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# Diagonalization of Quaternion Hermitian matrices

How do I go about diagonalizing such a matrix.

I ask because I need to sort out the following problem:

Let $D$ be the quaternion algebra over $\mathbb{Q}$ with $i^2 = -1, j^2 = -11, ij=-ji=k$.

Let $\mathfrak{O} = \langle 1,i,\frac{1+j}{2},\frac{i+k}{2}\rangle_{\mathbb{Z}}$ be a maximal order.

Consider the Hermitian matrix:

$A = \begin{pmatrix} 11 & -j-3k \\ j+3k & 11 \end{pmatrix}$.

I need to find all matrices $u\in M_2(\mathfrak{O})$ such that $uA\bar{u}^T = A$. This is proving difficult (there should be $48$ such matrices but extensive checks have only produced $8$ and those are easy diagonal/anti-diagonal ones!)

I think I can do it easily if I diagonalize $A$.

But how do I find $P\in$ $GL_2(D)$ such that $\bar{P}^T A P$ is diagonal with rational entries? Apparently such a matrix exists. Does the non-commutativity create problems with the usual construction?

-

Note that $U=\pmatrix{1&t\cr 0&1}$ and $X=\pmatrix{a&b\cr c&d}$ has $UX\bar U^T=\pmatrix{?&b+dt\cr c&d}\pmatrix{1&0\cr -t&1}=\pmatrix{?&b+dt\cr c-dt&?}$ (for $t$ with trace 0). So just make $t=(j+3k)/11$ in your example, killing the terms off the diagonal. Maybe I transposed this, but you can work it out.

-
Oh I can't believe how easy it was! I was too bogged down in theory to just try simple examples...thanks. –  fretty Jun 28 '13 at 11:28


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 Tr. Mat. Inst. Steklova, 2016, Volume 295, Pages 53–71 (Mi tm3747)

Degenerate billiards

S. V. Bolotin

Steklov Mathematical Institute of Russian Academy of Sciences, ul. Gubkina 8, Moscow, 119991 Russia

Abstract: In an ordinary billiard system, trajectories of a Hamiltonian system are elastically reflected after a collision with a hypersurface (scatterer). If the scatterer is a submanifold of codimension more than $1$, we say that the billiard is degenerate. We study those trajectories of degenerate billiards that have an infinite number of collisions with the scatterer. Degenerate billiards appear as limits of systems with elastic reflections or as small-mass limits of systems with singularities in celestial mechanics. We prove the existence of trajectories of such systems that shadow the trajectories of the corresponding degenerate billiards. The proofs are based on a version of the method of an anti-integrable limit.

 Funding Agency Grant Number Russian Science Foundation 14-50-00005 This work is supported by the Russian Science Foundation under grant 14-50-00005.

DOI: https://doi.org/10.1134/S037196851604004X

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English version:
Proceedings of the Steklov Institute of Mathematics, 2016, 295, 45–62

Bibliographic databases:

UDC: 531.01

Citation: S. V. Bolotin, “Degenerate billiards”, Modern problems of mechanics, Collected papers, Tr. Mat. Inst. Steklova, 295, MAIK Nauka/Interperiodica, Moscow, 2016, 53–71; Proc. Steklov Inst. Math., 295 (2016), 45–62

Citation in format AMSBIB
\Bibitem{Bol16} \by S.~V.~Bolotin \paper Degenerate billiards \inbook Modern problems of mechanics \bookinfo Collected papers \serial Tr. Mat. Inst. Steklova \yr 2016 \vol 295 \pages 53--71 \publ MAIK Nauka/Interperiodica \publaddr Moscow \mathnet{http://mi.mathnet.ru/tm3747} \crossref{https://doi.org/10.1134/S037196851604004X} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=3628514} \elib{http://elibrary.ru/item.asp?id=27643602} \transl \jour Proc. Steklov Inst. Math. \yr 2016 \vol 295 \pages 45--62 \crossref{https://doi.org/10.1134/S0081543816080046} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000395572400004} \scopus{http://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-85010767528}

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• https://doi.org/10.1134/S037196851604004X
• http://mi.mathnet.ru/eng/tm/v295/p53

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Related articles on Google Scholar: Russian articles, English articles

This publication is cited in the following articles:
1. Sergey V. Bolotin, “Degenerate Billiards in Celestial Mechanics”, Regul. Chaotic Dyn., 22:1 (2017), 27–53
2. S. V. Bolotin, V. V. Kozlov, “Topological approach to the generalized $n$-centre problem”, Russian Math. Surveys, 72:3 (2017), 451–478
3. J. Fejoz, A. Knauf, R. Montgomery, “Lagrangian relations and linear point billiards”, Nonlinearity, 30:4 (2017), 1326–1355
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### Author Topic: Final exam graded  (Read 4500 times)

#### Victor Ivrii

• Administrator
• Elder Member
• Posts: 2563
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##### Final exam graded
« on: December 14, 2014, 04:23:12 PM »
We graded the final exam. You can see your final exam mark and marks for each problem on BlackBoard

Your final grade is calculated as $0.6\max (Term\_Mark, FE\_Mark) + 0.4\min (Term\_Mark, FE\_Mark)$ but there will be some rounding up. Wait until you see it on ROSI (we will submit your marks electronically shortly, but they should be authorized)

You can see solutions (just posted) http://forum.math.toronto.edu/index.php?board=78.0

If you wrote exam in some special location (except EX340), you need to wait as we did not get your papers yet.

Reminding some rules

If you missed Final Exam you need to deal with the College Registrar. See
http://forum.math.toronto.edu/index.php?topic=222.0

About rechecking and regrading see
http://www.artsandscience.utoronto.ca/academichandbook.shtml#4.2

However, don't do it before you marks come up on ROSI

« Last Edit: December 14, 2014, 04:57:45 PM by Victor Ivrii »

#### Victor Ivrii

• Administrator
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• Posts: 2563
• Karma: 0
##### Re: Final exam graded
« Reply #1 on: December 15, 2014, 09:32:22 AM »
Access to FE marks disabled (hopefully temporarily)


text:
# Geographically, how large would a very populous city during the 16th century be?

A rich nation, constantly embroiled in warfare and more often than not being invaded, has built, about a couple decades ago, a grand city over the old capital.

It houses a very large population of 300 thousand, and it's very well strategically placed, about 50 kilometers from the city lies the coast, with another, albeit much smaller, similarly built city acting as port, this is great because the ocean in which the coast is set brings very rich trade, which made both cities major trade hubs in the continent.

The city is built on top of fertile plains and has a particularly thick river running through town. The technology available to the civilization is similar to that of the late 16th century and early 17th, along with the architectural / engineering knowledge of the Romans (not sure if the Romans would be able to rival that of the 16th century Europeans in this aspect, but the Rome was the first city to achieve a population of one million in Europe [133 B.C.], a feat only seen in Europe again after nearly two millennia [London 1810 A.C.], and they had a pretty awesome sewage system, which I intend to mimic in the city).

As already mentioned, the nation is constantly warring, so the city has huge, thick walls, with many bastions around it, similar to the walls of star fortress, made to deflect cannon fire, allowing the city to hold besiegers at bay if their only hope of conquest is through an invasion on the city.
The city is also built to withstand a long siege, it has huge underground storehouses that store food and supplies, and I thought of having large farms and estates inside the city, but I think that it would take up too much valuable space and would not be cost-effective or efficient. I also thought of some sort of vertical farming mumbo jumbo but I'm not sure of how that would work.

