution is a two-step process (Miller 2011: 85):
- The production and redistribution of variation, and
- Natural selection acts on variation resulting in differential reproduction among individuals in a population.
A population is a group of individuals from which mates are usually found. Sometimes this is referred to as a breeding population . All members of a population share a gene pool , which is all of the genes found among the individuals of the population.
Figure \(\PageIndex{2}\) - Graphic of microevolution. Modeled on University of California Museum of Paleontology's Understanding Evolution (evolution.berkeley.edu)
Scientists can study variation in a gene pool and differential reproduction in populations to determine allele frequencies , which is the frequency of an allele relative to that of other alleles of the same gene. Changing allele frequencies within a population from one generation to the next is the definition of evolution; specifically, this is microevolution. Macroevolution is change that occurs after many generations (hundred or thousands); e.g., speciation , which is the evolution of a species into a new species. “Macroevolution encompasses the grandest trends and transformation in evolution, such as the origin of mammals and the radiation of flowering plants (Evolution 101, 2015). Macroevolutionary events are reconstructed using data from living organisms, geology, and fossils. These are generally represented in a graphic form called a phylogentic tree (pictured below). Phylogenetic trees are a visual representation of proposed evolutionary relationships. They help researchers organize data. To learn more about phylogenetic trees check out Reading a Phylogenetic Tree.
Figure \(\PageIndex{3}\) - Simplified phylogenetic tree
Step 1: Production and Redistribution of Variation
As we know, production and redistribution of variation is the first step of evolution. There are several factors involved in the production and redistribution of variation: 1. Mutation 2. Genetic drift 3. Gene flow, and 4. Recombination. These factors, along with natural selection, which is the second step of evolution, are frequently referred to as the forces or mechanisms of evolution .
Mutation
Mutation is a random change in the genetic coding and is the only new source of genetic material. Mutations occur during the DNA replication process. In an evolutionary sense, it is only mutations that occur during meiosis (replication of sex cells) that are hereditary. Many mutations do not have detrimental effects to the individual, but some do. If the mutations are extreme, pregnancy often results in spontaneous abortion (miscarriage). However, mutations provide a necessary pool of genetic variation in a population.
Figure \(\PageIndex{4}\) - Illustration of a point mutation
There are different types of mutations. The first is a point mutation , which is a small change in the DNA sequence. Point mutations may be synonymous point mutations or silent mutations , which means that while the codon is altered the same amino acid is produced, or nonsynonymous point mutations , which result in the production of a different amino acid that can have a negative effect on the individual. For instance, sickle cell anemia results in a point mutation where “…a mutation on human chromosome 11 converts a GAG codon into a GTG codon. The GAG codon is encoded to produce the amino acid valine, whereas the GTG codon is encoded to produce glutamic acid. This substitution results in sickle cell anemia…” (Larsen 2008: 93).
Frameshift mutations occur when one or more nitrogenous bases are inserted into or deleted from the DNA sequence. The insertion or deletion causes the reading of the gene to be altered or stopped. It results in the production of a nonfunctioning protein.
Figure \(\PageIndex{5}\) - Illustration of a deletion mutation
Figure \(\PageIndex{6}\) - Illustration of an insertion mutation
Transposable elements are mutations that occur when pieces of DNA copy themselves into new areas of chromosomes. If this occurs in a non-coding segment of the DNA then there are little to no effects on the individual. If it transposes to a coding segment of DNA then serious consequences can occur. In extreme cases, entire chromosomes may be copied (a trisomy) or deleted (a monosomy).
Mutations may have no known cause. These are called spontaneous mutations . Mutations caused by environmental factors are called induced mutations .
Genetic Drift
Figure \(\PageIndex{7}\) - Illustration of founder effect
Genetic drift is the “…chance variation in allele frequency in small populations” (Gould 2006:1036). This can happen in a couple of ways: founder effect and bottleneck effect (or population bottleneck). Founder effect can occur if a small group separates from its parent group and settles elsewhere. Geographic isolation and endogamous marriage (human marriage pattern where mates are chosen from within a specific group, narrowing the gene pool) or mating practices restrict the gene pool, resulting in a lack of variation in comparison to the original parent population. If an allele is rare in the parent population, the small breeding population of the new group creates an environment whereby the rare allele can become common.
There are numerous examples of the founder effect among human populations. These include,
Figure \(\PageIndex{8\) - Map of blood group O
- Huntington’s Disease in Afrikaner populations in South Africa and among Venezuelans living near Lake Maraciabo,
- Prevalence of O blood type in Native Americans
- Ellis-van-Creveld syndrome (polydactyly) in Amish communities, and
- The Blue People of Kentucky (learn more about the Blue People www.indiana.edu/~oso/lessons/Blues/TheBlues.htm).
Bottleneck effect occurs when a population is drastically reduced in size due to some disaster, e.g., natural disaster, habitat destruction, or disease, leaving the surviving population with different allele frequencies as the original population. Four human population bottlenecks all related to climatic events have been identified. The most severe occurred about 70,000 years ago when a supervolcano erupted in Sumatra. It is estimated that only 5000 females survived in Africa (Ambrose 2003).
Figure \(\PageIndex{9}\) - Illustration of the bottleneck effect
Gene Flow
Figure \(\PageIndex{10}\) - Illustration of gene flow.
Gene flow , also called admixture or migration (in reference to people), is the exchange of genetic material between populations (Miller 2011). For humans, gene flow generally occurs when people move from one location to another; however, in the modern world it is not out of the realm of possibility that someone’s genetic material could move with the actual human being leaving their hometown. It should also be noted that migration does not always result in gene flow. For gene flow to occur interbreeding must happen. Human populations have exchanged genetic materials for millennia, helping to explain why there is only one human species.
Gene flow effects evolution in a couple of important ways (Evolution 101, 2015): • With the introduction and reintroduction of genes to a population its genetic variation is increased, and • The movement of genetic material can make distant populations genetically similar to one another, which reduces the chance of speciation.
Recombination
Recombination , or crossing over, can occur during meiosis. As the term implies, segments of chromosomes are exchanged between non-sister chromatids. This creates new combinations of genes that are not found in either parent (Speer 2003).
Step 2: Natural Selection
Figure \(\PageIndex{11}\) - Peppered moths.
Figure \(\PageIndex{12}\) - Types of selection.
Once variation is produced in some combination of the above processes, natural selection goes to work, causing directional change in allele frequency
related to environmental conditions that may result in evolutionary change, i.e., speciation. Shifts in allele frequency are called adaptation . In phrasing it this way, it sounds like natural selection works consciously, but it does not. Natural selection is a process that is mindless and mechanistic. Perfection and progress are not goals. It can only act on those variants that are present in a population.
Figure \(\PageIndex{13}\) - Male peacock display.
Darwin concluded, “…individuals with advantageous characteristics…survive in higher numbers and produce more offspring than members of a population lacking advantageous characteristics” (Larsen 2008: 96). This is referred to as differential reproduction or fitness . Specifically, fitness is defined by genotypes—some genotypes confer an advantage to an individual allowing that individual to reproduce at a greater frequency than another individual, characteristics that are inherited by their offspring. It could be something like beak shape that allows a bird to exploit resources better than other birds (think Darwin’s finches) or the color of wings that either hide or expose the peppered moth to predators (watch youtu.be/sVVldxxbWig for the full explanation) or a coloring, like the plumage of a male peacock, that attracts mates (this is referred to as sexual selection , which is a special case of natural selection which refers to an individuals ability to attract and copulate with a mate) that lead to differential reproduction.
There are three types of selection: disruptive, stabilizing, and directional selection. With disruptive selection, both homozygotes (the phenotypic extremes) are selected for while with directional selection only one of the homozygotes is selected for. Disrupting selection may result in a speciation event. Stabilizing selection refers to selection for the heterozygote, resulting in decreasing genetic diversity within the population.
If you would like a bit more on the types of selection, check out Bio.edu: Stabilizing selection ( http://www.brooklyn.cuny.edu/bc/ahp/LAD/C21/C21_Stabilizing.html ) Disruptive selection ( http://www.brooklyn.cuny.edu/bc/ahp/LAD/C21/C21_Disruptive.html ) Directional selection ( http://www.brooklyn.cuny.edu/bc/ahp/LAD/C21/C21_Directional.html )
Species and Speciation
Mayr proposed the most commonly referenced definition of species (1942; quoted in Provine 2004), “Species are a group of actually or potentially interbreeding natural populations, which are reproductively isolated from other such groups.” This definition is often referred to as the biological species concept . A key factor is that a species actually or potentially breeds in nature. However, this definition can be difficult to apply; e.g., organisms that reproduce asexually or organisms that form hybrids (carrion crows and hooded crows sometimes interbreed—are they one species or two species?). The biological species concept falls apart when applied to the fossil record, as it is impossible for scientists to observe the breeding behaviors of extinct organisms. Because of this, fossil species are called paleospecies and can cover long periods of time. Paleontologists and paleoanthropologists use comparative analysis with contemporary species that are closely related to determine whether variation seen in fossil specimens is intraspecific (differences within a species due to age and sex) or interspecific (differences due to reproductive isolation). It is important to remember that the species or paleospecies name is a classificatory tool used in the sciences. It helps scientists organize information and communicate more easily with one another.
Speciation is “…the process by which a new species evolves from an earlier species (Jurmain et al 2013: 105). This idea is one of Darwin’s five theories, specifically those of common descent and multiplicity of species, whereby new species arise from earlier species.
Simsonian Evolution
Just for fun, check out how The Simpsons explain evolution: www.youtube.com/watch?v=faRlFsYmkeY
Figure \(\PageIndex{14}\) - Darwin’s first tree.
Explore (optional)
If you would like to read more about the concepts introduced on this page, check out the following web sources:
What is evolution? A short essay by Laurence A. Moran exploring the definition of evolution. (bioinfo.med.utoronto.ca/Evolu...Evolution.html)
Evolution is a Fact and a Theory. Another short essay by Moran explaining how evolution is both a fact and a theory. (bioinfo.med.utoronto.ca/Evolu...nd_Theory.html)
Why is evolution controversial anyway? A short PBS video discussing the controversy of evolutionary theory. You will need Quicktime or RealPlayer to watch this video. This page has several videos on it—look for the video entitled Why is evolution controversial anyway? ( http://www.pbs.org/wgbh/evolution/educators/teachstuds/svideos.html )
References
- Ambrose S. Population bottleneck. In: Robinson R, editor. Genetics, Vol. 3. New York (NY): Macmillan Reference USA; 2003. p. 167-171.
- Artmann S. Neo-Darwinism, Origin of. In: Birx, HJ, editor. Encyclopedia of anthropology, Vol. 4. Thousand Oaks (CA): SAGE Reference; 2006. p. 1727-1732.
- Dusheck J. Population genetics. In: Robinson R, editor. Genetics, Vol. 3. New York (NY): Macmillan Reference, USA, 2003. p. 171-174.
- Ernst Mayr and the evolutionary synthesis [Internet]. Boston (MA): WGBH Educational Foundation and Clear Blue Sky Productions; c2001 [cited 2015 May 12]. Available from < http://www.pbs.org/wgbh/evolution/library/06/2/l_062_01.html >.
- Evol3000 [Internet]. 2013 November 13 [cited 2015 May 12]; Available from: wallace.genetics.uga.edu/groups/evol3000/wiki/fb221/Bottlenecks_and_Founder_Effects.html.
- Evolution 101 [Internet]. Berkeley (CA): University of California Museum of Paleontology; c2015 [cited 2015 May 12]. Available from evolution.berkeley.edu/evolibrary/article/evo_01.
- Ferrell K. The modern synthesis of evolutionary theory. In: Schlage N, Lauer J, editors. Science and its times, Vol. 6: 1900 to 1949. Detroit (MI): Gale; 2000. p. 104-107.
- Gould JL. Genetic drift. In: Birx HJ, editor. Encyclopedia of anthropology, Vol. 3. Thousand Oaks (CA): SAGE Reference; 2006. p. 1036.
- Heddle J. Mutation. In: Robinson R, editor. Genetics, Vol. 3. New York (NY): Macmillan Reference USA; 2003. p. 93-98.
- Jurmain R, Kilgore L, Trevathan W. Essentials of physical anthropology, 4th edition. Belmont (CA): Wadsworth, Cengage Learning; 2013. 437 p.
- Larsen CS. Our origins: discovering physical anthropology. New York (NY): W.W Norton & Company, Inc.; 2008. 430 p.
- Mayr, E. Systematics and the Origin of Species. New York (NY): Columbia University Press. Quoted in Provine, WB. Ernst Mayr: genetics and speciation. Genetics. 2004; 167(3): 1041-1046.
- Speer, MC. Crossing over. In: Robinson R, editor. Genetics, Vol. 3. New York (NY): Macmillan Reference USA; 2003. p. 194-197.
- Sykora, P. Darwinism, modern. In: Birx, HJ, editor. Encyclopedia of anthropology, Vol. 2. Thousand Oaks (CA): SAGE Reference; 2006. p. 692-696.