The city is also a knowledge hub, it has three universities, four military schools, an observatory, many schools and it houses the central government, the royal palace and the court, and contributing more to the scholarship of the city, the community is, as already mentioned, an economic center, attracting merchants from all over the world, and many markets and banks have been built on the city, attracting even more traders.

Last but not least, I'm aware that the city would be VERY expensive to built, but that won't be a problem, for the kingdom has since long been wanting to build a grandiose capital, and has hoarded money / goods / resources for centuries to build such city, and even then that nearly wasn't enough, the government almost got in debt and lost a lot of influence because of the construction, but eventually, it paid off, so finances won't be much of a problem here fellows.

So, with this information in mind, how large would be this city? Keep in mind that it was also built with a small "extra space" to allow growth.

P.S. - I'm an architecture illiterate, so please explain your ideas to me, and point out my mistakes. P.S. 2 - If anyone has a cool sounding name for the city, I'm all ears.

• Paris in 1550 had an estimated population of 275,000, Their defensive strategy was to use fortified outlying cities instead of the capital to tie down invaders. If the capital is besieged, where will the relieving army come from? Hoarding money may be the wrong idea - city earthwork defenses were often built over generations by the local labor, not contracted out...and many had to be rebuilt several times as technology changed. Standing armies and garrisons tend to be very, very expensive. Dec 29 '17 at 6:18
• What kind of food preservation will these vast storehouses use, and how will you keep pests from consuming all the stores? Dec 29 '17 at 6:21
• @user535733 About the outlying fortifications, the idea for the city here would be that around it, there would be many fortresses and castles, garrisoning relief armies, instead of cities. And about the hoarding money, sorry, I phrased myself bad, I meant also supplies and products, such as gold, in order to pay for the city's construction. About the contracted out, as I already mentioned, the city was made to be grand, it had to be built professionally to reflect the nation's will. And about the technology, I planned this world to be sort of "in amber" on military terms Dec 29 '17 at 6:35
• @user535733 By that I mean, military technology is progressing very slowly, and since long have cannons be used, so these defensive wall surely won't be outdated, and they're being orderly maintained too. About the standing armies, as I said, this country is constantly embroiled in war, and has developed a professional citizen army, much like the Romans, it's not terribly large, but it's very well-trained and capable, and the government dedicates much of their finances to the military, for the survival of the nation depends on it. Dec 29 '17 at 6:38
• @Jedboo Because transportation is still very elementary back in the 1700s, you can't really have an outlying fortress or castle without a city to support it. Also, if cannons are in play, don't waste your time with city walls. You'd be much better off putting those resources into mangling the terrain and fortifying the roads in and out. Dec 29 '17 at 6:41

# Large cities of ~1500

Wikipedia conveniently has a list, and it is well sourced from Chandler, Modelski, Morris and others in the field. Lets run down some of the largest cities from 1500-1600. I rank all cities above 200,000 by their median estimate in the time range. I skipped pre-Columbian estimates for New World cities.

### Beijing 700,000

Beijing's city walls were completed in 1553, right in the middle of this time period. It had two sets of walls, the 'Inner Wall' and 'Outer Wall.' Interestingly, the Inner Wall was not inside the Outer Wall, but directly north of it. The inner city wall was 24 km long, and the outer wall was 28 km long. Together, I estimate they enclosed about 50 km$$^2$$. This makes the total density within the walls about 14,000 per km$$^2$$.

### Istanbul 600,000

The Theodosian walls still stood as the boundary of the city in the 1500s. There was some suburban development outside the walls, and a lot of development across the Golden Horn in Galata or Pera. The modern area inside the walls is the Fatih district; combined with Beyoglu to the north to represent the suburban areas, that gives an area of 22 km$$^2$$ for a density of about 27,000 per km$$^2$$.

### Vijayanagar 500,000

The modern Hampi archaeological site is 41 km$$^2$$. The inner city walls were somewhat smaller than this, but since there were more wells outside the walls than inside, this larger number is likely more accurate. This gives a density of about 12,000 per km$$^2$$. Medieval travelers claimed that there was a second outer wall that enclosed a large area of farmland, around 500-750 km$$^2$$. So an alternate lower density could be calculated for this city.

### Cairo 300,000

Old Cairo has largely been buried under the new, so there isn't a lot of good information on how big the old city was. According to (unsourced) Wikipedia, by the end of the 15th century, Cairo was composed of Old Cairo, Downtown Cairo, Bulaq, and Azbakeya. Measuring these areas on Google Maps, I get a total area of around 20 km$$^2$$; for a density of 15,000 per km$$^2$$.

### Kyoto 300,000

Wikipedia has a 1696 map of Kyoto (a little after our time period), where it covers what is the modern day Kamigyo, Nakagyo and Simogyo wards. These wards have a total area of 21 km$$^2$$. Since we measured the city in 1700, we ought to use the 1700 population of 350,000, giving us a density of 16,000 per km$$^2$$.

### Paris 250,000

Paris' city walls for this time period were finished in 1383 (the Wall of Charles V) and covered 4.4 km$$^2$$. Maps made in this period (like this one from 1572) show some development outside the walls. If we estimate the actual city size as the 1st through 6th arrondissements of modern Paris, we get an area of 10 km$$^2$$ for a density of 25,000 per km$$^2$$.

### Hangzhou 250,000

Several sources list Hangzhou as being this big, but it had been much larger in the past, up to a million people and probably the largest city in the world for most of the 1200s. It was sacked by the Mongols in 1276 and declined, though it was still bigger than Beijing when Marco Polo and Ibn Battuta visited in roughly 1290 and 1345, respectively. The problem is, smaller 1500s Hangzhou with only a quarter of the population certainly didn't fill out its old size, and it is difficult to estimate the occupied area of the city was at this point. There must have been substantial ruins surrounding the occupied parts of the city center.

### Tabriz 250,000

I can't find any useful information on Medieval or Early Modern Tabriz.

### Naples 200,000

I couldn't find much information on the old walls of the city, but based on the location of the gates, I estimate that the old city occupied what is now San Ferdinando (which was explicitly developed by the Spanish in the 16th century) and the 2nd municipality (which includes the old port town along with Avvocata, another area developed by the Spanish). Summed together, these areas are about 6 km$$^2$$, for a density of 33,000 per km$$^2$$.

### Agra 200,000

This is another one that is tough to adjudicate. Agra was only founded in 1504. Agra fort, a 0.4 km$$^2$$ walled city was built on the location at that time, and served as the capital of Sikandar Lodi's empire until the death of his son in battle in 1526. Then there were years of chaos and the fortress was ruined until Akbar re-united Northern India and made it his capital again, completing a rebuild of the fort in 1573. But then Akbar moved his palace to Fatehpur Sikri, about 35 km to the southwest, in 1571 and stayed at that site until the local water supply ran out in 1585. After that, the emperors finally moved back to Agra. The conclusion is, the population was probably in a lot of flux over time, and there were never proper city walls, just a small fort.

# Conclusion: 10-30 km$$^2$$

Our density estimates for the largest cities of the 16th century range from 12,000 to 33,000 per km$$^2$$. Thus, if you want a city of 300,000 people, it should have a geographical area anywhere from 10 to 30 km$$^2$$, based on real world precedents.


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# Quantum phase transition

In physics, a quantum phase transition (QPT) is a phase transition between different quantum phases (phases of matter at zero temperature). Contrary to classical phase transitions, quantum phase transitions can only be accessed by varying a physical parameter—such as magnetic field or pressure—at absolute zero temperature. The transition describes an abrupt change in the ground state of a many-body system due to its quantum fluctuations. Such a quantum phase transition can be a second-order phase transition.[1] Quantum phase transition can also be represented by the topological fermion condensation quantum phase transition, see e.g. strongly correlated quantum spin liquid. In case of three dimensional Fermi liquid, this transition transforms the Fermi surface into a Fermi volume. Such a transition can be a first-order phase transition, for it transforms two dimensional structure (Fermi surface) into three dimensional. As a result, the topological charge of Fermi liquid changes abruptly, since it takes only one of a discrete set of values.

## Classical description

To understand quantum phase transitions, it is useful to contrast them to classical phase transitions (CPT) (also called thermal phase transitions).[2] A CPT describes a cusp in the thermodynamic properties of a system. It signals a reorganization of the particles; A typical example is the freezing transition of water describing the transition between liquid and solid. The classical phase transitions are driven by a competition between the energy of a system and the entropy of its thermal fluctuations. A classical system does not have entropy at zero temperature and therefore no phase transition can occur. Their order is determined by the first discontinuous derivative of a thermodynamic potential. A phase transition from water to ice, for example, involves latent heat (a discontinuity of the internal energy ${\displaystyle U}$ ) and is of first order. A phase transition from a ferromagnet to a paramagnet is continuous and is of second order. (See phase transition for Ehrenfest's classification of phase transitions by the derivative of free energy which is discontinuous at the transition). These continuous transitions from an ordered to a disordered phase are described by an order parameter, which is zero in the disordered and nonzero in the ordered phase. For the aforementioned ferromagnetic transition, the order parameter would represent the total magnetization of the system.

Although the thermodynamic average of the order parameter is zero in the disordered state, its fluctuations can be nonzero and become long-ranged in the vicinity of the critical point, where their typical length scale ξ (correlation length) and typical fluctuation decay time scale τc (correlation time) diverge:

${\displaystyle \xi \propto |\epsilon |^{-\nu }\,\,=\left({\frac {|T-T_{c}|}{T_{c}}}\right)^{-\nu }}$
${\displaystyle \tau _{c}\propto \xi ^{z}\propto |\epsilon |^{-\nu z},}$

where

${\displaystyle \epsilon ={\frac {T-T_{c}}{T_{c}}}}$

is defined as the relative deviation from the critical temperature Tc. We call ν the (correlation length) critical exponent and z the dynamical critical exponent. Critical behavior of nonzero temperature phase transitions is fully described by classical thermodynamics; quantum mechanics does not play any role even if the actual phases require a quantum mechanical description (e.g. superconductivity).

## Quantum description

Diagram of temperature (T) and pressure (p) showing the quantum critical point (QCP) and quantum phase transitions.

Talking about quantum phase transitions means talking about transitions at T = 0: by tuning a non-temperature parameter like pressure, chemical composition or magnetic field, one could suppress e.g. some transition temperature like the Curie or Néel temperature to 0 K.

As a system in equilibrium at zero temperature is always in its lowest-energy state, a QPT cannot be explained by thermal fluctuations. Instead, quantum fluctuations, arising from Heisenberg's uncertainty principle, drive the loss of order characteristic of a QPT. The QPT occurs at the quantum critical point (QCP), where quantum fluctuations driving the transition diverge and become scale invariant in space and time.

Although absolute zero is not physically realizable, characteristics of the transition can be detected in the system's low-temperature behavior near the critical point. At nonzero temperatures, classical fluctuations with an energy scale of kBT compete with the quantum fluctuations of energy scale ħω. Here ω is the characteristic frequency of the quantum oscillation and is inversely proportional to the correlation time. Quantum fluctuations dominate the system's behavior in the region where ħω > kBT, known as the quantum critical region. This quantum critical behavior manifests itself in unconventional and unexpected physical behavior like novel non Fermi liquid phases. From a theoretical point of view, a phase diagram like the one shown on the right is expected: the QPT separates an ordered from a disordered phase (often, the low temperature disordered phase is referred to as 'quantum' disordered).

At high enough temperatures, the system is disordered and purely classical. Around the classical phase transition, the system is governed by classical thermal fluctuations (light blue area). This region becomes narrower with decreasing energies and converges towards the quantum critical point (QCP). Experimentally, the 'quantum critical' phase, which is still governed by quantum fluctuations, is the most interesting one.


text:
Last edited by Jujind
Thursday, July 16, 2020 | History

3 edition of Just for fraction circles found in the catalog.

Just for fraction circles

Carol Alexander

# Just for fraction circles

## by Carol Alexander

Written in English

Subjects:
• Fractions,
• Mathematics,
• Problem solving in children,
• Study and teaching (Elementary)

• The Physical Object
FormatUnknown Binding
Number of Pages42
ID Numbers
Open LibraryOL12122847M
ISBN 101561079014
ISBN 109781561079018
OCLC/WorldCa37045190

Using an Area Model to Add/Subtract Fractions. Using an area model is also helpful in adding and subtracting fractions with unlike denominators. To start, kids need each fraction to have the same number of parts (again, a “like denominator”). They can set up area models for each fraction just as we did before to compare.   Montessori fraction insets are fabulous. They're also expensive and take up a lot of shelf space. For Montessori homeschools, that's not very practical. I've found DIY Montessori fraction materials that will give children a hands-on experience of fractions at a fraction of the price. (Couldn't resist saying that!) I want to keep this simple by featuring basic Montessori fraction materials that.

In a previous school, circle time followed potty time and children who had already gone potty played with puzzles at circle until everyone had had their turn in the bathroom. You don’t need to use an independent activity but having a predictable routine going into circle time like a song you play or sing to signal it’s time to come together. This math fractions ClipArt gallery offers 24 illustrations of a circle, or pie, broken into sixth slices. This includes images of individual slices, as well as the pie with missing slices. Fraction .

While both Mathematical Circles (Russian Experience) and BMC booksBMC books are organized byare organized by topics, one of the forthcoming books, Sergei Dorichenko’s Moscow Math Circle Curriculum in Day-by-Day Sets of Problems has a distinctly different structure. As suggested by the title it consists (mostly) ofAs suggested by the title, it consists (mostly) of. 7 fraction tips and tricks to help with fractions. Tips include multiplying fractions,finding the LCM,dividing fractions and more Circle the left column of prime ly the numbers you circle to find your Many people just learn the trick, but don't really understand what they are doing. Tip 2. Adding fractions without using a.

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### Just for fraction circles by Carol Alexander Download PDF EPUB FB2

Download Resource. Contents: Our free printable fraction circles are designed to make it easy to teach children fractions and fraction printables are available in color and black and white. I recommend you print them on colored cardstock paper and laminate for durability.

Fractions Circles in Pieces ~ Fractions Sectors in Bold Primary Colors. This is a set of circle fractions from the whole to twelfths. Colors included: red, blue, yellow, gray and white fills. And a black outline version with no fill.

Two options are included: with inner lines and without. Plea. Today I want to focus on books to teach fractions, because obviously there is a wonderful connection between division and fractions. In fact, many of the books I included in the book list for division could be used to explore fractions.

*Please Note: Some of the following links are affiliate links and help support the work of this site. These activities can be used to follow the teaching episodes based on, Book 7, Fraction Circles and are for those students who are able to use the associated number properties.

Attachments. 58 KB. Reading fun, adventure-filled, and visually engaging books as part of your lessons will help children remember the math concepts far longer, and with smiles.