5.4: Add and Subtract Rational Expressions
A fraction is a proportion. The fraction communicates the number of parts out of the whole.
\(\dfrac{a}{b} \dfrac{\textcolor{blue}{\longleftarrow \text{The part is the numerator.}}}{\textcolor{blue}{\longleftarrow \text{The whole is the denominator.}}}\)
For example, the egg carton pictured below contains \(10\) eggs. Part of the eggs are brown (\(7\) eggs) while the rest are white (\(3\) eggs). A fraction \(\dfrac{a}{b}\) quickly communicates the proportion of eggs that are brown or are white. Proportions can also be given as a decimal or percentage.
Proportion of brown eggs: \(\dfrac{7}{10} = 7 ÷ 10 = 0.7 = 70 \%\)
Proportion of white eggs: \(\dfrac{3}{10} = 3 ÷ 10 = 0.3 = 30 \%\)
The proportion of brown and white eggs describes all of the eggs. Intuitively, we know this proportion is \(100 \%\). The math should match our intuition, so let’s investigate.
\(\begin{array} &70 \% + 30 \% &= 100 \% \\ \dfrac{7}{10} + \dfrac{3}{10} &= \dfrac{10}{10} \end{array}\)
Our investigation leads us to deduce that we add the numerators, but not the denominators.
- Add or subtract the numerators.
- Keep the same denominator.
- Reduce if necessary.
\[\dfrac{a}{c} + \dfrac{b}{c} = \dfrac{a+b}{c}\]
What works for real numbers also works for algebraic quantities. Algebra follows the same rules, which is an excellent anchor from which to perform your own algebraic investigations.
Add \(\dfrac{2x}{x+3} + \dfrac{x}{x+3}\)
Solution
The denominators match. Add the two numerators, but keep the denominator.
\(\dfrac{2x}{x+3} + \dfrac{x}{x+3} = \dfrac{2x+x}{x+3} = \dfrac{3x}{x+3}\)
Since the result cannot be further simplified, the answer is \(\dfrac{3x}{x+3}\)
Subtract \(\dfrac{y^2+2y+8}{y^2−2y−15} − \dfrac{y^2+2}{y^2−2y−15}\)
Solution
The denominators are the same. Subtract the numerators. Remember that each numerator is a quantity.
\(\begin{array} &\dfrac{y^2+2y+8}{y^2−2y−15} − \dfrac{y^2+2}{y^2−2y−15} &= \dfrac{y^2+2y+8−(y^2+2)}{y^2−2y−15} &\text{Numerator: subtract the quantity \((y^2 + 2)\).} \\ & = \dfrac{y^2+2y+8−y^2−2}{y^2−2y−15} &\text{Each term of the quantity is subtracted.} \\ & = \dfrac{2y+6}{y^2−2y−15} &\text{The subtraction is complete.} \\& = \dfrac{2(y+3)}{(y−5)(y+3)} &\text{Factor numerator and denominator. Cancel.} \\ &= \dfrac{2}{y−5} &\text{The answer is most simplified.} \end{array}\)
Addition and Subtraction with Different Denominators
In order to add rational numbers or rational expressions, we need to find a common denominator. Create common denominators using equivalent fractions. That is, \(\dfrac{a \cdot \textcolor{red}{d}}{b \cdot \textcolor{red}{d}} = \dfrac{ad}{bd}\). Then use the process for adding fractions with common denominators.
Add \(\dfrac{7}{75a^2b} + \dfrac{13}{180ab}\)
Solution
To find the least common denominator, we play a game of “who has more?” for each denominator. Break down denominators into prime factors. Imagine that each denominator is a set of cards, and each card is a factor of the denominator. Because we have two denominators, we have two players who will play the game, “who has more?”. Whichever player has the most of a factor “wins,” and they place those cards into the winning pile. Ready to play?
Player #\(1\): \(75a^2 \longrightarrow\)
Player #\(2\): \(180ab \longrightarrow\)
| Who has more? | Player #\(1\) | Player #\(2\) | Winning Pile |
| \(2\)'s | \(\checkmark\) | \(2 \cdot 2\) | |
| \(3\)'s | \(\checkmark\) | \(3 \cdot 3\) | |
| \(5\)'s | \(\checkmark\) | \(5 \cdot 5\) | |
| \(a\)'s | \(\checkmark\) | \(a \cdot a\) | |
| \(b\)'s | TIE | TIE | \(b\) |
The Winning Pile \(= 2^2 \cdot 3^2 \cdot 5^2 \cdot a^2 \cdot b\)
The Least Common Denominator \(= 900a^2b\)
Use the least common denominator to create a common denominator. Determine what factors are missing in each denominator. Multiply the missing factors to the denominator and also to the numerator. Then add.
\(\dfrac{7}{75a^2b} + \dfrac{13}{180ab} = \dfrac{7 \textcolor{red}{\cdot 12}}{75a^2b \textcolor{red}{\cdot 12}} + \dfrac{13 \textcolor{red}{\cdot 5a}}{180ab \textcolor{red}{\cdot 5a}} = \dfrac{84}{900a^2b} + \dfrac{65a}{900a^2b} = \dfrac{84+65a}{900a^2b} \).
The answer is: \(\dfrac{84+65a}{900a^2b}\)
Add \(\dfrac{2y+1}{(4y+1)^2(y+6)} + \dfrac{5}{(4y+1)(y+6)}\)
Solution
In the game of “who has more?” the quantities \((4y + 1)\) and \((y + 6)\) act as single numbers.
Player #\(1\): \((4y + 1)^2 (y + 6) \longrightarrow\)
Player #\(2\): \((4y + 1)(y + 6) \longrightarrow\)
Winning Pile = \((4y + 1)^2 (y + 6)\) = Least Common Denominator
\(\begin{array} &\dfrac{2y+1}{(4y+1)^2(y+6)} + \dfrac{5}{(4y+1)(y+6)} &= \dfrac{2y+1}{(4y+1)^2(y+6)} + \dfrac{5 \textcolor{red}{\cdot (4y+1)}}{(4y+1)(y+6) \textcolor{red}{\cdot (4y+1)}} \\ &= \dfrac{2y+1+(20y+5)}{(4y+1)^2(y+6)} &\text{Simplify the numerator, but leave the denominator in factored form.} \\ &= \dfrac{22y+6}{(4y+1)^2(y+6)} &\text{The numerator is simplified. This is the answer!} \end{array}\)
Tip: The numerator can be factored using GCF \(= 2\): \(22y + 6 = 2(11y + 3)\). However, this factorization does not lead to a cancellation. Therefore, the stated numerator is preferred.
Subtract \(\dfrac{u^2+2u}{u^2+3u−10} − \dfrac{2u}{3u−6}\)
Solution
To compare the factors of each denominator, we need to factor.
\( \begin{array} &&\dfrac{u^2+2u}{u^2+3u−10} − \dfrac{2u}{3u−6} = \dfrac{u^2+2u}{(u+5)(u−2)} − \dfrac{2u}{3(u−2)} &\text{The denominators are factored.} \end{array}\)
Play the game of “who has more?” to find the Least Common Denominator. Quantities act as single numbers.
Player #\(1\):
Player #\(2\):
Winning Pile \(= 3(u + 5)(u − 2) =\) Least Common Denominator
\(\begin{array} &\dfrac{u^2+2u}{(u+5)(u−2)} − \dfrac{2u}{3(u−2)} &= \dfrac{\textcolor{red}{3}(u^2+2u)}{\textcolor{red}{3}(u+5)(u−2)} − \dfrac{2u \textcolor{red}{(u+5)}}{3(u−2) \textcolor{red}{(u+5)}} &\text{Determine the missing factors & multiply.} \\ &= \dfrac{3u^2+6u}{3(u+5)(u−2)} − \dfrac{2u^2+10u}{3(u+5)(u−2)} &\text{Simplify numerators.} \\ &= \dfrac{3u^2+6u−(2u^2+10u)}{3(u+5)(u−2)} &\text{Numerator: subtract the quantity.} \\ &= \dfrac{3u^2+6u−2u^2−10u}{3(u+5)(u−2)} &\text{Subtract each term of the quantity.} \\ &= \dfrac{u^2−4u}{3(u+5)(u−2)} &\text{The numerator is simplified. This is the answer!} \end{array}\)
Tip: The numerator can be factored using GCF \(= u\): \(u^2 − 4u = u(u − 4)\). However, this factorization does not lead to a cancellation. Therefore, the stated numerator is preferred.
Subtract \(\dfrac{3c}{49−c^2} − \dfrac{1}{c^2−7c}\)
Solution
To compare the factors of each denominator, we need to factor.
\(\begin{array} &&\dfrac{3c}{(7−c)(7+c)} − \dfrac{1}{c(c−7)} &\text{The denominators are factored.} \end{array}\)
Play the game of “who has more?” to find the Least Common Denominator. Quantities act as single numbers.
Player #\(1\):
Player #\(2\):
Note: \(7 − c\) and \(c − 7\) are opposites. Since they are opposites, one quantity can be made to “look like” the other.
\(7 − c = −(c − 7)\)
Next, bounce the negative to the numerator. Voilà! This is how we make a Least Common Denominator.
Winning Pile \(= c(c + 7)(c − 7) =\) Least Common Denominator
\(\begin{array} &\dfrac{3c}{(7−c)(7+c)} − \dfrac{1}{c(c−7)} &= \dfrac{3c}{\textcolor{red}{−}(c−7)(7+c)} −\dfrac{1}{c(c−7)} &\text{Replace \(7 − c\) with \(−(c − 7)\).} \\& = \dfrac{\textcolor{red}{−}3c}{(c−7)(7+c)} − \dfrac{1}{c(c−7)} &\text{Since \(\dfrac{a}{−b} = \dfrac{-a}{b}\), put the negative on top.} \\ & = \dfrac{−3c \cdot \textcolor{red}{c}}{\textcolor{red}{c}(c−7)(c+7)} − \dfrac{1 \textcolor{red}{\cdot (c+7)}}{c(c−7) \textcolor{red}{(c+7)}} &\text{Construct the LCD. Note: \(7 + c = c + 7\).} \\ & = \dfrac{−3c^2−(c+7)}{c(c−7)(c+7)} &\text{Numerator: subtract the quantity.} \\ &= \dfrac{−3c^2−c−7}{c(c−7)(c+7)} &\text{The numerator has been simplified.} \\ &= \dfrac{− 3c^2+c+7}{c(c−7)(c+7)} &\text{Factor out \(−1\).} \end{array}\)
Either of the last two lines could be your final answer. Which do you prefer? It’s your choice!
After simplifying the numerator, pause and ask yourself, “what more can I do to simplify?” If the numerator is factorable, check to see if the factors are the same as those in the denominator. If so, cancel the common factors. Keep the denominator factored to remind yourself (and your reader) what factors do not cancel.
Try It! (Exercises)
For #1-20, add or subtract the rational expressions as indicated.
- \(\dfrac{5x}{3y^2} + \dfrac{x}{3y^2}\)
- \(\dfrac{14a}{5b} − \dfrac{4a}{5b}\)
- \(\dfrac{4r}{9qp} + \dfrac{1}{3q^2p}\)
- \(\dfrac{1−n}{6n} + \dfrac{2n+1}{3n}\)
- \(\dfrac{n+2}{12m^3} − \dfrac{2}{3m^3n}\)
- \(\dfrac{t+6}{t+5} + \dfrac{t+4}{t+5}\)
- \(\dfrac{6c+1}{(5c+2)^3} − \dfrac{c−1}{(5c+2)^3}\)
- \(\dfrac{2}{x−3} + \dfrac{3}{10}\)
- \(\dfrac{4}{y+2} + \dfrac{2}{3y+6}\)
- \(\dfrac{3}{x+3} − \dfrac{3}{x−3}\)
- \(\dfrac{h−2}{(h+1)^2} + \dfrac{2}{3h+3}\)
- \(\dfrac{5}{y−1} − \dfrac{3}{y^2−1}\)
- \(\dfrac{−2}{a^2−5a+6} + \dfrac{2}{a−3}\)
- \(\dfrac{4}{d−3} − \dfrac{5d−12}{d^2−6d+9}\)
- \(\dfrac{5}{v^2−5v} − \dfrac{v}{5v−25}\)
- \(\dfrac{1−n}{n^2−4} − \dfrac{n−2}{n^2−2n−8}\)
- \(3 + \dfrac{x}{ x+2}\)
- \(\dfrac{3p−1}{p^2+5p+6} + \dfrac{2p−3}{p^2+3p+2}\)
- \(\dfrac{3}{q+3} + \dfrac{2q}{q^2+8q+15}\)
- \(\dfrac{w+2}{4w+16} − \dfrac{2}{w^2+4w}\)
For #21-30, opposite quantities occur. Add or subtract the rational expressions as indicated.
- \(\dfrac{3}{8} + \dfrac{5}{−8}\)
- \(\dfrac{y^2}{y−5} + \dfrac{4y+5}{5−y}\)
- \(\dfrac{6}{w−2} + \dfrac{w+3}{2−w}\)
- \(\dfrac{2a−7}{5a−8} + \dfrac{6+10a}{8−5a}\)
- \(\dfrac{u^2−2}{2u^2−u−3} + \dfrac{u−2}{3−2u}\)
- \(\dfrac{b−23}{b^2−b −20} − \dfrac{2}{5−b}\)
- \(\dfrac{t+1}{t−1} − \dfrac{3t^2−1}{1−t^2}\)
- \(\dfrac{2}{6−3x} + \dfrac{x}{x^2−4}\)
- \(\dfrac{5r}{1−2r} − \dfrac{2r}{2r+1} + \dfrac{3}{4r^2−1}\)
- \(\dfrac{2y}{y^2−9} − 1 + \dfrac{4}{3−y}\)


4.2: Converting with Base Systems
- Convert another base to base 10.
- Write numbers in different base systems.
- Convert base 10 to other bases.
- Determine errors in converting between bases.
In our system of numbers, we use base 10, but using base 10 was not a given within other systems. There were other systems that used bases other than 10, as we saw with the Mayans and the Babylonians. The base 10 system comes down to grouping objects in sets of 10, but grouping in sets of 10 only happens if the culture values grouping by that many. We feel 10 is natural because we have 10 fingers. There are other systems using other grouping values, such as 4 or 20.
One good reason for examining other bases is to remind ourselves how we had to learn arithmetic when we were young, memorizing rules for our base 10 system. We had to learn why those arithmetic rules made sense, such as why 1+1=2 and 1+2=3. Another good reason for learning other base systems is due to computers; their circuitry instead uses base 2.
In this section, we explore other base systems and how to convert between them.
Conversion of Another Base into Base 10 and Other Bases
We saw in Hindu-Arabic Positional System that our Hindu-Arabic system uses base 10 , which is a system using place values of digits that depend on powers of 10 (or, are based on powers of 10). We’ve already worked with bases other than base 10: The Babylonian system was base 60, while the Mayan system was base 20.
To explore how our base 10 system is used, answer the following question: What’s the following quantity: 4,572? You probably said four thousand five hundred seventy-two (no, there is no “and” between hundred and seventy). But why do you think that 4 means four thousand? A very young person when learning their numbers might say that’s a four five seven and two. But you added the context of thousands to the four. Why?
Place value, that’s why. You learned early on that where the numeral was gave it different meanings. Ten thousands, thousands, hundreds, tens, and ones. So, you translate that symbol string (4,572) into “four thousand five hundred seventy-two.” As we saw in Hindu-Arabic Positional System, expanding a Hindu-Arabic number involved writing the number using each digit times its appropriate power of 10. So, we could write 4,572 as .
One possible reason we use base 10 is that we have 10 fingers, and in the cultures where the Hindu-Arabic system developed, that became the standard. Other cultures may have used other ways of organizing numbers, perhaps using 20 by including toes, or using 60 because 60 has many divisors. Mathematically though, base 10 is an awkward base to work in since 10 has limited divisors. But we think it is easy and simple because that’s what we’ve been taught to use.
Using a base 10 system means we need 10 symbols to make our numbering system work: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Now imagine that we all only had 6 fingers instead of 10 and our counting system was based on those 6 fingers. We would be counting in groups of 6, not groups of 10. How would this change how we work with quantity?
First, we’d need only six symbols. Let’s use 0, 1, 2, 3, 4, 5. Second, our place values would be based on powers of 6, not powers of 10. For instance, the number 3,024 in base 6 would be . That is how you can translate a base 6 number into a base 10 number. When we calculate that expression we get .
This means the base 6 number 3,024 is equal to the base 10 number 664.
From now on, if we are using a base 6 number, we will follow it with the subscript 6, like the following: 3,024 6 means the number is in base 6.
A base 10 number gets no subscript (it’s the standard). So, 3,024 is a base 10 number. A base 13 number would be 4,672 13 .
So, a base 6 system uses only the symbols 0, 1, 2, 3, 4, and 5. Also, the place values use powers of 6. However, we still don’t know how to count in base 6. In order to do so, we’d have to know how to represent the quantities larger than five in base 6. Let’s review how our base 10 system works by counting from 0 to 100, which shows how larger values are represented.
In writing the base 10 numbers, you start with these first 10 values:
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
But you’ve run out of symbols. So, we use two digits:
| 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 |
The 1 out front means you’ve run out of digits one time.
But now you’ve run out twice. Continuing with those numbers gives:
| 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 |
And so on,
| 30 | 31 | 32 | 33 | 34 | etc.… |
Eventually, you hit the 90s,
| 90 | 91 | 92 | 93 | 94 | 95 | 96 | 97 | 98 | 99 |
And you’ve run out of the digits again! So, we say we’ve run out of digits in the tens place one time, hence:
| 100 | 101 | 102 | 103 | 104 | 105 | 106 | 107 | 108 | 109 |
That’s the pattern we use in base 10. We write out the symbols until we’ve used all the symbols, then add a digit in front that counts how many times we’ve used the digits. Knowing the numbers, or being able to count higher and higher, is necessary to understand how all the arithmetic works, as it all goes back to counting.
The counting pattern is the same for any other base, including base 6. So, let’s start:
| 0 | 1 | 2 | 3 | 4 | 5 |
But we’ve run out of symbols! Just like in base 10, we use a second digit, where the first digit will tell us we’ve run out of symbols one time.
| 10 | 11 | 12 | 13 | 14 | 15 |
And we use the same pattern:
| 20 | 21 | 22 | 23 | 24 | 25 |
| 30 | 31 | 32 | 33 | 34 | 35 |
| 40 | 41 | 42 | 43 | 44 | 45 |
| 50 | 51 | 52 | 53 | 54 | 55 |
But we’ve run out of symbols for that front digit. So, we indicate it the same way as in base 10…by adding a third digit in front, indicating we’ve run out of symbols once in the second place:
| 100 | 101 | 102 | 103 | 104 | 105 |
| 110 | 111 | etc. |
The symbol pattern is the same, but truncated. We only use the six symbols. So that is how we represent base 6. Being able to write out these numbers is important when working with addition in the base.
When using a base larger than 10, though, we need more symbols. Instead of creating new symbols, we use capital letters, with A representing the digit for "10," B representing the digit for "11," and so on.
What are the digits used for base 7?
- Answer
-
Since this is base 7, we need only 7 symbols: 0, 1, 2, 3, 4, 5, 6.
What are the digits for base 4?
What are the digits used for base 14?
- Answer
-
Since this is base 14, we need 14 symbols. We don’t have single character numbers for 10, 11, 12, and 13, so, in a fit of inspired creativity, we use capital letters A, B, C, D to represent those quantities. So, the digits in base 14 are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D.
What are the digits used for base 12?
As mentioned in the text, working in base 10 is mathematically awkward. Ten has only two natural number divisors: 2 and 5. This means dividing into groups is not easy. However, 12, or a dozen, has more divisors: 2, 3, 4, and 6. The Dozenal Society recognizes this more mathematically pleasant detail. It advocates for a switch to using base 12 for numbers. Their argument is based on the divisibility of the number 12. But has there ever been a society that used such a system? The answer is yes. A dialect of the Gwandara language in Nigeria uses the base 12 system. It is unlikely, though, that the Dozenal Society will achieve their goal, as the base 10 system is so entrenched in our society.
Convert 3,601 7 into base 10.
- Answer
-
In base 7, the place values are powers of 7. Since there are four digits, the highest power of 7 that is used is 3. This yields .
Convert Base 7 to Base 10
Convert 421 6 into base 10.
Convert 4B7 14 into base 10.
- Answer
-
In base 14, the place values are powers of 14. Since there are three digits, the highest power of 14 is 2. Also recall that in base 14, 10 is represented by A, 11 is represented by B, 12 is represented by C, and 13 is represented by D. Using that, we convert to base 10:
4 B 7 14 = 4 × 14 2 + B × 14 1 + 7 × 14 0 = 4 × 196 + 11 × 14 + 7 × 1 = 784 + 154 + 7 = 945 4 B 7 14 .= 4 × 14 2 + B × 14 1 + 7 × 14 0 = 4 × 196 + 11 × 14 + 7 × 1 = 784 + 154 + 7 = 945
Convert A3C 14 into base 10.
Convert A16 12 into base 10.
- Answer
-
In base 12, the place values are powers of 12. Since there are three digits, the highest power of 12 is 2. Also recall that in base 12, 10 is represented by A and 11 is represented by B. Using that, we convert to base 10:
A 16 14 = 10 × 12 2 + 1 × 12 1 + 6 × 12 0 = 10 × 144 + 1 × 12 + 6 × 1 = 1,440 + 12 + 6 = 1, 458 A 16 14 .= 10 × 12 2 + 1 × 12 1 + 6 × 12 0 = 10 × 144 + 1 × 12 + 6 × 1 = 1,440 + 12 + 6 = 1,458
Convert 5AB 12 into base 10.
Convert 1011 2 into base 10.
- Answer
-
In base 2, the place values are powers of 2. Since there are four digits, the highest power of 2 is 3. Using that, we convert to base 10:
.
Convert 11011 2 into base 10.
Before Napoleon’s France, which adopted the base 10 system, a modified base 12 system was often used in Europe. Twelve is easily divisible into groups of 2, 3, 4, and 6, which makes it easier to work with. Even our numbering system retains a bit of this. You have likely noticed that we use the words thirteen , fourteen , fifteen , and so on to indicate 10 and 3, 10 and 4, 10 and 5, and so one. Even the 20s reinforce this idea, as in twenty-one, and twenty-two. However, two numbers don’t follow this pattern, namely 11 and 12. If they followed the same rules, they’d be one teen and two teen. We even have a special word for 12; that is, a dozen. However, etymologically speaking, the words eleven and twelve are likely derived by referencing the number 10. These two numbers may date back to the Old English words endleofan and twelf , which can be traced back further to ain lif and twa lif . The word lif here may be the base word for “to leave.” This would suggest ain lif is one left after 10, and twa lif is two left after 10, or, 11 and 12.
Write the numbers in base 7 up to 100 7 .
- Answer
-
Step 1: Using the patterns we indicated earlier, we begin with the first seven digits.
0, 1, 2, 3, 4, 5, 6
Step 2: Since we’ve run out of digits, we start with 10, indicating we’ve run out of symbols once.
10, 11, 12, 13, 14, 15, 16
Step 3: Continuing in the same way, we get:
20, 21, 22, 23, 24, 25, 26
30, 31, 32, 33, 34, 35, 36
40, 41, 42, 43, 44, 45, 46
50, 51, 52, 53, 54, 55, 56
60, 61, 62, 63, 64, 65, 66
Now, all the digits have been used in the leading digits. Since the digits have all been used in that leading digit, we use 100, as in base 10.
100
Write the numbers of base 4 up to 100 4 .
Write the numbers in base 14 up to 100 14 .
- Answer
-
Step 1: Using the patterns we indicated earlier, we begin with the first 14 digits.
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D
Step 2: Since we’ve run out of digits, we start with 10, indicating we’ve run out of symbols once.
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 1C, 1D
Step 3: Continuing in the same way, we get:
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 2A, 2B, 2C, 2D
30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 3A, 3B, 3C, 3D
40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 4A, 4B, 4C, 4D
50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 5A, 5B, 5C, 5D
60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 6A, 6B, 6C, 6D
70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 7A, 7B, 7C, 7D
80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 8A, 8B, 8C, 8D
90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 9A, 9B, 9C, 9D
A0, A1, A2, A3, A4, A5, A6, A7, A8, A9, AA, AB, AC, AD
B0, B1, B2, B3, B4, B5, B6, B7, B8, B9, BA, BB, BC, BD
C0, C1, C2, C3, C4, C5, C6, C7, C8, C9, CA, CB, CC, CD
D0, D1, D2, D3, D4, D5, D6, D7, D8, D9, DA, DB, DC, DD
100
Write the numbers of base 12 up to 100 4 .
Base 2 is important in the digital age, as it is the system used by computers. It is the simplest base to work with, but has the drawback that the numbers in base 2 may use many, many digits. In Addition and Subtraction in Base Systems and Multiplication and Division in Base Systems, we will look at base 2 in each situation.
Write the numbers in base 2 up to 100 2 .
- Answer
-
Base 2 uses only two symbols: 0 and 1. Following the pattern established previously, the numbers in base 2 up to 100 2 are 0, 1, 10, 11, and 100.
Write the numbers in base 3 up to 100 3 .
Before the British arrived in Hawaii, people there used a system that combined two different bases. Objects were initially grouped into collections of four, and a collection of four was referred to as kauna . A person could have two kauna and three “ones” (in Hindu-Arabic, 11). Or they could have eight kauna and one “ones” (In Hindu-Arabic, 33). However, sets of four were grouped in collections of 10. A set of 10 kauna was ka’au . The collections of ka’au were grouped by 10 also. Which meant that 10 ka’au (this is 40 in Hindu-Arabic) would be lau (or 400 in Hindu-Arabic). What this shows is that the Hawaiian culture developed a system that used base 4 combined with base 10.