Here’s a collection of favourite books that all feature aspects of math related to circles, including 2d and 3d shapes, fractions, and pi. 1/8 One Eighth 1/8 One Eighth 1/8 One Eighth 1/8 One Eighth 1/8 One Eighth 1/8 One Eighth 1/8 One Eighth 1/8 One Eighth 1/7 One Seventh 1/7 One Seventh. If so, you can build this Fraction Flip Book with your students as a reference tool.

The flip book can be placed into their math folders for a quick reference. If they get stuck on a problem or just need a visual representation of a fraction, a fraction number line, or a shaded fraction circle this tool is readily available for them.

The black and white fraction circles can be used as a manipulative to compare fractions. Photocopy the worksheet onto an overhead projection slide. Use a pencil to lightly color the appropriate circle to represent the first fraction on the paper copy. Use a non-permanent overhead pen to color the appropriate circle to represent the second fraction.

Distribute cutouts of four circles, or have the students cut the circles out. Guide the students through the process of folding the circles to create differing fractions, including one whole, two halves, four fourths, and eight eighths.

Have students draw lines along folds and label the backs of the fractions. Fractions Flip Book -Common Core Aligned- Use this flip book as an introduction to fractions or as practice.

It can also be used as a reference throughout a fractions unit. This flip book includes fractions from wholes to eighths. It allows for students to practice writing fractions in numbers and.

Step 2: Introducing Fractions Circles. Before my students begin their artistic creations, we spend some time so they can fold and cut fractions from some “practice” circles. We discuss reliable ways to cut exact fourths and eighths. I challenge them to fold and cut the smallest fractional pieces possible.

The kids thrill at cutting 32 tiny. Create a Fractions Flip Book. Have your students create a mystic rose. A fun way to explore fractions around a circle by using a compass to measure the fraction out.

Other Fraction Teaching Resources. Mindfulness is a superpower that your students can master with just five-minutes practice a day. Check out this list of 18 quick and. In geometry, circle packing is the study of the arrangement of circles (of equal or varying sizes) on a given surface such that no overlapping occurs and so that no circle can be enlarged without creating an overlap.

The associated packing density, η, of an arrangement is the proportion of the surface covered by the lisations can be made to higher dimensions – this is called.

At the outset, the book shows how a mere circle, divided like a pie - in halves, quarters and eighths - can be used to make pictures of all kinds. The pictures range from complex and visually striking patterns to simple but adorable animals - a cute mouse, for example, is created using four different one-eighth pieces of a pie (with a tail.

This is well achieved by use of worksheets with fraction circles or manipulative. Ensure to first work with like denominators then shift to unlike denominators when the students have learned the basic concept of fractions.

There are several sets of fraction worksheets that teachers can use to educate students about fractions. Kids will learn to make friends with fractions in these engaging and interactive fractions games.

From working with a number line to comparing fraction quantities, converting mixed numbers, and even using fractions in addition and subtraction problems, the fractions games below introduce your students to their next math challenge as they play to rack up points and win the game.

Fraction Fun is a crazy looking book about fractions being used in a variety of ways. The first part of the book is just an intro about fractions, but the rest are filled with lessons that can be done in the class to help students understand fractions in more settings than just a pizza.

Recommended grade level: /5(42). SpriteGru is an experienced teaching aids producer and aiming to provide the best quality and user experience. With this Magnetic Fraction Tiles you got: All your needs: for basic Fraction & Percentages education.

This compact set comes with 12 circles and 78pcs RectangleReviews: 1. Fill the Circle game. The purpose of this fraction game is to introduce fraction vocabulary and increase the children’s ability to visualize fraction pieces relative to a whole object.

Preparation: Trace and cut out the whole circles from a fraction circle set. To do this just trace one, then cut out 3 or 4 at a time from lightweight card. of fraction lessons includes hands-on explorations and activities that invoke problem solving, reasoning and proving, communicating, connecting, and representing fractions.

Cross-curricular activities link fractions to language arts, music, science, art, and social studies. We hope that as you use the ideas in this book in your classroom, your.

Great book to use when introducing fractions. Illustrates how fractions can look different. Would open up discussion with students on having them describe fractions they see (half a sandwich). Not just fractions of a whole item, but fractions of a set.4/5(62).In the image below, the circle has been divided into three parts of equal size.

Each part represents $\frac{1}{3}$ of the circle. This type of model is called a fraction circle. Other shapes, such as rectangles, can also be used to model fractions.fraction circle settings, and poses an additional related question.

• The second contains ﬁ ve follow-up, hands-on practice questions for students to answer using the fraction circles. Worksheet problems can be completed with or without use of the fraction circles, depending on grade and ability level.


text:
# A wire whose cross-sectional area is 4mm 2 is stretched by 0.1 mm by a certain load. if a...

## Question:

A wire whose cross-sectional area is 4mm{eq}^2 {/eq} is stretched by 0.1 mm by a certain load. if a similar wire double the area of cross-section is under the same load, then find the longation.

## Elongation of the wire

According to Hooke's law, the stress is directly proportional to the strain. The strain is defined as the ratio of the change in the length to the original length of the object or wire. And the proportionality constant is known as Young Modulus.

Given

Area of the wire {eq}A = 4 \ mm^{2} {/eq}

Length of the elongation {eq}\delta = 0.1 \ mm {/eq}

Now, the elongation of the wire is given by

{eq}\delta = \dfrac{PL}{AE} {/eq}

Where

• P is the applied load
• L is the original length of the wire
• E is the young modulus of the wire

Now, if the area has been doubled, then

{eq}\delta_{2} = \dfrac{PL}{A_{2}E} \\ \delta_{2} = \dfrac{PL}{2AE} \\ \delta_{2} = \dfrac{\delta}{2} \\ \delta_{2} = 0.05 \ mm {/eq}


text:
Open access peer-reviewed chapter

# General Stability in Viscoelasticity

Written By

Salim A. Messaoudi

Submitted: December 4th, 2015 Reviewed: May 12th, 2016 Published: September 21st, 2016

DOI: 10.5772/64217

From the Edited Volume

## Viscoelastic and Viscoplastic Materials

Edited by Mohamed Fathy El-Amin

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## Abstract

In this chapter, we consider a problem which describes the motion of a viscoelastic body and investigate the effect of the dissipation induced by the viscoelastic (integral) term on the solution. Precisely, we show that, under reasonable conditions on the relaxation function, the system stabilizes to a stationary state. We also obtain a general decay estimate from which the usual exponential and polynomial decay rates are only special cases.

### Keywords

• general decay
• memory
• relaxation function
• stability
• viscoelasticity

## 1. Introduction

Elastic materials, when subjected to a suddenly applied loading state held constant thereafter, respond instantaneously with a state of deformation which remains constant. On the other hand, Newtonian viscous fluids respond to a suddenly applied state of uniform shear stress by a steady flow process. However, there exist materials for which any suddenly applied and maintained state of uniform stress produces an instantaneous deformation followed by a flow process which might or might not be limited in magnitude as time grows. Such materials exhibit both instantaneous elasticity effects and creep characteristics. Obviously, such a behavior cannot be described by either an elasticity theory or a viscosity theory only but it combines features of each. The most interesting examples of such materials are polymers, which can display all the intermediate range of properties (glassy, brittle solid or an elastic rubber or a viscous liquid) depending on temperature and the experimentally chosen time scale. Such materials are said to possess memories.