Conversion of Base 10 into Another Base
Converting from base 10 into another base uses repeated division, recording the remainder at each step. Then, the number in the new base is the remainder starting from the last remainder found. To be accurate in what we’re saying, we need to remind ourselves of some terminology associated with division. When integers are divided, the one being divided is the dividend , and the one that is dividing the dividend is the divisor . The quotient is the largest natural number that can be multiplied by the divisor where the product is less than the dividend.
When the integer is divided by the integer , is called the dividend and is the divisor.
To convert a base 10 number into base , we divide by , recording the remainder. Then we divide the quotient from that step by the base , and record the remainder again. We continue this process until the quotient is 0. Then, the base number has digits that start with the last remainder and use each remainder in reverse order.
Convert 298 to base 6.
- Answer
-
We divide 298 by 6, and record the remainder. Then we divide the quotient from that step by 6, and record the remainder again. We continue this process until the quotient is 0. Then, the base 6 number has digits that start with the last remainder and use each remainder in reverse order.
Step 1: When we divide 298 by 6, we get
4 9 r 4 6 298 . The quotient is 49 and the remainder is 4.4 9 r 4 6 298 Step 2: Now we divide the quotient, 49, by 6. This gives
8 r 1 6 49 . The quotient is 8 and the remainder is 1.8 r 1 6 49 Step 3: Repeating, we get
1 r 2 6 8 . The quotient is 1 and the remainder is 2.1 r 2 6 8 Step 4: Finally, we perform the operation on the quotient 1,
0 r 1 6 1 giving us a quotient of 0 and a remainder of 1.0 r 1 6 1 Step 5: The base 6 number has digits equal to the remainders in reverse order, 1214 6 . So, 298 in base 10 when converted to base 6 is 1214 6 .
Convert 693 to base 7.
Converting from Base 10 to Another Base
Convert 45,134 to base 13.
- Answer
-
We divide 45,134 by 13, and record the remainder. Then we divide the quotient from that step by 13, and record the remainder again. We continue this process until the quotient is 0. Then, the base 13 number has digits that start with the last remainder and use each remainder in reverse order. It is at this step that we’ll convert to the base 13 digits, which are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C.
Step 1: When we divide 45,134 by 13, we get
471 r 11 13 45 ,134 . The quotient is 3,471 and the remainder is 11.3, 471 r 11 13 45 ,134 Step 2: Now we divide the quotient, 3,471, by 13. This gives
267 r 0 13 3 , 471 267 r 0 13 3 . The quotient is 267 and the remainder is 0.,471 Step 3: Repeating, we get
20 r 7 13 267 . The quotient is 20 and the remainder is 7.20 r 7 13 267 Step 4: Again, and we get
1 r 7 13 20 . The quotient is 1 and the remainder is 7.1 r 7 13 20 Step 5: Finally, we get
0 r 1 13 1 , with quotient 0 and a remainder 1.0 r 1 13 1 Step 6: The base 13 number has digits equal to the remainders in reverse order, which were 1, 7, 7, 0, and 11. The 11 is written as B in base 13. So, 45,134 in base 10 when converted to base 13 is 1770B 13 .
Convert 9,275 to base 12.
Convert 100 to base 2.
- Answer
-
Following the pattern above:
Step 1: We divide 100 by 2, and record the remainder.
Step 2: Then we divide the quotient from that step by 2, and record the remainder again.
Step 3: We continue this process until the quotient is 0.
Step 4: Following this process, the remainders are, in order, 0, 0, 1, 0, 0, 1, 1. Writing those in reverse order gives the number in base 2, 1100100 2 .
Notice that 100 in base 2 took seven digits.
Convert 137 to base 2.
Converting from Hindu-Arabic Numbers to Mayan Numbers
To convert from a Hindu-Arabic number to a Mayan number involves two distinct processes. First, the number must be converted to base 20, using the process described and demonstrated previously. Next, that base 20 number has to be written using Mayan numerals. For reference, the Mayan numerals and their values are below.
Convert the following into Mayan numbers.
- 51
- 653
- Answer
-
-
The Mayan system is base 20, so we must use 20 in the process from above. The first division has a quotient of 2 and remainder of 11. The 11 serves as the “ones” digit. Dividing that quotient, 2, by 20 has a quotient of 0 with a remainder of 2. The 2 becomes the “twenties” digit of the number. So, in base 20, the number would be 2 followed by 11. The Mayan symbols for 2 and 11 are
and
. Writing these vertically, with the “ones” digit on top, as appropriate for Mayan numbers, results in:
-
The Mayan system is base 20, so we must use 20 in the process from above. The first division, 673 divided by 20, has a quotient of 32 and remainder of 13. Dividing that quotient, 32, by 20 has a quotient of 1 with a remainder of 12. Dividing that quotient, 1, by 20 has a quotient of 0 and a remainder of 1. Since there are three remainders here, this is a three-digit number. The 1 is the “20-squared” digit, the 12 is the “twenties” digit, and the 13 is the “ones” digit. So, in base 20, the number would be 1 followed by 12 followed by 13.. The Mayan symbols for 1, 12 and 13 are
,
, and
. Writing these vertically, as appropriate for Mayan numbers, would result in:
-
The Mayan system is base 20, so we must use 20 in the process from above. The first division has a quotient of 2 and remainder of 11. The 11 serves as the “ones” digit. Dividing that quotient, 2, by 20 has a quotient of 0 with a remainder of 2. The 2 becomes the “twenties” digit of the number. So, in base 20, the number would be 2 followed by 11. The Mayan symbols for 2 and 11 are
and
. Writing these vertically, with the “ones” digit on top, as appropriate for Mayan numbers, results in:
Convert the following into Mayan numbers.
1372.
2,171
There have been base systems that use bases other than 10. Some bases used were 20, 12, and 27! Visit this site to see more on the languages that used other bases .
Errors in Converting Between Bases
There are some common errors that are made when converting between bases. Often, it comes down to using an “illegal” symbol in the new base.
A base 10 number is converted to base 7 and the result was 2081 7 . Was an error committed? How do you know?
- Answer
-
The result has the digit 8 in it. In base 7, 8 is an illegal symbol. Based on that, an error was committed.
A base 10 number is converted to base 4 and the result was 3702 4 . Was an error committed? How do you know?
When converting from base 10 to another base, an illegal symbol will be used if a mistake was made in the division process used to find the number in the new base. Since the digits are based on the remainders, any remainder that is an illegal symbol would indicate an error.
When changing from base 10 to base 8, the division process resulted in the following remainders: 1, 0, 9, 2, 4. Was an error committed? How do you know?
- Answer
-
The remainders include 9, which in base 8 is an illegal symbol.
When changing from base 10 to base 6, the division process resulted in the following remainders: 5, 0, 0, 10. Was an error committed? How do you know?
Another possible way to detect an error in converting between bases is to count the number of digits. When converting from a higher base to a lower base, the number of digits cannot get smaller. Similarly, when converting from a lower base to a higher base, the number of digits cannot get bigger. So, if a base 10 number is converted to a base 3 number, the number of digits in the new base 3 numbers cannot be less than the number of digits in the base 10 number. Similarly, if a base 7 number is converted to base 10, the number of digits in the base 10 number cannot be more than the number of digits in the original base 7 number.
A five-digit base 10 number is converted to a base 5 number. The base 5 number has four digits. Was an error committed? How do you know?
- Answer
-
Since 10 is larger than 5, the base 5 number cannot have less digits than the base 10 number. Since it did, we know an error has been made.
A six-digit base 12 number is converted into a base 10 number. The base 10 number has five digits. Was an error committed? How do you know?
Check Your Understanding
- In base 25, how many symbols would be necessary?
- In base 18, what would the place value of the 4 be in the number 348 18 be?
- Convert 2304 5 into base 10.
- When counting in base 9, what number would follow 38 9 ?
- Convert 329 into base 8.
- Convert ABC 14 into base 10.
- How do you know a mistake was made when converting from base 10 to base 4 and the result is 152 4 ?


13.2: Properties
The following facts outline some relationships countability and the set operations. They can be used to more easily prove that a set is countable or uncountable using the already-known countability or uncountability of a related set.
- Every subset of \(\mathbb{N}\) is countable.
- If there exists an injection \(A \hookrightarrow \mathbb{N}\text{,}\) then the set \(A\) is countable.
- Suppose \(A \subseteq B\text{.}\) If \(B\) is countable, then so is \(A\text{.}\)
- Suppose \(A \subseteq B\text{.}\) If \(A\) is un countable, then so is \(B\text{.}\)
- If \(A\) and \(B\) are countable, then \(A\cup B\) and \(A\cap B\) are both countable.
- Proof outline.
-
- Assume \(A \subseteq \mathbb{N}\text{.}\) If \(A\) is finite, then it is countable by definition. So assume that \(\vert A \vert = \infty\text{.}\) We can construct a sequence \(\{a_k\}\) that contains each element of \(A\) exactly once as follows.
\begin{align*} a_0 & = \text{ smallest number in } A, \\ a_1 & = \text{ next smallest number in } A, \\ a_2 & = \text{ next smallest number in } A, \\ & \; \vdots \end{align*}
Therefore, \(A\) is countable.- If \(f: A \hookrightarrow \mathbb{N}\) is injective, then \(f: A \rightarrow f(A)\) is a bijection, so that \(A\) and its image \(f(A)\) have the same size. But \(f(A)\) is countable by Statement 1, so using the definition of countable along with Fact 12.3.2 , conclude that \(A\) is countable.
- If \(B\) is countable, then by definition there exists a bijection \(f: B \rightarrow \mathbb{N}\text{.}\) Then \(f\vert _A: A \rightarrow \mathbb{N}\) is an injection. Apply Statement 2.
- This is the contrapositive of Statement 3, under the common assumption \(A \subseteq B\text{.}\)
- For \(A \cap B\text{,}\) consider \(A \cap B \subseteq A\) and apply Statement 3.
Now consider \(A \cup B\text{.}\) For simplicity, we will assume \(A \cap B = \emptyset\text{,}\) so that \(A \cup B = A \sqcup B\text{.}\) Since \(A\) and \(B\) are countable, we can write their elements as sequences:
\begin{align*} A & = \{ \, a_0, \, a_1, \, a_2, \, \ldots \, \}, & B & = \{ \, b_0, \, b_1, \, b_2, \, \ldots \, \} \text{.} \end{align*}
We can then write the elements of \(A \sqcup B\) in a sequence by interleaving these two sequences:\begin{equation*} A \sqcup B = \{ \, a_0, \, b_0, \, a_1, \, b_1, \, a_2, \, b_2, \, \ldots \, \} \text{.} \end{equation*}
Checkpoint \(\PageIndex{1}\)
Prove \(A \cup B\) is countable even in the case \(A \cap B \ne \emptyset\text{.}\)
- Hint.
-
Consider the sets
\begin{align*} A' & = A \smallsetminus (A\cap B), & B' & = B \smallsetminus (A\cap B), & C & = A' \sqcup B'. \end{align*}
Then \(A\cup B\) is the disjoint union of \(C\) and \(A\cap B\text{.}\)
The set of prime numbers is countable, since it is a subset of \(\mathbb{N}\text{.}\)
The unit interval \((0,1)\) on the real number line is uncountable because it contains the uncountable subset \(\scr{C}\) from Lemma 13.1.1 .
Set \(\mathbb{R}\) is uncountable.
- Proof
-
This follows from Lemma 13.1.1 and Statement 4 of Proposition \(\PageIndex{1}\).
The Cartesian product set \(\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}\) is uncountable because it has an uncountable subset: the \(x\)-axis has the same size as \(\mathbb{R}\text{.}\)


6: Thermochemistry
Useful forms of energy are also available from a variety of chemical reactions other than combustion. For example, the energy produced by the batteries in a cell phone, car, or flashlight results from chemical reactions. This chapter introduces many of the basic ideas necessary to explore the relationships between chemical changes and energy, with a focus on thermal energy.
-
- 6.1: Prelude to Thermochemistry
- Useful forms of energy are also available from a variety of chemical reactions other than combustion. For example, the energy produced by the batteries in a cell phone, car, or flashlight results from chemical reactions. This chapter introduces many of the basic ideas necessary to explore the relationships between chemical changes and energy, with a focus on thermal energy.
-
- 6.2: Energy Basics
- Energy is the capacity to do work (applying a force to move matter). Heat is energy that is transferred between objects at different temperatures; it flows from a high to a low temperature. Chemical and physical processes can absorb heat (endothermic) or release heat (exothermic). The SI unit of energy, heat, and work is the joule (J). Specific heat and heat capacity are measures of the energy needed to change the temperature of a substance or object.
-
- 6.4: Calorimetry
- Calorimetry is used to measure the amount of thermal energy transferred in a chemical or physical process. This requires careful measurement of the temperature change that occurs during the process and the masses of the system and surroundings. These measured quantities are then used to compute the amount of heat produced or consumed in the process using known mathematical relations. Calorimeters are designed to minimize energy exchange between the system and its surroundings.
-
- 6.6: Enthalpy
- If a chemical change is carried out at constant pressure and the only work done is caused by expansion or contraction, q for the change is called the enthalpy change with the symbol ΔH. Examples of enthalpy changes include enthalpy of combustion, enthalpy of fusion, enthalpy of vaporization, and standard enthalpy of formation. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess’s law.
Contributors and Attributions
-
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>).


6.7: Inverse and Implicit Functions. Open and Closed Maps
I. "If \(f \in C D^{1}\) at \(\vec{p},\) then \(f\) resembles a linear map (namely \(d f )\) at \(\vec{p}."\) Pursuing this basic idea, we first make precise our notion of "\(f \in C D^{1}\) at \(\vec{p}\)."
A map \(f : E^{\prime} \rightarrow E\) is continuously differentiable, or of class \(C D^{1}\) (written \(f \in C D^{1}),\) at \(\vec{p}\) iff the following statement is true:
\[\begin{array}{l}{\text { Given any } \varepsilon>0, \text { there is } \delta>0 \text { such that } f \text { is differentiable on the }} \\ {\text { globe } \overline{G}=\overline{G_{\vec{p}}(\delta)}, \text { with }} \\ {\qquad\|d f(\vec{x} ; \cdot)-d f(\vec{p} ; \cdot)\|<\varepsilon \text { for all } \vec{x} \in \overline{G}.}\end{array}\]
By Problem 10 in §5, this definition agrees with Definition 1 §5, but is no longer limited to the case \(E^{\prime}=E^{n}\left(C^{n}\right).\) See also Problems 1 and 2 below.
We now obtain the following result.
Let \(E^{\prime}\) and \(E\) be complete. If \(f : E^{\prime} \rightarrow E\) is of class \(C D^{1}\) at \(\vec{p}\) and if \(d f(\vec{p} ; \cdot)\) is bijective (§6), then \(f\) is one-to-one on some globe \(\overline{G}=\overline{G_{\vec{p}}}(\delta).\)
Thus \(f\) "locally" resembles df \((\vec{p} ; \cdot)\) in this respect.
- Proof
-
Set \(\phi=d f(\vec{p} ; \cdot)\) and
\[\left\|\phi^{-1}\right\|=\frac{1}{\varepsilon}\]
(cf. Theorem 2 of §6).
By Definition 1, fix \(\delta>0\) so that for \(\vec{x} \in \overline{G}=\overline{G_{\vec{p}}(\delta)}\).
\[\|d f(\vec{x} ; \cdot)-\phi\|<\frac{1}{2} \varepsilon.\]
Then by Note 5 in §2,
\[(\forall \vec{x} \in \overline{G})\left(\forall \vec{u} \in E^{\prime}\right) \quad|d f(\vec{x} ; \vec{u})-\phi(\vec{u})| \leq \frac{1}{2} \varepsilon|\vec{u}|.\]
Now fix any \(\vec{r}, \vec{s} \in \overline{G}, \vec{r} \neq \vec{s},\) and set \(\vec{u}=\vec{r}-\vec{s} \neq 0.\) Again, by Note 5 in §2,
\[|\vec{u}|=\left|\phi^{-1}(\phi(\vec{u}))\right| \leq\left\|\phi^{-1}\right\||\phi(\vec{u})|=\frac{1}{\varepsilon}|\phi(\vec{u})|;\]
so
\[0<\varepsilon|\vec{u}| \leq|\phi(\vec{u})|.\]
By convexity, \(\overline{G} \supseteq I=L[\vec{s}, \vec{r}],\) so (1) holds for \(\vec{x} \in I, \vec{x}=\vec{s}+t \vec{u}, 0 \leq t \leq 1\).
Noting this, set
\[h(t)=f(\vec{s}+t \vec{u})-t \phi(\vec{u}), \quad t \in E^{1}.\]
Then for \(0 \leq t \leq 1\),
\[\begin{aligned} h^{\prime}(t) &=D_{\vec{u}} f(\vec{s}+t \vec{u})-\phi(\vec{u}) \\ &=d f(\vec{s}+t \vec{u} ; \vec{u})-\phi(\vec{u}). \end{aligned}\]
(Verify!) Thus by (1) and (2),
\[\begin{aligned} \sup _{0 \leq t \leq 1}\left|h^{\prime}(t)\right| &=\sup _{0 \leq t \leq 1}|d f(\vec{s}+t \vec{u} ; \vec{u})-\phi(\vec{u})| \\ & \leq \frac{\varepsilon}{2}|\vec{u}| \leq \frac{1}{2}|\phi(\vec{u})|. \end{aligned}\]
(Explain!) Now, by Corollary 1 in Chapter 5, §4,
\[|h(1)-h(0)| \leq(1-0) \cdot \sup _{0 \leq t \leq 1}\left|h^{\prime}(t)\right| \leq \frac{1}{2}|\phi(\vec{u})|.\]
As \(h(0)=f(\vec{s})\) and
\[h(1)=f(\vec{s}+\vec{u})-\phi(\vec{u})=f(\vec{r})-\phi(\vec{u}),\]
we obtain (even if \(\vec{r}=\vec{s})\)
\[|f(\vec{r})-f(\vec{s})-\phi(\vec{u})| \leq \frac{1}{2}|\phi(\vec{u})| \quad(\vec{r}, \vec{s} \in \overline{G}, \vec{u}=\vec{r}-\vec{s}).\]
But by the triangle law,
\[|\phi(\vec{u})|-|f(\vec{r})-f(\vec{s})| \leq|f(\vec{r})-f(\vec{s})-\phi(\vec{u})|.\]
Thus
\[|f(\vec{r})-f(\vec{s})| \geq \frac{1}{2}|\phi(\vec{u})| \geq \frac{1}{2} \varepsilon|\vec{u}|=\frac{1}{2} \varepsilon|\vec{r}-\vec{s}|\]
by (2).
Hence \(f(\vec{r}) \neq f(\vec{s})\) whenever \(\vec{r} \neq \vec{s}\) in \(\overline{G};\) so \(f\) is one-to-one on \(\overline{G},\) as claimed.\(\quad \square\)
Under the assumptions of Theorem 1, the maps \(f\) and \(f^{-1}\) (the inverse of \(f\) restricted to \(\overline{G}\)) are uniformly continuous on \(\overline{G}\) and \(f[\overline{G}],\) respectively.
- Proof
-
By (3),
\[\begin{aligned}|f(\vec{r})-f(\vec{s})| & \leq|\phi(\vec{u})|+\frac{1}{2}|\phi(\vec{u})| \\ & \leq|2 \phi(\vec{u})| \\ & \leq 2\|\phi\||\vec{u}| \\ &=2\|\phi\||\vec{r}-\vec{s}| \quad(\vec{r}, \vec{s} \in \overline{G}). \end{aligned}\]
This implies uniform continuity for \(f\). (Why?)
Next, let \(g=f^{-1}\) on \(H=f[\overline{G}]\).
If \(\vec{x}, \vec{y} \in H,\) let \(\vec{r}=g(\vec{x})\) and \(\vec{s}=g(\vec{y});\) so \(\vec{r}, \vec{s} \in \overline{G},\) with \(\vec{x}=f(\vec{r})\) and \(\vec{y}=f(\vec{s}).\) Hence by (4),
\[|\vec{x}-\vec{y}| \geq \frac{1}{2} \varepsilon|g(\vec{x})-g(\vec{y})|,\]
proving all for \(g,\) too.\(\quad \square\)
Again, \(f\) resembles \(\phi\) which is uniformly continuous, along with \(\phi^{-1}\).
II. We introduce the following definition.
A map \(f :(S, \rho) \rightarrow\left(T, \rho^{\prime}\right)\) is closed (open) on \(D \subseteq S\) iff, for any \(X \subseteq D\) the set \(f[X]\) is closed (open) in \(T\) whenever \(X\) is so in \(S.\)
Note that continuous maps have such a property for inverse images (Problem 15 in Chapter 4, §2).
Under the assumptions of Theorem 1, \(f\) is closed on \(\overline{G},\) and so the set \(f[\overline{G}]\) is closed in \(E.\)
Similarly for the map \(f^{-1}\) on \(f[\overline{G}]\).
- Proof for \(E^{\prime}=E=E^{n}\left(C^{n}\right)\) (for the general case, see Problem 6)
-
Given any closed \(X \subseteq \overline{G},\) we must show that \(f[X]\) is closed in \(E.\)
Now, as \(\overline{G}\) is closed and bounded, it is compact (Theorem 4 of Chapter 4, §6).
So also is \(X\) (Theorem 1 in Chapter 4, §6), and so is \(f[X]\) (Theorem 1 of Chapter 4, §8).
By Theorem 2 in Chapter 4, §6, \(f[X]\) is closed, as required.\(\quad \square\)
For the rest of this section, we shall set \(E^{\prime}=E=E^{n}\left(C^{n}\right)\).
If \(E^{\prime}=E=E^{n}\left(C^{n}\right)\) in Theorem 1, with other assumptions unchanged, then \(f\) is open on the globe \(G=G_{\vec{p}}(\delta),\) with \(\delta\) sufficiently small.
- Proof
-
We first prove the following lemma.
\(f[G]\) contains a globe \(G_{\vec{q}}(\alpha)\) where \(\vec{q}=f(\vec{p})\).
- Proof
-
Indeed, let
\[\alpha=\frac{1}{4} \varepsilon \delta,\]
where \(\delta\) and \(\varepsilon\) are as in the proof of Theorem 1. (We continue the notation and formulas of that proof.)
Fix any \(\vec{c} \in G_{\vec{q}}(\alpha);\) so
\[|\vec{c}-\vec{q}|<\alpha=\frac{1}{4} \varepsilon \delta.\]
Set \(h=|f-\vec{c}|\) on \(E^{\prime}.\) As \(f\) is uniformly continuous on \(\overline{G},\) so is \(h\).
Now, \(\overline{G}\) is compact in \(E^{n}\left(C^{n}\right);\) so Theorem 2(ii) in Chapter 4, §8, yields a point \(\vec{r} \in \overline{G}\) such that
\[h(\vec{r})=\min h[\overline{G}].\]
We claim that \(\vec{r}\) is in \(G\) (the interior of \(\overline{G})\).
Otherwise, \(|\vec{r}-\vec{p}|=\delta ;\) for by (4),
\[\begin{aligned} 2 \alpha=\frac{1}{2} \varepsilon \delta=\frac{1}{2} \varepsilon|\vec{r}-\vec{p}| & \leq|f(\vec{r})-f(\vec{p})| \\ & \leq|f(\vec{r})-\vec{c}|+|\vec{c}-f(\vec{p})| \\ &=h(\vec{r})+h(\vec{p}). \end{aligned}\]
But
\[h(\vec{p})=|\vec{c}-f(\vec{p})|=|\vec{c}-\vec{q}|<\alpha;\]
and so (7) yields
\[h(\vec{p})<\alpha<h(\vec{r}),\]
contrary to the minimality of \(h(\vec{r})\) (see (6)). Thus \(|\vec{r}-\vec{p}|\) cannot equal \(\delta\).
We obtain \(|\vec{r}-\vec{p}|<\delta,\) so \(\vec{r} \in G_{\vec{p}}(\delta)=G\) and \(f(\vec{r}) \in f[G].\) We shall now show that \(\vec{c}=f(\vec{r}).\)
To this end, we set \(\vec{v}=\vec{c}-f(\vec{r})\) and prove that \(\vec{v}=\overrightarrow{0}.\) Let
\[\vec{u}=\phi^{-1}(\vec{v}),\]
where
\[\phi=d f(\vec{p} ; \cdot),\]
as before. Then
\[\vec{v}=\phi(\vec{u})=d f(\vec{p} ; \vec{u}).\]
With \(\vec{r}\) as above, fix some
\[\vec{s}=\vec{r}+t \vec{u} \quad(0<t<1)\]
with \(t\) so small that \(\vec{s} \in G\) also. Then by formula (3),
\[|f(\vec{s})-f(\vec{r})-\phi(t \vec{u})| \leq \frac{1}{2}|t \vec{v}|;\]
also,
\[|f(\vec{r})-\vec{c}+\phi(t \vec{u})|=(1-t)|\vec{v}|=(1-t) h(\vec{r})\]
by our choice of \(\vec{v}, \vec{u}\) and \(h.\) Hence by the triangle law,
\[h(\vec{s})=|f(\vec{s})-\vec{c}| \leq\left(1-\frac{1}{2} t\right) h(\vec{r}).\]
(Verify!)
As \(0<t<1,\) this implies \(h(\vec{r})=0\) (otherwise, \(h(\vec{s})<h(\vec{r}),\) violating (6)).