Many scientists, such as Maxwell, Kelvin, Voigt, and Boltzmann, have contributed in modeling these phenomena. Boltzmann, in 1874, supplied the first formulation of a three-dimensional theory of isotropic viscoelasticity. He elaborated the model of a “linear” viscoelastic solid on a basic assumption which states that at any (fixed) point x of the body, the stress at any time t depends on the strain at all the proceeding times. In addition, if the strain at all preceding times is in the same direction, then the effect is to decrease the corresponding stress. The influence of a previous strain on the stress depends on the time elapsed since that strain occurred and is weaker than those strains that occurred long ago. Such properties make the model of solid, elaborated by Boltzmann, a material with (fading) memory. These memory effects are expressed by the dependence on the deformation gradient. Therefore, for these “viscoelastic” materials, the stress at each point and at each instant does not depend only on the present value of the deformation gradient but on the entire temporal prehistory of the motion. In addition, Boltzmann made the assumption that a superposition of the influence of previous strains holds, which means that the stress-strain relation is linear. Mathematically, this is interpreted by the time convolution of a “relaxation” function with the Laplacian of the solution. As a consequence, a subtle damping effect is produced. The types of equations we intend to discuss in this chapter are of the form:

E1

where is a bounded domain with regular boundary, g is a nonincreasing positive function, referred to as the relaxation function which describes the viscoelastic material in consideration, f is an external force, and u(xt) is the position of a point x “in the reference configuration” at a time t.

In early 1970s, Dafermos [1, 2] discussed a one-dimensional viscoelastic model, where he proved, for smooth monotonically decreasing relaxation functions, various existence and asymptotic stability results. However, no rate of decay has been given. After that, viscoelastic problems have attracted the attention of many researchers and many results of existence and long-time behavior have been established. To the best of our knowledge, the first work that studied the uniform decay of solutions was presented by Dassios and Zafirapoulos [3]. In their work, Dassios and Zafirapoulos presented a viscoelastic problem in ℝ3 and proved a polynomial decay for exponentially decaying kernels. In 1994, Muñoz Rivera [4] considered, in ℝn and in bounded domains, equations for linear isotropic homogeneous viscoelastic solids, with exponentially decaying memory kernels and showed that, in the absence of body forces, solutions decay exponentially for the bounded-domain case, whereas, for the whole space case, the decay is of a polynomial rate. After that, Cabanillas and Muñoz Rivera [5] studied problems, where the kernels are of algebraic (but not exponential) decay rates and showed that the decay of solutions is algebraic at a rate which can be determined by the rate of the decay of the relaxation function and the regularity of solutions. This result was later improved by Barreto et al. [6], where equations related to linear viscoelastic plates were treated. For viscoelastic systems with localized frictional dampings, Cavalcanti et al. [7] considered the following problem:

E2000

where is a bounded domain of ℝn (n ≥ 1) with a smooth boundary ∂g is a positive nonincreasing function satisfying, for two positive constants, the conditions:

- ξ 1 g t g t - ξ 2 g t , t 0 , E2

and a(x) ≥ a0 > 0 in a subdomain ω ⊂ , with meas(ω) > 0 and satisfying some geometry restrictions. They established an exponential rate of decay. Berrimi and Messaoudi [8] improved Cavalcanti’s result by weakening the conditions on both a and g. In particular, the function a can vanish on the whole domain and consequently the geometry condition is no longer needed. This result has been later extended to a situation, where a source is competing with the viscoelastic dissipation, by Berrimi and Messaoudi [9]. Also, Cavalcanti et al. [10] have studied a quasilinear equation, in a bounded domain, of the form:

| u t | ρ u t t - u - u t t + 0 t g t - τ u τ d τ - γ u t = 0 , E3

with ρ > 0, and a global existence result for γ ≥ 0, as well as an exponential decay for γ > 0, have been established. Messaoudi and Tatar [11,12] discussed the situation when γ = 0 and established polynomial and exponential decay results in the presence, as well as in the absence, of a nonlinear source term. Fabrizio and Polidoro [13] studied a homogeneous viscoelastic equation in the presence of a linear frictional damping (aut, a > 0) and showed that the exponential decay of the relaxation function g is a necessary condition for the exponential decay of the solution energy of the solution. In other words, the presence of the memory term, with a non-exponentially decaying relaxation function, may prevent the exponential decay even if the frictional damping is linear. He also obtained a similar result for the polynomial decay case.

For more general decaying kernels, Messaoudi [14,15] considered

u t t Δ u + 0 t g t τ Δ u τ d τ = b u m 2 u , in × 0 , u x t = 0 , x , t 0 u x , 0 = u 0 x , u t x 0 = u 1 x , x E1.1

with b = 0 and b = 1 and for relaxation functions satisfying

g ( t )  ξ ( t )g( t ), t  0,   E1.2

where ξ: ℝ+→ℝ+ is a nonincreasing differentiable function. He showed that the rate of the decay of the energy is exactly the rate of decay of g, which is not necessarily of exponential or polynomial decay type. After that, a series of papers using Eq. (1.2) have appeared. See for instance, Han and Wang [16], Liu [17,18], Park and Park [19], and Xiaosen and Mingxing [20].

In this work, we intend to study the following problem:

u t t Δ u + 0 t g t τ Δ u τ d τ + a u t m 2 u t = 0 , i n × 0 , u x t = 0 , x , t 0 u x , 0 = u 0 x , u t x 0 = u 1 x , x E1.3

where is a bounded and regular domain of ℝn, a > 0 is a constant, and g is a positive nonincreasing function satisfying Eq. (1.2). We will establish some general decay results depending on the behavior of g and m.

## 2. Preliminary

In this section, we present some material needed in the proof of our result and state a global existence result which can be proved using the well-known Galerkin method. See, for example, [2,3]. In order to prove our main result, we make the following assumptions:

(A1) g: ℝ+ → ℝ+ is a bounded differentiable function such that

g0>0,gt-γtgt,1-0gsds=l>0,E7

where γ(t) is a differentiable function satisfying

γt>0,γt0,and0+γtdt=+.E8

(A2) Concerning the nonlinearity in the damping, we assume that

1<m2nn-2,ifn>2andm>1,ifn=1,2E9

Remark 2.1. Examples of functions satisfying (A1) are

g1t=a1+tv,v>1,g2t=ae-bt+1p0<p1E10
g3t=a1+tln1+tv,v>1,E11

for a and b constants to be chosen properly.

Proposition 2.1. Let u0u1H01×L2 be given. Assume that (A1), (A2) hold. Then problem (1.3) has a unique global solution:

uC0;H01
utC0;L2Lm×0.E2.1

Proposition 2.2. [21] Let E: ℝ+ → ℝ+ be a non-increasing function and φ: ℝ+ → ℝ+ be an increasing C2-function such that

φ0=0andφt+ast+.E14

Assume that there exist q ≥ 0 and A > 0 such that

s+Eq+1tφtdtAES,0S<+,E15

then we have

E( t )cE( 0 ) ( 1+φ( t ) ) 1 q , t0, if q>0,E16
E( t )cE( 0 ) e ωφ( t ) , t0, if q=0,E17

where c and ω are positive constants independent of the initial energy E(0).