Thus, indeed,
\[|\vec{v}|=|f(\vec{r})-\vec{c}|=0,\]
i.e.,
\[\vec{c}=f(\vec{r}) \in f[G] \quad \text { for } \vec{r} \in G.\]
But \(\vec{c}\) was an arbitrary point of \(G_{\vec{q}}(\alpha).\) Hence
\[G_{\vec{q}}(\alpha) \subseteq f[G],\]
proving the lemma.\(\quad \square\)
Proof of Theorem 2. The lemma shows that \(f(\vec{p})\) is in the interior of \(f[G]\) if \(\vec{p}, f, d f(\vec{p} ; \cdot),\) and \(\delta\) are as in Theorem 1.
But Definition 1 implies that here \(f \in C D^{1}\) on all of \(G\) (see Problem 1).
Also, \(d f(\vec{x} ; \cdot)\) is bijective for any \(\vec{x} \in G\) by our choice of \(G\) and Theorems 1 and 2 in §6.
Thus \(f\) maps all \(\vec{x} \in G\) onto interior points of \(f[G];\) i.e., \(f\) maps any open set \(X \subseteq G\) onto an open \(f[X],\) as required.\(\quad \square\)
Note 1. A map
\[f :(S, \rho) \underset{\text { onto }}{\longleftrightarrow} (T, \rho^{\prime})\]
is both open and closed ("clopen") iff \(f^{-1}\) is continuous - see Problem 15(iv)(v) in Chapter 4, §2, interchanging \(f\) and \(f^{-1}.\)
Thus \(\phi=d f(\vec{p} ; \cdot)\) in Theorem 1 is "clopen" on all of \(E^{\prime}\).
Again, \(f\) locally resembles \(d f(\vec{p} ; \cdot)\).
III. The Inverse Function Theorem. We now further pursue these ideas.
Under the assumptions of Theorem 2, let \(g\) be the inverse of \(f_{G}\left(f \text { restricted to } G=G_{\vec{p}}(\delta)\right)\).
Then \(g \in C D^{1}\) on \(f[G]\) and \(d g(\vec{y} ; \cdot)\) is the inverse of \(d f(\vec{x} ; \cdot)\) whenever \(\vec{x}=g(\vec{y}), \vec{x} \in G.\)
Briefly: "The differential of the inverse is the inverse of the differential."
- Proof
-
Fix any \(\vec{y} \in f[G]\) and \(\vec{x}=g(\vec{y}) ;\) so \(\vec{y}=f(\vec{x})\) and \(\vec{x} \in G.\) Let \(U=d f(\vec{x} ; \cdot).\)
As noted above, \(U\) is bijective for every \(\vec{x} \in G\) by Theorems 1 and 2 in §6; so we may set \(V=U^{-1}.\) We must show that \(V=d g(\vec{y} ; \cdot).\)
To do this, give \(\vec{y}\) an arbitrary (variable) increment \(\Delta \vec{y},\) so small that \(\vec{y}+\Delta \vec{y}\) stays in \(f[G]\) (an open set by Theorem 2).
As \(g\) and \(f_{G}\) are one-to-one, \(\Delta \vec{y}\) uniquely determines
\[\Delta \vec{x}=g(\vec{y}+\Delta \vec{y})-g(\vec{y})=\vec{t},\]
and vice versa:
\[\Delta \vec{y}=f(\vec{x}+\vec{t})-f(\vec{x}).\]
Here \(\Delta \vec{y}\) and \(\vec{t}\) are the mutually corresponding increments of \(\vec{y}=f(\vec{x})\) and \(\vec{x}=g(\vec{y}).\) By continuity, \(\vec{y} \rightarrow \overrightarrow{0}\) iff \(\vec{t} \rightarrow \overrightarrow{0}.\)
As \(U=d f(\vec{x} ; \cdot)\),
\[\lim _{\vec{t} \rightarrow \overline{0}} \frac{1}{|\vec{t}|}|f(\vec{x}+\vec{t})-f(\vec{t})-U(\vec{t})|=0,\]
or
\[\lim _{\vec{t} \rightarrow \overrightarrow{0}} \frac{1}{|\vec{t}|}|F(\vec{t})|=0,\]
where
\[F(\vec{t})=f(\vec{x}+\vec{t})-f(\vec{t})-U(\vec{t}).\]
As \(V=U^{-1},\) we have
\[V(U(\vec{t}))=\vec{t}=g(\vec{y}+\Delta \vec{y})-g(\vec{y}).\]
So from (9),
\[\begin{aligned} V(F(\vec{t})) &=V(\Delta \vec{y})-\vec{t} \\ &=V(\Delta \vec{y})-[g(\vec{y}+\Delta \vec{y})-g(\vec{y})]; \end{aligned}\]
that is,
\[\frac{1}{|\Delta \vec{y}|}|g(\vec{y}+\Delta \vec{y})-g(\vec{y})-V(\Delta \vec{y})|=\frac{|V(F(\vec{t}))|}{|\Delta \vec{y}|}, \quad \Delta \vec{y} \neq \overrightarrow{0}.\]
Now, formula (4), with \(\vec{r}=\vec{x}, \vec{s}=\vec{x}+\vec{t},\) and \(\vec{u}=\vec{t},\) shows that
\[|f(\vec{x}+\vec{t})-f(\vec{x})| \geq \frac{1}{2} \varepsilon|\vec{t}|;\]
i.e., \(|\Delta \vec{y}| \geq \frac{1}{2} \varepsilon|\vec{t}|.\) Hence by (8),
\[\frac{|V(F(\vec{t}))|}{|\Delta \vec{y}|} \leq \frac{|V(F(\vec{t}) |}{\frac{1}{2} \varepsilon|\vec{t}|}=\frac{2}{\varepsilon}\left|V\left(\frac{1}{|\vec{t}|} F(\vec{t})\right)\right| \leq \frac{2}{\varepsilon}\|V\| \frac{1}{|\vec{t}|}|F(\vec{t})| \rightarrow 0 \text { as } \vec{t} \rightarrow \overrightarrow{0}.\]
Since \(\vec{t} \rightarrow \overrightarrow{0}\) as \(\Delta \vec{y} \rightarrow \overrightarrow{0}\) (change of variables!), the expression (10) tends to 0 as \(\Delta \vec{y} \rightarrow \overrightarrow{0}.\)
By definition, then, \(g\) is differentiable at \(\vec{y},\) with \(d g(\vec{y};)=V=U^{-1}\).
Moreover, Corollary 3 in §6, applies here. Thus
\[\left(\forall \delta^{\prime}>0\right)\left(\exists \delta^{\prime \prime}>0\right) \quad\|U-W\|<\delta^{\prime \prime} \Rightarrow\left\|U^{-1}-W^{-1}\right\|<\delta^{\prime}.\]
Taking here \(U^{-1}=d g(\vec{y})\) and \(W^{-1}=d g(\vec{y}+\Delta \vec{y}),\) we see that \(g \in C D^{1}\) near \(\vec{y}.\) This completes the proof.\(\quad \square\)
Note 2. If \(E^{\prime}=E=E^{n}\left(C^{n}\right),\) the bijectivity of \(\phi=d f(\vec{p} ; \cdot)\) is equivalent to
\[\operatorname{det}[\phi]=\operatorname{det}\left[f^{\prime}(\vec{p})\right] \neq 0\]
(Theorem 1 of §6).
In this case, the fact that \(f\) is one-to-one on \(G=G_{\vec{p}}(\delta)\) means, componentwise (see Note 3 in §6), that the system of \(n\) equations
\[f_{i}(\vec{x})=f\left(x_{1}, \ldots, x_{n}\right)=y_{i}, \quad i=1, \ldots, n,\]
has a unique solution for the \(n\) unknowns \(x_{k}\) as long as
\[\left(y_{1}, \ldots, y_{n}\right)=\vec{y} \in f[G].\]
Theorem 3 shows that this solution has the form
\[x_{k}=g_{k}(\vec{y}), \quad k=1, \ldots, n,\]
where the \(g_{k}\) are of class \(C D^{1}\) on \(f[G]\) provided the \(f_{i}\) are of class \(C D^{1}\) near \(\vec{p}\) and det \(\left[f^{\prime}(\vec{p})\right] \neq 0.\) Here
\[\operatorname{det}\left[f^{\prime}(\vec{p})\right]=J_{f}(\vec{p}),\]
as in §6.
Thus again \(f\) "locally" resembles a linear map, \(\phi=d f(\vec{p} ; \cdot)\).
IV. The Implicit Function Theorem. Generalizing, we now ask, what about solving \(n\) equations in \(n+m\) unknowns \(x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{m}?\) Say, we want to solve
\[f_{k}\left(x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{m}\right)=0, \quad k=1,2, \ldots, n,\]
for the first \(n\) unknowns (or variables) \(x_{k},\) thus expressing them as
\[x_{k}=H_{k}\left(y_{1}, \ldots, y_{m}\right), \quad k=1, \ldots, n,\]
with \(H_{k} : E^{m} \rightarrow E^{1}\) or \(H_{k} : C^{m} \rightarrow C\).
Let us set \(\vec{x}=\left(x_{1}, \ldots, x_{n}\right), \vec{y}=\left(y_{1}, \ldots, y_{m}\right),\) and
\[(\vec{x}, \vec{y})=\left(x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{m}\right)\]
so that \((\vec{x}, \vec{y}) \in E^{n+m}\left(C^{n+m}\right)\).
Thus the system of equations (11) simplifies to
\[f_{k}(\vec{x}, \vec{y})=0, \quad k=1, \ldots, n\]
or
\[f(\vec{x}, \vec{y})=\overrightarrow{0},\]
where \(f=\left(f_{1}, \ldots, f_{n}\right)\) is a map of \(E^{n+m}\left(C^{n+m}\right)\) into \(E^{n}\left(C^{n}\right) ; f\) is a function of \(n+m\) variables, but it has \(n\) components \(f_{k};\) i.e.,
\[f(\vec{x}, \vec{y})=f\left(x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{m}\right)\]
is a vector in \(E^{n}\left(C^{n}\right)\).
Let \(E^{\prime}=E^{n+m}\left(C^{n+m}\right), E=E^{n}\left(C^{n}\right),\) and let \(f : E^{\prime} \rightarrow E\) be of class \(C D^{1}\) near
\[(\vec{p}, \vec{q})=\left(p_{1}, \ldots, p_{n}, q_{1}, \ldots, q_{m}\right), \quad \vec{p} \in E^{n}\left(C^{n}\right), \vec{q} \in E^{m}\left(C^{m}\right).\]
Let \([\phi]\) be the \(n \times n\) matrix
\[\left(D_{j} f_{k}(\vec{p}, \vec{q})\right), \quad j, k=1, \ldots, n.\]
If \(\operatorname{det}[\phi] \neq 0\) and if \(f(\vec{p}, \vec{q})=\overrightarrow{0},\) then there are open sets
\[P \subseteq E^{n}\left(C^{n}\right) \text { and } Q \subseteq E^{m}\left(C^{m}\right),\]
with \(\vec{p} \in P\) and \(\vec{q} \in Q,\) for which there is a unique map
\[H : Q \rightarrow P\]
with
\[f(H(\vec{y}), \vec{y})=\overrightarrow{0}\]
for all \(\vec{y} \in Q;\) furthermore, \(H \in C D^{1}\) on \(Q\).
Thus \(\vec{x}=H(\vec{y})\) is a solution of (11) in vector form.
- Proof
-
With the above notation, set
\[F(\vec{x}, \vec{y})=(f(\vec{x}, \vec{y}), \vec{y}), \quad F : E^{\prime} \rightarrow E^{\prime}.\]
Then
\[F(\vec{p}, \vec{q})=(f(\vec{p}, \vec{q}), \vec{q})=(\overrightarrow{0}, \vec{q}),\]
since \(f(\vec{p}, \vec{q})=\overrightarrow{0}\).
As \(f \in C D^{1}\) near \((\vec{p}, \vec{q}),\) so is \(F\) (verify componentwise via Problem 9(ii) in §3 and Definition 1 of §5).
By Theorem 4, §3, \(\operatorname{det}\left[F^{\prime}(\vec{p}, \vec{q})\right]=\operatorname{det}[\phi] \neq 0\) (explain!).
Thus Theorem 1 above shows that \(F\) is one-to-one on some globe \(G\) about \((\vec{p}, \vec{q}).\)
Clearly \(G\) contains an open interval about \((\vec{p}, \vec{q}).\) We denote it by \(P \times Q\) where \(\vec{p} \in P, \vec{q} \in Q ; P\) is open in \(E^{n}\left(C^{n}\right)\) and \(Q\) is open in \(E^{m}\left(C^{m}\right).\)
By Theorem 3, \(F_{P \times Q}\) (\(F\) restricted to \(P \times Q)\) has an inverse
\[g : A \underset{\text { onto }}{\longleftrightarrow} P \times Q,\]
where \(A=F[P \times Q]\) is open in \(E^{\prime}\) (Theorem 2), and \(g \in C D^{1}\) on \(A.\) Let the map \(u=\left(g_{1}, \ldots, g_{n}\right)\) comprise the first \(n\) components of \(g\) (exactly as \(f\) comprises the first \(n\) components of \(F )\).