Next, we introduce the “modified energy”:

Et:121-0tgsdsut22+12ut22+12gout,E2.2

where

govt=0tgt-τvt-vτ22dτ.E19

Remark 2.2. By multiplying Eq. (1.3) by ut and integrating over Ω, using integration by parts and hypotheses (A1), (A2), we get, after some manipulations, as in [3,20],

Et-autmdx-12gout+12gtut2-autmdx+12gout0.E2.3

## 3. Decay of solutions

In order to state and prove our main result, we set

Ftεt+ε1Ψt+ε2XtE3.1

where ε1 and ε2 are positive constants to be specified later and

Ψtuutdx,χt-ut0tgt-τut-uτdτdx.E22

Lemma 3.1. For ε1 and ε2 so small, we have

α1FtEtα2FtE3.2

holds for two positive constants α1 and α2.

Proof. It is straightforward to see that

F(t)E(t)+ε12|ut|2dx+ε12|u|2dx+ε22|ut|2dx+ε22(0tg(tτ)(u(t)u(τ))dτ)2dxE(t)+ε12|ut|2dx+ε12Cp|u|2dx+ε22|u2|2dx+ε22Cp(1l)(gou)(t)α2E(t),E24

where Cp is the Poincaré constant. In the other hand,

F(t)E(t) ε1 2| ut |2dx ε1 2|u|2dx ε2 2| ut |2dx ε2 2Cp( 1l)( gou)(t)12lu( t)22+12ut22+12( gou)(t) +1 γ+2u γ+2 γ+2[ ε1+ε2 2]| ut |2dx ε1 2Cp| u|2dx ε2 2Cp( 1l)( gou)(t)α1E(t), E25

for ε1 and ε2 small enough. Thus, Eq. (3.2) is established.

Lemma 3.2. Assume that m ≥ 2 and assumptions (A1), (A2) hold. Then, the functional Ψ(t) satisfies, along the solution of Eq. (1.3), the estimate:

Ψtut2dx-14u2dx+1-l2lgout+Cutmdx,E3.3

where C is a “generic” positive constant independent of t.

Proof. By using Eq. (1.3), we easily see that

Ψt=ut2dx-u2dx-autm-2utudx+ut0tgt-τuτdτdx.E3.4

We now estimate the third term of the right-hand side of Eq. (3.4), using Young’s inequality and (A2). Thus, we get

|ut|m2utudxδ|u|mdx+cδ|ut|mdxδCu2m2u22+cδ|ut|mdxδCEm22(0)u22+cδ|ut|mdxE3.5

where cδ is a constant depending on δ. For the fourth term of the right-hand side of Eq. (3.4), we get

u(t)t0g(tτ)u(τ)dτdx12|u(t)|2dx+12(t0g(tτ)|u(τ)|dτ)2dx12|u(t)|2dx+12Ω(t0g(tτ)(|u(τ)u(t)|+|u(t)|)dτ)2dxE3.6

We then use Cauchy-Schwarz inequality, Young’s inequality, and the fact that

0tgτdτ0+gτdτ=1-l,E3.7aaaa

to obtain, for any η > 0,

Ω(0tg(tτ)(|u(τ)u(t)|+|u(t)|)dτ)2dxΩ(0tg(tτ)|u(τ)u(t)|dτ)2dx+Ω(0tg(tτ)|u(t)|dτ)2dx+2Ω(0tg(tτ)|u(τ)u(t)|dτ)dx(0tg(tτ)|u(t)|dτ)dx(1+η)Ω(0tg(tτ)|u(t)|dτ)2dx+(1+1η)Ω(0tg(tτ)|u(τ)u(t)|dτ)2dx(1+1η)Ω0tg(tτ)dτ0tg(tτ)|u(τ)u(t)|2dτdx+(1+η)Ω|u(t)|2(0tg(tτ)dτ)2dx(1+η)(1l)2Ω|u(t)|2dx+(1+1η)(1l)Ω0tg(tτ)|u(τ)u(t)|2dτdxE3.7

By combining Eqs. (3.4)(3.7), we arrive at

Ψ(t)ut2dx12|u(t)|2dx+δCEm22(0)u22+12(1+η)(1l)2|u(t)|2dx+cδ|ut|mdx+12(1+η)(1l)0tg(tτ)|u(τ)u(t)|2dτdxut2dx12[1(1+η)(1l)22δCEm22(0)]|u(t)|2dx+12(1+η)(1l)(gou)(t)+cδ|ut|mdx. E32

By choosing η = l/(1 − l) and

(0), estimate (3.3) is established.

Lemma 3.3. Assume that m ≥ 2 and assumptions (A1), (A3) hold. Then, the functional Xt (t) satisfies, along the solution of Eq. (1.3) and for any δ > 0, the estimate

X(t)δ[1+2(1l)2]u22+cδ(gou)(t)+cδ|ut|mdx+(δ0tg(s)ds)ut2dx+g(0)4δCp((gou)(t)).E3.8

Proof: By using Eq. (1.3), we easily see that

X(t)=u(t)(0tg(tτ)(u(t)u(τ))dτ)dx(0tg(tτ)u(τ)dτ)(0tg(tτ)(u(t)u(τ))dτ)dx+a|ut|m2ut(t)0tg(tτ)(u(t)u(τ))dτdxut0tg(tτ)(u(t)u(τ))dτdx(0tg(s)ds)ut2dxE3.9

Similarly to Eq. (3.4), we estimate the right-hand side terms of Eq. (3.9). So for any δ > 0, we have

-ut0tgt-τut-uτdτdxδu2dx+1-l4δgut,E3.10
Ω(0tg(ts)u(s)ds)(0tg(ts)(u(t)u(s))ds)dxδΩ(0tg(ts)|u(s)|ds)2dx+14δΩ(0tg(ts)|u(t)u(s)|ds)2dxδΩ(0tg(ts)(|u(t)u(s)|+|u(t)|)ds)2dx+14δ(0tg(ts)ds)Ω0tg(ts)|u(t)u(s)|2ds dx2δΩ(0tg(ts)|u(t)u(s)|ds)2dx+2δ(1l)2Ω|u|2dx+14δ(1l)(gu)(t)(2δ+14δ)(1l)(gu)(t)+2δ(1l)2Ω|u|2dx, E3.11
Ω|ut|m2ut(t)0tg(tτ)(u(t)u(τ))dτdxδΩ|0tg(tτ)(u(t)u(τ))dτ|mdx+cδΩ|ut|mdxδ(0tg(τ)dτ)m1Ω0tg(tτ)|u(t)u(τ)|mdτdx+cδΩ|ut|mdxδ(1l)m1C0tg(tτ)u(t)u(τ)2mdτ+cδΩ|ut|mdxδ(1l)m1Cp(2E(0)l)m220tg(tτ)u(t)u(τ)22dτ+cδΩ|ut|mdxδ(1l)m1Cp(2E(0)l)m22(gu)(t)+cδΩ|ut|mdx, E3.12
-ut0tgt-τut-uτdτdxδut2dx+g04δCp0t-gt-sut-us2dsdx.E3.13

A combination of Eqs. (3.9)(3.13), then, yields Eq. (3.8).

Theorem 3.4. Let u0u1H01× L2 (Ω) be given. Assume that (A1), (A2) hold. Then, for any t0 > 0, there exist positive constants K and λ such that the solution of Eq. (1.3) satisfies

EtKe-λt0tγsdstt0,ifm2E3.14
EtK1+γt-2m-22-m,tt0,if1<m<2.E3.15

Proof: We start with the case m ≥ 2. Since g(0) > 0, then there exists t0 > 0 such that

0tgsds0t0gsds=g0>0,tt0.E41

By using Eqs. (2.3), (3.1), (3.3), and (3.8), we obtain

F(t)(1(ε1+ε2)cδ)a|ut|mdx[ε2(g0δ)ε1]ut2dx[ε1l4ε2δ(1+2(1l)2)]u22+[12ε2g(0)4δCp](gou)(t)+[ε11l2l+ε2cδ](gou)(t).E42

At this point, we choose δ so small that

g0-δ>12g0,4lδ1+21-l2<14g0.E43

Whence δ is fixed, the choice of any two positive constants ε1 and ε2 satisfying

14g0ε2<ε1<12g0ε2E3.16

makes

k1=ε2g0-δ-ε1>0,k2=ε1l4-ε2δ1+21-l2>0.E45

We then pick ε1 and ε2 so small that Eqs. (3.2) and (3.16) remain valid and, further,

1-ε1+ε2cδ>0,12-ε2g04δCp>0.E46

Therefore, we arrive at

Ft-βEt+Cgout,tt0,E3.17

for two constants cβ > 0. We multiply (3.17) by γ(t) and use Eqs. (1.2) and (2.3), to get

γtFt-βγtEt-cgout-βγtEt-cEt,tt0.E48

This implies that

γtFt+cEt-βγtEt,tt0.E49

Hence,

γtFt+cEt-γtFt-βγtEt,tt0.E50

Again, by using the fact that γ′(t) ≤ 0, letting

Lt=γtFt+cEt,E51

and noting that L~E, we arrive at

Lt-βγtEt-λγtLt,tt0.E3.18

A simple integration of Eq. (3.18) over (t0t) leads to

LtLt0e-λt0tγsds,tt0.E53

We obtain, then, Eq. (3.14) by virtue of equivalence of E and L.

To establish Eq. (3.15), we re-estimate Eqs. (3.5) and (3.12), for m < 2, as follows

utm-2utudxδu2dx+cδut2m-2dxδCu22+cδCutmdx2m-2m.E3.19

Similarly, we have

utm-2utt0tgt-τut-uτdτdxδCgout+cδutmdx2m-2m.E3.20

By repeating all above steps and using Eqs. (3.19), (3.20) instead of Eqs. (3.5), (3.12), we arrive at

Ft-βEt+cgout+c1utmdx2m-2m,tt0;

E56

which gives

Et-β1Ft+cgout+c1utmdx2m-2m,tt0E3.21

By multiplying Eq. (3.21) by γ(t)Eq(t), for q > 0 to be specified later, and using (A1), Eq. (2.3), and Young’s inequality, we get

γ(t)Eq+1(t)β2γ(t)FqF(t)+cEq(t)(E(t))+cγ(t)Eq(t)(E(t))2m2mβ2γ(t)q+1dFq+1(t)dtc2dEq+1(t)dt+μEqm2m(t)+cμ(E(t)).E3.22

By choosing q = (2 − m)/(2m − 2) (hence, qm/(2 − m) = q + 1] and taking μ small enough, Eq. (3.22) yields

γ(t)Eq+1(t)β2q+1γ(t)dFq+1(t)dtc2dEq+1(t)dt+c(E(t))β2q+1ddt(γ(t)Fq+1(t))+β2γ(t)q+1cq+1dEq+1(t)dt+c(E(t)).E3.23

By recalling that γ ′ (t) ≤ 0 and integrating (3.23) over (ST), S ≥ t0, we get

STγtEq+1tdtβ2q+1γSFq+1S+cq+1Eq+1S+2ESAESE3.24

for some positive constant A. Therefore, Proposition 2.2 gives (3.15). This completes the proof.

Remark 3.1. Estimates (3.14) and (3.15) also hold for t ∈ [0, t0] by virtue of continuity and boundedness of and γ.

Remark 3.2. This result generalizes and improves many results in the literature. In particular, it allows some relaxation functions which satisfy

gt-agρt,1ρ<2,E61

instead of the usual assumption 1 ≤ ρ < 3/2.

Remark 3.3. Note that the exponential and the polynomial decay estimates, given in early works, are only particular cases of Eq. (3.14). More precisely, we obtain exponential decay for γ(t)≡a and polynomial decay for γ(t) = a(1 + t)− 1, where a > 0 is a constant.

## Acknowledgments

The author would like to express his sincere thanks to King Fahd University of Petroleum and Minerals for its support. This work has been funded by KFUPM under project # IN151002.

## References

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2. 2. Dafermos C.M., On abstract Volterra equations with applications to linear viscoelasticity, J. Differ. Equations 7 (1970), 554–569.
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4. 4. Munoz Rivera J.E, Asymptotic behavior in linear viscoelasticity, Quart. Appl. Math. 52(4) (1994), 628–648.
5. 5. Cabanillas E.L. and Munoz Rivera J.E., Decay rates of solutions of an anisotropic inhomogeneous n-dimensional viscoelastic equation with polynomial decaying kernels, Comm. Math. Phys. 177 (1996), 583–602.
6. 6. Barreto R., Lapa E.C., and Munoz Rivera J.E., Decay rates for viscoelastic plates with memory, J. Elasticity 44(1) (1996), 61–87.
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Written By

Salim A. Messaoudi

Submitted: December 4th, 2015 Reviewed: May 12th, 2016 Published: September 21st, 2016


text:
## Section2.1Signed Numbers

### SubsectionTypes of Numbers

Numbers greater than zero are called positive numbers, and numbers less than zero are negative numbers. We use a number line to illustrate relationships among numbers.

To the right of zero we mark the numbers 1, 2, 3, 4, ... at evenly spaced intervals. These numbers are called the natural or counting numbers. The whole numbers include the natural numbers and zero: 0, 1, 2, 3, ... .

The natural numbers, zero, and the negatives of the natural numbers are called the integers:

\begin{equation*} \cdots -3,~ {-2},~ {-1},~ 0,~ 1,~ 2,~ 3 \cdots \end{equation*}

Fractions, such as $\dfrac{2}{3}, -\dfrac{5}{4}$ and $3.6\text{,}$ lie between the integers on the number line.

###### 1.

What are integers?

As we move from left to right on a number line, the numbers increase. In the figure below, the graph of $-6$ lies to the left of the graph of $-2\text{.}$ Therefore, $-6$ is less than $-2\text{,}$ or, equivalently, $-2$ is greater than $-6\text{.}$

We use special symbols to indicate order:

\begin{equation*} \begin{aligned} \blert{\lt} \amp ~~~~ \text{means}~~~~ \blert{\text{is less than}}\\ \blert{\gt} \amp ~~~~ \text{means}~~~~ \blert{\text{is greater than}} \end{aligned} \end{equation*}

For example,

\begin{align*} -6 \lt -2 \amp ~~~~ \text{means}~~~~ {-6}~~ \text{is less than}~~ {-2}\\ -2 \gt -6 \amp ~~~~ \text{means}~~~~ {-2}~~ \text{is greater than}~~ {-6} \end{align*}

The small ends of the symbols $\lt$ and $\gt$ always point to the smaller number.

###### Example2.1.
1. Which is the lower altitude, $-81$ feet or $-94$ feet?
2. Express the relationship using one of the order symbols.
Solution
1. Negative altitudes correspond to feet below sea level, and 94 feet is farther below sea level than 81 feet. Therefore, $-91$ feet is the lower altitude.
2. $-94 \lt -81\text{,}$ or $-81 \gt -94$

### SubsectionAdding Two Numbers with the Same Sign

We'll use number lines to review operations on signed numbers.

###### Case 1: The sum of two positive numbers.

Illustrate the sum $5+3$ on a number line.