Then
\[g(\vec{x}, \vec{y})=(u(\vec{x}, \vec{y}), \vec{y})\]
exactly as \(F(\vec{x}, \vec{y})=(f(\vec{x}, \vec{y}), \vec{y}).\) Also, \(u : A \rightarrow P\) is of class \(C D^{1}\) on \(A,\) as \(g\) is (explain!).
Now set
\[H(\vec{y})=u(\overrightarrow{0}, \vec{y});\]
here \(\vec{y} \in Q,\) while
\[(\overrightarrow{0}, \vec{y}) \in A=F[P \times Q],\]
for \(F\) preserves \(\vec{y}\) (the last \(m\) coordinates). Also set
\[\alpha(\vec{x}, \vec{y})=\vec{x}.\]
Then \(f=\alpha \circ F\) (why?), and
\[f(H(\vec{y}), \vec{y})=f(u(\overrightarrow{0}, \vec{y}), \vec{y})=f(g(\overrightarrow{0}, \vec{y}))=\alpha(F(g(\overrightarrow{0}, \vec{y}))=\alpha(\overrightarrow{0}, \vec{y})=\overrightarrow{0}\]
by our choice of \(\alpha\) and \(g\) (inverse to \(F).\) Thus
\[f(H(\vec{y}), \vec{y})=\overrightarrow{0}, \quad \vec{y} \in Q,\]
as desired.
Moreover, as \(H(\vec{y})=u(\overrightarrow{0}, \vec{y}),\) we have
\[\frac{\partial}{\partial y_{i}} H(\vec{y})=\frac{\partial}{\partial y_{i}} u(\overrightarrow{0}, \vec{y}), \quad \vec{y} \in Q, i \leq m.\]
As \(u \in C D^{1},\) all \(\partial u / \partial y_{i}\) are continuous (Definition 1 in §5); hence so are the \(\partial H / \partial y_{i}.\) Thus by Theorem 3 in §3, \(H \in C D^{1}\) on \(Q.\)
Finally, \(H\) is unique for the given \(P, Q;\) for
\[\begin{aligned} f(\vec{x}, \vec{y})=\overrightarrow{0} & \Longrightarrow(f(\vec{x}, \vec{y}), \vec{y})=(\overrightarrow{0}, \vec{y}) \\ & \Longrightarrow F(\vec{x}, \vec{y})=(\overrightarrow{0}, \vec{y}) \\ & \Longrightarrow g(F(\vec{x}, \vec{y}))=g(\overrightarrow{0}, \vec{y}) \\ & \Longrightarrow(\vec{x}, \vec{y})=g(\overrightarrow{0}, \vec{y})=(u(\overrightarrow{0}, \vec{y}), \vec{y}) \\ & \Longrightarrow \vec{x}=u(\overrightarrow{0}, \vec{y})=H(\vec{y}). \end{aligned}\]
Thus \(f(\vec{x}, \vec{y})=\overrightarrow{0}\) implies \(\vec{x}=H(\vec{y});\) so \(H(\vec{y})\) is the only solution for \(\vec{x}. \quad \square\)
Note 3. \(H\) is said to be implicitly defined by the equation \(f(\vec{x}, \vec{y})=\overrightarrow{0}.\) In this sense we say that \(H(\vec{y})\) is an implicit function, given by \(f(\vec{x}, \vec{y})=\overrightarrow{0}\).
Similarly, under suitable assumptions, \(f(\vec{x}, \vec{y})=\overrightarrow{0}\) defines \(\vec{y}\) as a function of \(\vec{x}.\)
Note 4. While \(H\) is unique for a given neighborhood \(P \times Q\) of \((\vec{p}, \vec{q}),\) another implicit function may result if \(P \times Q\) or \((\vec{p}, \vec{q})\) is changed.
For example, let
\[f(x, y)=x^{2}+y^{2}-25\]
(a polynomial; hence \(f \in C D^{1}\) on all of \(E^{2}).\) Geometrically, \(x^{2}+y^{2}-25=0\) describes a circle.
Solving for \(x,\) we get \(x=\pm \sqrt{25-y^{2}}.\) Thus we have two functions:
\[H_{1}(y)=+\sqrt{25-y^{2}}\]
and
\[H_{2}(y)=-\sqrt{25-y^{2}}.\]
If \(P \times Q\) is in the upper part of the circle, the resulting function is \(H_{1}.\) Otherwise, it is \(H_{2}.\) See Figure 28.
V. Implicit Differentiation. Theorem 4 only states the existence (and uniqueness) of a solution, but does not show how to find it, in general.
The knowledge itself that \(H \in C D^{1}\) exists, however, enables us to use its derivative or partials and compute it by implicit differentiation, known from calculus.
(a) Let \(f(x, y)=x^{2}+y^{2}-25=0,\) as above.
This time treating \(y\) as an implicit function of \(x, y=H(x),\) and writing \(y^{\prime}\) for \(H^{\prime}(x),\) we differentiate both sides of (x^{2}+y^{2}-25=0\) with respect to \(x,\) using the chain rule for the term \(y^{2}=[H(x)]^{2}\).
This yields \(2 x+2 y y^{\prime}=0,\) whence \(y^{\prime}=-x / y\).
Actually (see Note 4), two functions are involved: \(y=\pm \sqrt{25-x^{2}};\) but both satisfy \(x^{2}+y^{2}-25=0;\) so the result \(y^{\prime}=-x / y\) applies to both.
Of course, this method is possible only if the derivative \(y^{\prime}\) is known to exist. This is why Theorem 4 is important.
(b) Let
\[f(x, y, z)=x^{2}+y^{2}+z^{2}-1=0, \quad x, y, z \in E^{1}.\]
Again \(f\) satisfies Theorem 4 for suitable \(x, y,\) and \(z\).
Setting \(z=H(x, y),\) differentiate the equation \(f(x, y, z)=0\) partially with respect to \(x\) and \(y.\) From the resulting two equations, obtain \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).


14.1: Groundwater and Aquifers
Groundwater is stored in the open spaces within rocks and within unconsolidated sediments. Rocks and sediments near the surface are under less pressure than those at significant depth and therefore tend to have more open space. For this reason, and because it’s expensive to drill deep wells, most of the groundwater that is accessed by individual users is within the first 100 meters of the surface. Some municipal, agricultural, and industrial groundwater users get their water from greater depth, but deeper groundwater tends to be of lower quality than shallow groundwater, so there is a limit as to how deep we can go.
Porosity is the percentage of open space within an unconsolidated sediment or a rock. Primary porosity is represented by the spaces between grains in a sediment or sedimentary rock. Secondary porosity is porosity that has developed after the rock has formed. It can include fracture porosity—space within fractures in any kind of rock. Some volcanic rock has a special type of porosity related to vesicles, and some limestone has extra porosity related to cavities within fossils.
Porosity is expressed as a percentage calculated from the volume of open space in a rock compared with the total volume of rock. The typical ranges in porosity of a number of different geological materials are shown in Figure \(\PageIndex{1}\). Unconsolidated sediments tend to have higher porosity than consolidated ones because they have no cement, and most have not been strongly compressed. Finer-grained materials (e.g., silt and clay) tend to have greater porosity—some as high as 70%—than coarser materials (e.g., gravel). Primary porosity tends to be higher in well-sorted sediments compared to poorly sorted sediments, where there is a range of smaller particles to fill the spaces made by the larger particles. Glacial till, which has a wide range of grain sizes and is typically formed under compression beneath glacial ice, has relatively low porosity.
Consolidation and cementation during the process of lithification of unconsolidated sediments into sedimentary rocks reduces primary porosity. Sedimentary rocks generally have porosities in the range of 10% to 30%, some of which may be secondary (fracture) porosity. The grain size, sorting, compaction, and degree of cementation of the rocks all influence primary porosity. For example, poorly sorted and well-cemented sandstone and well-compressed mudstone can have very low porosity. Igneous or metamorphic rocks have the lowest primary porosity because they commonly form at depth and have interlocking crystals. Most of their porosity comes in the form of secondary porosity in fractures. Of the consolidated rocks, well-fractured volcanic rocks and limestone that has cavernous openings produced by dissolution have the highest potential porosity, while intrusive igneous and metamorphic rocks, which formed under great pressure, have the lowest.
Porosity is a measure of how much water can be stored in geological materials. Almost all rocks contain some porosity and therefore contain groundwater. Groundwater is found under your feet and everywhere on the planet. Considering that sedimentary rocks and unconsolidated sediments cover about 75% of the continental crust with an average thickness of a few hundred meters, and that they are likely to have around 20% porosity on average, it is easy to see that a huge volume of water can be stored in the ground.
Porosity is a description of how much space there could be to hold water under the ground, while permeability describes how those pores are shaped and interconnected. This determines how easy it is for water to flow from one pore to the next. Larger pores mean there is less friction between flowing water and the sides of the pores. Smaller pores mean more friction along pore walls, and also more twists and turns for the water to have to flow-through. A permeable material has a greater number of larger, well-connected pores spaces, whereas an impermeable material has fewer, smaller pores that are poorly connected. Permeability is the most important variable in groundwater. Permeability describes how easily water can flow through the rock or unconsolidated sediment and how easy it will be to extract the water for our purposes. The characteristic of permeability of a geological material is quantified by geoscientists and engineers using a number of different units, but the most common is the hydraulic conductivity . The symbol used for hydraulic conductivity is K . Although hydraulic conductivity can be expressed in a range of different units, in this book we will always use meters per second.
As shown on Figure \(\PageIndex{2}\) there is a wide range of permeability in geological materials from 10 −12 meters per second (0.000000000001 meters per second) to around 1 metre per second. Unconsolidated materials are generally more permeable than the corresponding rocks (compare sand with sandstone, for example), and the coarser materials are much more permeable than the finer ones. The least permeable rocks are unfractured intrusive igneous and metamorphic rocks, followed by unfractured mudstone, sandstone, and limestone. The permeability of sandstone can vary widely depending on the degree of sorting and the amount of cement that is present. Fractured igneous and metamorphic rocks, and especially fractured volcanic rocks, can be highly permeable, as can limestone that has been dissolved along fractures and bedding planes to create solutional openings.
Why is clay porous but not permeable?
Both sand and clay deposits (and sandstone and mudstone) are quite porous (30% to 50% for sand and 40% to 70% for silt and clay), but while sand can be quite permeable, clay and mudstone are not.
The surface of most silicate mineral grains has a slight negative charge due to imperfections in the mineral structure. Water (H 2 O) is a polar molecule. This means that while it has no overall electrical charge, one side of the molecule has a slight positive charge (the side with the two hydrogens), compared to a slight negative charge on the other side. Water is strongly attracted to all mineral grains and water within that bound water layer (a few microns around each grain) is not able to move and flow along with the rest of the groundwater.
In the two diagrams of Figure \(\PageIndex{3}\) the bound water is represented by medium blue lines around each grain, and the water that is not bound, and can flow, is shown in light blue. In the sand, there is still a lot of water that is able to move through the sediment, but in the clay deposit almost all of the water is held tightly to the grains and this reduces the permeability.
We have now seen that there is a wide range of porosity in geological materials and an even wider range of permeability. Groundwater exists everywhere there is porosity. However, whether that groundwater is able to flow in significant quantities depends on the permeability. An aquifer is defined as a body of rock or unconsolidated sediment that has sufficient permeability to allow water to flow through it. Unconsolidated materials like gravel, sand, and even silt make relatively good aquifers, as do rocks like sandstone. Other rocks can be good aquifers if they are well fractured. An aquitard is a body that does not allow transmission of a significant amount of water, such as a clay, a till, or a poorly fractured igneous or metamorphic rock. These are relative terms, not absolute, and are usually defined based on someone’s desire to pump groundwater; what is an aquifer to someone who does not need a lot of water, may be an aquitard to someone else who does. An aquifer that is exposed at the ground surface is called an unconfined aquifer . An aquifer where there is a lower permeability material between the aquifer and the ground surface is known as a confined aquifer , and the aquitard separating ground surface and the aquifer is known as the confining layer .
Figure \(\PageIndex{4}\) shows a cross-section of a series of rocks and unconsolidated materials, some of which might serve as aquifers and others as aquitards or confining layers. The granite is much less permeable than the other materials, and so is an aquitard in this context. The yellow layer is very permeable and would make an ideal aquifer. The overlying grey layer is a confining layer.
The upper buff-colored layer ( K = 10 −2 meters per second) does not have a confining layer and is an unconfined aquifer. The yellow layer ( K = 10 -1 meters per second) is “confined” by the confining layer ( K = 10 −4 meters per second), and is a confined aquifer. The confined aquifer gets most of its water from the upper part of the hill where it is exposed at the surface, and relatively little by seepage through the fine silt layer.