1. Graph the first number, $+5\text{,}$ as shown below.
2. Move $3$ units in the positive direction, or to the right.
3. This brings us to the sum, which is $+8\text{.}$ Thus, $5+3=8\text{.}$
###### Case 2: The sum of two negative numbers.

Illustrate the sum $(-5)+(-3)$ on a number line.

1. Graph the first number, $-5\text{,}$ as shown below.
2. Move $3$ units in the negative direction, or to the left.
3. This brings us to the sum, $-8\text{,}$ as expected. Thus, $(-5)+(-3)=-8\text{.}$
###### Look Closer.

In the examples above, we see that:

\begin{equation*} \begin{aligned} \amp \blert{\text{The sum of two positive numbers is positive.}}\\ \amp \blert{\text{The sum of two negative numbers is negative.}} \end{aligned} \end{equation*}

Thus, the sign of the sum is the same as the sign of the two terms. To find the value of the sum, we add the absolute values, or unsigned parts, of the two terms.

###### Example2.2.

Add $(-4)+(-7)$

Solution

The sum is negative. We add the absolute values of $-4$ and $-7$ (the numbers 4 and 7 without their signs) to get 11, then make the sum negative: $(-4)+(-7)=-11$

### SubsectionAdding Two Numbers with Opposite Signs

Suppose you have debts of $9, and assets of$6. (Your "assets" are your possessions.) What is your net worth? We can model this situation by the sum

\begin{equation*} -9+(+6) \end{equation*}

where debts are represented by negative numbers and assets by positive numbers. If you use your $6 to pay off part of your debt, you will still owe$3, so your net worth is $-3\text{.}$ It makes sense that if your debts are greater than your assets, your net worth is negative.

Illustrate the sum $(-9)+(+6)$ on a number line.

1. Graph the first number, $-9\text{,}$ as shown below.
2. Move $6$ units in the positive direction, or to the right.
3. This brings us to the sum, $-3\text{.}$ Thus, $(-9)+(+6)=-3\text{.}$
###### Look Closer.

In the example above, we see that the sum has the same sign as the number with the larger absolute value. (Can you explain this in terms of debts and assets?) When the two numbers have opposite signs, we find the value of the sum by subtracting the absolute values of the two numbers.

###### Example2.3.
1. Add $(-5)+(+9)$
2. Add $(+5)+(-9)$
Solution
1. The two numbers have opposite signs, so we subtract 5 from 9 to get 4, then make the answer positive, because $+9$ has a larger absolute value than $-5\text{.}$ Thus, $(-5)+(+9)=4.$
2. We subtract 5 from 9 to get 4, then make the answer negative, because $-9$ has a larger absolute value than $+5\text{.}$ Thus,$(+5)+(-9)=-4.$

###### 2.

What is the absolute value of a number?

###### 3.

When we add two numbers with opposite signs, we their absolute values.

We now have two rules for adding signed numbers.

1. To add two numbers with the same sign, add their absolute values. The sum has the same sign as the numbers.
2. To add two numbers with opposite signs, subtract their absolute values. The sum has the same sign as the number with the larger absolute value.

### SubsectionSubtracting a Positive Number

Let's compare the two problems:

\begin{equation*} 9+(-5)~~~~\text{and}~~~~9-(+5) \end{equation*}

Illustrate the sum $9+(-5)$ on a number line.

1. Plot $9$ on the number line.
2. Move $5$ units in the negative direction, or to the left.
3. The sum is $4\text{.}$

Illustrate the sum $9-(+5)$ on a number line.

1. Plot $9$ on the number line.
2. Move $5$ units to the left, to indicate subtraction.
3. The result is $4\text{.}$ The picture is identical to the graph above.
###### Look Closer.

We see that subtracting a positive number has the same result as adding the negative number with the same absolute value. Thus, to subtract a positive number, we add its opposite.

### SubsectionSubtracting a Negative Number

When we add a negative number we move to the left on the number line. There are only two directions we can move on a number line, so when we subtract a negative number we must move to the right.

Compare the graphs for adding $-3$ and subtracting $-3$

• Addition: $~~~~5+(-3)~~~~$ Move 3 units to the left, to get 2.
• Subtraction: $~~~~5-(-3)~~~~$ Move 3 units to the right, to get 8.
###### Look Closer.

Notice that

\begin{equation*} 5-(-3)=8 ~~~~~ \text{and} ~~~~~5+(+3)=8 \end{equation*}

Subtracting a negative number has the same result as adding the positive number with the same absolute value. Thus, to subtract a negative number, we add its opposite.

###### 4.

True or False: $-x$ always represents a negative number. Give an example to support your answer.

###### 5.

Subtracting a negative number is the same as with the same absolute value.

From the examples above, we see that to subtract any number, positive or negative, we add its opposite. We can do this in steps as follows.

###### Rules for Subtracting Integers.

To subtract $b$ from $a\text{:}$

1. Change the sign of $b\text{.}$
2. Change the subtraction to addition.


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# What is the relationship between conductance and resistance?

## What is the relationship between conductance and resistance?

Conductance is the opposite of resistance: the measure of how easy it is for electric current to flow through something. Conductance is symbolized with the letter “G” and is measured in units of mhos or Siemens. Mathematically, conductance equals the reciprocal of resistance: G = 1/R.

## What is the difference between conductance and resistance?

Conductance is proportional to how much flow occurs for a given pressure, and resistance is proportional to how much pressure is required to achieve a given flow.

What is the formula for total resistance?

If you know the total current and the voltage across the whole circuit, you can find the total resistance using Ohm’s Law: R = V / I. For example, a parallel circuit has a voltage of 9 volts and total current of 3 amps.

Is conductance the reciprocal of resistance?

Conductance is the reciprocal of resistance. The unit of conductance is the siemens (S). You can analyze parallel resistors by describing each resistor as a conductance.

### What is the relation between conductance and conductivity?

Conductance is the extrinsic property, whereas the inherent property is conductivity. This suggests that conductance is an object’s property depending on its quantity/mass or physical form and scale, whereas conductivity is the object’s intrinsic property of the substance.

### What is conductance and resistance in relation to density of mobile charge carriers?

The permeability to current of a conductor is called conductance (formula sign G) and, hence, is inversely proportional to the resistance. A conductor has a resistance of 1W if a voltage of 1 V drops when a current of 1 A passes this conductor….2.3. Resistance and Conductance.

R resistance
A cross-sectional area of the conductor

What is the difference between conductance and conductivity?

What law does v IR represent?

Ohm’s law states that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance, and directly proportional to the resistance of the circuit. The formula for Ohm’s law is V=IR.

## What do VI and R stand for in Ohm’s law?

where I is the current through the conductor in units of amperes, V is the voltage measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms.

## What happens to the total resistance?

Adding more parallel resistances to the paths causes the total resistance in the circuit to decrease. As you add more and more branches to the circuit the total current will increase because Ohm’s Law states that the lower the resistance, the higher the current.

What is conductance and specific conductance?

Conductance means the flow of current through the liquid conductor. Specific conductance is defined as the conducting capacity of a solution of the dissolved electrolyte and the whole solution is being placed between two electrodes are 1 sq. cm and length 1 cm. Also, read more about Molar Conductance, here.

What is difference between conductor and conductance?

Conductance is a property of the component but conductivity is a property of the material. Conductance depends on the dimensions of the conductor, but conductivity does not depend on the dimensions. Conductance is measured in Siemens while conductivity is measured in Siemens per meter.