Image Descriptions
| Solid rocks | Percent porosity (Unfractured) | Percent porosity (Fractured) |
|---|---|---|
| Volcanic Rock | 0% to roughly 20% | Roughly 20% to 50% |
| Intrusive igneous and metamorphic rocks | 0% to 5% | 5% to 10% |
| Limestone with cavernous openings | N/A | 0% to 50% |
| Limestone without cavernous openings | 0% to 20% | N/A |
| Mudstone | 0% to 30% | N/A |
| Sandstone | 0% to roughly 10% | Roughly 10% to 30% |
| Conglomerate | 10% to 30% | N/A |
| Unconsolidated Materials | Percent porosity |
|---|---|
| Glacial Till | 11% to 20% |
| Silt and clay | 36% to 70% |
| Sand | 30% to 50% |
| Gravel | 24% to 40% |
[Return to Figure \(\PageIndex{1}\)]
| Material type | Hydraulic conductivity (meters per second) |
|---|---|
| Limestone with solutional openings | 10 to the power of negative 5 to 10 to the power of negative 1.2 (10 −5 to 10 −1.2 ) |
| Fractured volcanic rock | 10 to the power of negative 6 to 10 to the power of negative 1.7 (10 −6 to 10 −1.7 ) |
| Fractured igneous and metamorphic rock | 10 to the power of negative 7.5 to 10 to the power of negative 3.5 (10 −7.5 to 10 −3.5 ) |
| Limestone | 10 to the power of negative 8.4 to 10 to the power of negative 4.9 (10 −8.4 to 10 −4.9 ) |
| Sandstone | 10 to the power of negative 8.9 to 10 to the power of negative 5 (10 −8.9 to 10 −5 ) |
| Mudstone | 10 to the power of negative 11.3 to 10 to the power of negative 8 (10 −11.3 to 10 −8 ) |
| Unfractured igneous and metamorphic rock | 10 to the power of negative 12 to 10 to the power of negative 8.6 (10 −12 to 10 −8.6 ) |
| Material Type | Hydraulic conductivity (meters per second) |
|---|---|
| Clay | 10 to the power of negative 10.6 to 10 to the power of negative 7.7 (10 −10.6 to 10 −7.7 ) |
| Glacial till | 10 to the power of negative 10.4 to 10 to the power of negative 5.2 (10 −10.4 to 10 − 5.2 ) |
| Silt | 10 to the power of negative 8 to 10 to the power of negative 4.3 (10 −8 to 10 −4.3 ) |
| Silty sand | 10 to the power of negative 6 to 10 to the power of negative 2.7 (10 −6 to 10 −2.7 ) |
| Sand | 10 to the power of negative 5.2 to 10 to the power of negative 1.7 (10 −5.2 to 10 −1.7 ) |
| Gravel | 10 to the power of negative 2.7 to 10 to the power of negative 0.1 (10 −2.7 to 10 −0.1 ) |
[Return to Figure \(\PageIndex{2}\)]
Figure \(\PageIndex{4}\) image description: The relative permeability of various sedimentary layers. Permeability is denoted by hydraulic conductivity (K = meters per second). The buff-colored, top layer is an unconfined aquifer with K = 10 −2 . The grey-colored second layer is a confining layer with K = 10 −4 . The yellow-colored third layer is a confined aquifer with K = 10 −1 . The fourth layer is unlabelled, with K = 10 −4 . The bottom layer is granite, K = 10 −10 . [Return to Figure \(\PageIndex{4}\)]
Media Attributions
- Figure \(\PageIndex{1}\), 14.1.2, 14.1.3, 14.1.4: © Steven Earle. CC BY.


Section 2.3: Promoting Community-Driven Change in Family and Community Systems to Support Girls’ Holistic Development in Senegal
The Big Picture
This is the story of the Girls’ Holistic Development Program designed and implemented by the non-profit organization, Grandmother Project – Change through Culture, in Southern Senegal starting in 2010 and evaluated on several occasions by outside researchers. The case study describes this innovative program, the results, and the lessons learned that are relevant to other African contexts, and to other collectivist cultures in the Global South or Global North. This author is the co-founder and Executive Director of the Grandmother Project initiative.
The Grandmother Project’s Mission
…is to improve the health and well-being of women, children, and families
in countries in the Global South, by empowering communities to drive their own development
by building on their own experience, resources, and cultural realities.
Senegal Community Context
Background of the Girls’ Holistic Development Program in Senegal
Preparatory Phase
Participatory and Rapid Qualitative Assessment
“We should go back to our roots. We need to recognize what is positive within our culture and hold on to it jealously.”
Demba, NGO community development worker
“If we lose our cultural values, we will be forced to replace them with other people’s values.”
Abdoulaye, Teacher
Series of Forum-Dialogues
Bassirou, Community Health Volunteer
“Even though we didn’t go to school, we understood everything, we shared our knowledge and everyone appreciated our ideas.”
Fatamata, Grandmother Leader
Traditionally in community meetings, there was no open communication between men and women. The inclusive nature of the forums, with men and women of different ages and statuses within the community, was appreciated by almost all community members. However, a few of the elders said that they felt uncomfortable being in the same meeting with people much younger than themselves.
“There is often a constraint in community discussions because different categories of community members do not openly speak up. It is good to bring together men and women of different social classes and ages so that everyone can learn from each other.”
Mballo, Former National Parliamentarian
“In other workshops, we grandmothers were criticized for our traditional ideas. That’s why, before coming, we were afraid. But we are happy that we could contribute to the discussion without being criticized.”
Oumou, Grandmother Leader
Partnership with the Ministry of Education (MOE)
Implementation of the Girls’ Holistic Development Program
A Holistic Approach for Systemic Change
GHD Model
Program Goal and Objectives
Holistic Focus on Girls’ Rights and Needs
Building Communication Relationships
Involvement of Community Leaders
Building on Cultural and Religious Roles and Values
Program Components
Change through Culture Process of Change
Community Dialogue for Consensus-Building for Change
These key activities are briefly described below in terms of its purpose and participants.
“The intergenerational forums are very important as they help to re-establish dialogue between elders, parents, and youth. In the recent past, elders and adults didn’t listen to young people’s ideas and underestimated them. Thanks to the forums there is now more communication between older and younger community members.”
Grandmother Leader
“The forums have increased women’s confidence in themselves. Before, they didn’t dare express themselves in front of men. During community meetings, only men were allowed to speak. But now men know that women also have good ideas and encourage them to speak up.”
Young mother
“Before there was not enough discussion between men and women in families. It was a problem that separated them and was the root of frequent misunderstandings and arguments. The intergenerational forums have helped to solve this problem.”
Young adolescent boy
“We never before had the opportunity to sit all together and discuss like this although it is the best way to promote the development of our community elder leaders”
“This is a very important day because we are here to honor the grandmothers who are the teachers of young couples and of children. Before this project, the grandmothers were practically dead in the village and now they have been revived. It is since the grandmothers have resumed their role that teen pregnancies have greatly decreased.”
Mamadou, Village Elder
“Since we participated in the grandmother leadership training, the relationship between us grandmothers has changed. Now there is a permanent dialogue between us. Whenever one person has an idea of what we should do regarding our girls, we get together to discuss. Since the training, the relationships and communication between us have been strengthened”.
Grandmother Leader
“Thanks to these training sessions, I have become more confident. I no longer hesitate when there is something that needs to be said or done. I no longer bow my head when speaking before a group of men.
Grandmother Leader
“Before we used to scold our granddaughters all the time and they were rather afraid of us. Through the training, we realized that that is not a good way to communicate with them. Now we talk softly to the young girls and they listen to our advice with regards to sexuality and other things.”
Village headman’s wife
“These workshops support national priorities to increase children’s understanding and adoption of positive cultural values and to strengthen teacher-community relationships and communication”.
Amadou Lamine Wade, District Education Office Director
“Teachers alone do not have all of the knowledge that children need to learn. They also need to learn about positive cultural values and behavior. I don’t know of anyone else in the community who is more knowledgeable regarding the values that children should acquire. That is what justifies the presence of the grandmothers here today. And increased communication between teachers and grandmothers is very beneficial to children, especially to girls.”
Mr. Ba, Supervisor, District Education Office
“We are honored to have been invited to participate in this workshop along with teachers and school directors. We are going to work together with the teachers to encourage all children, those in school and not, to learn the values that are important in our culture.”
Maimouna, Grandmother
“There is a change in our relationships with our grandmothers. Before, we preferred to go to dancing parties or to watch television instead of being with them. Now, we spend more time with the grandmothers, listening to their stories that teach us about important values”.
Adolescent girl
”We are closer to our grandmothers now. If we have questions related to sexuality we can discuss them with our grandmothers, more easily than with our mothers. Now we are comfortable talking to the grandmothers”.
Adolescent girl
“This meeting has been very useful because it has allowed us to discuss important issues with others from our same village and with people from other villages. In our community we plan to organize meetings with all generations, to discuss what we can do together to prevent child marriage and FGM.”
Moussa, Village Headman
“These meetings are very beneficial because they encourage communication and understanding between us. During this meeting, I realized that FGM is not recommended by Islam. Many Imams were present and none of them support the practice”.
Cissé, Grandmother Leader


6.4: Addressing Challenging Behavior
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Preschool children rely on guidance from their teachers to practice and eventually master the complex skills required to navigate social interactions with peers and adults. With guidance from caring adults in their environment, preschool children can develop a sense of competence, learn developmentally appropriate ways to socialize with peers, and resolve conflicts. [1]
As the adults that work with children on a daily basis and who form positive relationships with them, teachers have a critical role in addressing challenging behaviors. Teachers should:
Teachers also need support from their programs. Programs should:
Conflict is a natural part of social interactions. Teachers can help children navigate conflict successfully by facilitating conflict resolution. Teachers should:
[5]
Include the children directly involved and use these steps
1. Acknowledge there is a problem or conflict: What happened? How do you feel?
2. Ask for solution ideas
3. Give it a try
4. Follow up with children, use PDA (Positive, Descriptive Acknowledgment)
It’s important to see that the issue is resolved or that the children move on.
Challenging behavior often indicates that a child is experiencing stress from a number of factors: internal stress from fatigue, poor nutrition, illness, pain or discomfort; external stress from a mismatch in the classroom environment or expectations, poor relationships with the teachers or children, overly difficult or overly simple tasks, limited social skills, trauma in the home environment; or a combination of these factors. A consistent and supportive teacher or other important adult can provide support for the child during short periods of stress. If challenging behavior continues over longer periods of time, it may be necessary to examine possible contributors to the behavior in the classroom and from other sources. A tiered intervention framework such as the Teaching Pyramid may be an appropriate response to challenging behavior.
The teaching pyramid model (Fox et al. 2003) describes a primary level of universal practices—classroom preventive practices that promote the social and emotional development of all children built on a foundation of positive relationships; secondary interventions that address specific social and emotional learning needs of children at risk for challenging behavior; and development of individualized interventions (tertiary level) for children with persistent problem behavior (see the diagram “The Teaching Pyramid”). [6]
Universal classroom practices include developmentally appropriate, child-centered classroom environments that promote children’s developing independence, successful interactions, and engagement in learning. While universal practices may be enough to promote the development of social competence in the majority of children in the classroom, teachers may find that there are children whose lack of social and emotional skills or whose challenging behavior requires more focused attention.
According to the California Collaborative on Social and Emotional Foundations for Early Learning (CA CSEFEL), fundamental to promoting social and emotional competence in young children is guiding children in their efforts to build positive relationships with adults and peers and creating supportive social and emotional learning environments for all children. For children at risk of developing behavior problems, targeted social and emotional strategies may be necessary, and for those children who display very persistent and severe challenging and behavior problems, individualized intensive interventions are required, when the children do not respond to typical preventive practices, child guidance procedures, or social–emotional teaching strategies that would normally work with most children (CA CSEFEL Teaching Pyramid, Promoting Social and Emotional Competence, Module 3a, 2013).
|
Old Way |
New Way |
|---|---|
|
General intervention for all behavior problems |
Intervention matched to purpose of the behavior |
|
Intervention is reactive |
Intervention is proactive |
|
Focus on behavior reduction |
Focus on teaching the child new skills |
|
Quick fix |
Long-term interventions |
[1] California Preschool Program Guidelines by the California Department of Education is used with permission
[2] Content by Jennifer Paris is licensed under CC BY 4.0 ;
California Preschool Program Guidelines by the California Department of Education is used with permission
[3] Content by Jennifer Paris is licensed under CC BY 4.0 ;
California Preschool Program Guidelines by the California Department of Education is used with permission
[4] Content by Jennifer Paris is licensed under CC BY 4.0 ;
California Preschool Program Guidelines by the California Department of Education is used with permission
[5] Supporting Inclusive Early Learning: Working Together for Inclusion & Belonging by Supporting Inclusive Early Learning is used with permission
[6] California Preschool Program Guidelines by the California Department of Education is used with permission (pg. 143-144)
[7] Graphic by the California Preschool Program Guidelines by the California Department of Education is used with permission
[8] California Preschool Program Guidelines by the California Department of Education is used with permission